Equilibria

In this section, you will learn about pH and pOH — important ideas that help us understand acids and bases. You will discover how the concentration of ions in solutions affects how acidic or basic it is. You will also be able to:

· use the pH scale to tell if a solution is an acid or a base · do simple calculations of pH or pOH for strong and weak acids and bases · analyse how buffers work to keep pH steady in solutions · explain why pH is important in our bodies, the environment, and in industries.

KEY IDEAS

· Buffer action is how a buffer solution works to prevent changes in pH when small amount of acid or base is added.

· Buffer solution is a solution that keeps the pH constant, even when small amounts of acid or base are added.

· Ionic product of water is the product of the concentrations of hydrogen ions and hydroxide ions in pure water at a given temperature.

· pH tells us how acidic or basic a solution is.

· The pH scale is a numerical scale used to measure how acidic or basic (alkaline) a solution is.

· pOH measures the amount of hydroxide ions (OH-) in a solution. It helps show how basic or acidic a solution is.

pH and pOH Have you ever tasted something sour like a lemon? Or touched something slippery like soap? That is because some of these substances around us are acids, and others are bases. Scientists use a special number called pH to tell how acidic or basic or neutral a solution is. There is also another number called pOH that helps us understand how basic a liquid is. pH shows how active the tiny hydrogen ions (H+) are in a solution.

If there are more H+ ions, the solution is acidic (pH less than 7).

If there are more OH- ions (called hydroxide ions), the solution is basic (pH more than 7).

If there are equal H+ and OH- ions, the solution is neutral (pH is 7).

These numbers help us know:

1. if something is safe to touch

2. if it is good for plants

3. if it is right to taste, or even if it is right to drink The word pH stands for “potency of hydrogen”. This tells us how many hydrogen ions are in a solution. To check how acidic or basic a liquid is, we use a special scale called the pH scale. So, the pH scale helps scientists know how acidic or basic something is by checking the balance between H+ and OH- ions.

pH is defined as the negative logarithm to the base ten of the hydrogen ion concentration of a substance.

pH = − log[ H+] pH = log( 1_ [ H+]) [H+] = 10⁻ᵖᴴThis means that:

1. when a liquid has more hydrogen ions (H+), its pH number is low (more acidic).

2. when a liquid has fewer hydrogen ions, its pH number is high (less acidic or more basic).

This helps scientists show even tiny amounts of hydrogen ions using simple small numbers on the pH scale.

A change of 1 pH unit represents a 10-fold change in the hydrogen ion concentration, [H+] pOH is the potential of hydroxyl ions in solution. It is defined as the negative logarithm to the base ten of the hydroxyl ion concentration of a substance.

pOH = − log[ OH−] pOH = log( ¹_ [ OH−]) [ O H−] = 10⁻ᵖᴼᴴTable 1.1: A table showing the pH and its corresponding [H+] and [OH-] pH [H+] (mol/dm⁻³) [OH-] (mol/ dm-3) 0 1.0 × 10⁰= 1 1.0 × 10⁻¹⁴= 0.00000000000001 1 1.0 × 10⁻¹= 0.1 1.0 × 10⁻¹³= 0.0000000000001 2 1.0 × 10⁻²= 0.01 1.0 × 10⁻¹²= 0.000000000001 3 1.0 × 10⁻³= 0.001 1.0 × 10⁻¹¹= 0.00000000001 4 1.0 × 10⁻⁴= 0.0001 1.0 × 10⁻¹⁰= 0.0000000001 5 1.0 × 10⁻⁵= 0.00001 1.0 × 10⁻⁹= 0.000000001 6 1.0 × 10⁻⁶= 0.000001 1.0 × 10⁻⁸= 0.00000001 7 1.0 × 10⁻⁷= 0.0000001 1.0 × 10⁻⁷= 0.0000001 8 1.0 × 10⁻⁸= 0.00000001 1.0 × 10⁻⁶= 0.000001 9 1.0 × 10⁻⁹= 0.000000001 1.0 × 10⁻⁵= 0.00001 10 1.0 × 10⁻¹⁰= 0.0000000001 1.0 × 10⁻⁴= 0.0001 11 1.0 × 10⁻¹¹= 0.00000000001 1.0 × 10⁻³= 0.001 12 1.0 × 10⁻¹²= 0.000000000001 1.0 × 10⁻²= 0.01 13 1.0 × 10⁻¹³= 0.0000000000001 1.0 × 10⁻¹= 0.1 14 1.0 × 10⁻¹⁴= 0.00000000000001 1.0 × 10⁰= 1 pH Scale The pH scale is a scale that is used to distinguish between acidic, neutral and basic (alkaline) a solution. The pH scale ranges from zero (0) to fourteen (14).

pH Scale Chart (0–14) pH Value Type of

Solution

H+ and OH- Balance Examples

0 – 6 Acidic More H+ ions than OH- ions Lemon juice, vinegar 7 Neutral Equal H+ and OH- ions Pure water 8 – 14 Basic (Alkaline) More OH- ions than H+ ions Soap, baking soda water The pH scale

Figure 1.1: The pH scale showing the pH of household items Significance of the Values of pH in Everyday Life pH tells us how sour (acidic) or soapy (alkaline/basic) things are. This helps us make safe choices in farming, health, food, water, and cleaning!

1. Farming (Agriculture)

a. Soil pH affects how plants grow.

b. Most plants thrive in soil that is slightly acidic or neutral (around 6–7).

c. Farmers check soil pH and add:

i. Lime (to make it less acidic)

ii. Sulphur (to make it more acidic)

2. Health & Medicine

a. Human blood is slightly basic, about pH 7.4—Slight changes in this pH can make you very sick.

b. Your stomach has strong acid (pH 1–2) to help digest food.

c. Doctors check pH in body fluids (like urine) to diagnose health issues.

3. Food & Cooking

a. pH affects taste, freshness, and safety of food.

b. Vinegar is very acidic and keeps food from spoiling.

c. Baking needs just the right pH (7 – 8) so cakes rise and taste good. This is done by usually adding baking soda

4. Water Safety

a. Drinking water should be about pH 7 (neutral).

b. If water is too acidic or too basic, it can:

· Damage pipes · Harm fish and plants

c. Water plants adjust pH during purification process to make water safe for consumption.

5. Cleaning Products

a. Some cleaners (like vinegar) need to be acidic to remove hard water stains.

b. Others are alkaline (like soap and bleach) to remove grease.

c. Choosing the right pH cleaner makes cleaning better and safer.

Activity 1.1 Comparing pH of Household Substances Option A Materials needed pH strips or indicator, transparent plastic cup or test tube, spoon, spatula, water samples, lemon juice, vinegar, milk, toothpaste solution, baking soda

solution, soap solution.

Procedure

1. Use the link to watch the Video: https://youtu.be/V5Mq_cL9Bck

2. Carry out the following activities and record the colour and the pH of the various solutions in the Table below.

S/n Test Colour change observed pH

a. Dip a pH strip in 10 cm³distilled water in a test tube

b. Dip a new pH strip in 10 cm³tap water in a test tube

c. Dip a new pH strip in small lemon juice in a test tube

d. Dip a new pH strip in small orange juice in a test tube

e. Dip a new pH strip in small vinegar in a test tube

f. Dip a new pH strip in small baking soda

solution in a test tube

g. Dip a new pH strip in small soap solution in a test tube

h. Dip a new pH strip in small milk solution in a test tube

i. Dip a new pH strip in small toothpaste

solution in a test tube

3. Compare results, rank the solutions from most acidic to most basic.

Option B Materials needed Beetroot juice (natural pH indicator) or universal indicator, pH scale chart (0–14), transparent plastic cups or test tubes, dropper or spoon, marker, lemon juice, vinegar, orange juice, baking soda solution, milk, soap solution, toothpaste solution, tap water Procedure

1. Create an indicator from red beetroot juice (Beetroot juice is a natural indicator which changes colour with pH. It is red in acid and purple in base).

2. Boil chopped beetroot in water, cool, and use the juice.

3. Label each cup or test tube with the name of the household substance.

4. Pour a small amount of each substance into its corresponding cup.

5. Add a few drops of red beetroot juice to each sample.

6. Observe and record the colour change in each cup.

7. Match the colour to the pH scale chart to estimate the pH value.

8. Record the pH value in a table.

9. Classify each substance as acidic (pH < 7), neutral (pH = 7), or basic (pH > 7).

Observation Table

Substance Colour Change Estimated pH Acidic, Neutral or Basic?

Lemon juice Vinegar Orange juice Baking soda Milk Soap solution Toothpaste

solution

Tap water Discussion

1. Which substance was the most acidic?

2. Which one was the most basic?

3. Were there any surprises in your results?

4. Why is it important to know the pH of substances we use or consume daily?

Meaning of Ionic Product of Water

Ionic product of water (Kw) is the product of the concentrations of hydrogen ions [H+] and hydroxide ions [OH−] in pure water at a particular temperature. It shows how much water dissociates (breaks) into hydrogen and hydroxide ions.

Even though water mostly stays as a molecule (H₂O), a tiny amount splits like this: H₂ O(l) ⇌ H(aq) + + OH(aq) − It expresses the product of the molar concentrations of hydrogen and hydroxide ions at equilibrium:

K_(c) = [H+][OH−]_ [H₂ O] K_(c) [H₂ O] = [H+][OH−] K_(c) [H₂ O] = K_(w) K_(w) = [H+][OH−] K_(w) is temperature dependent.

At a temperature of 25°C (298K);

K_(w) = [H+][OH−] = 1.0 × 10⁻¹⁴mol²dm⁻⁶In pure water, [H+] = [OH-].

Thus:

[H+][H+] = 1 × 10⁻¹⁴[H+]²= 1 × 10⁻¹⁴[H+] = √1 × 10⁻¹⁴[H+] = 1 × 10⁻⁷mol/dm3 ∴ [OH−] = 1 × 10⁻⁷mol/dm3 As temperature increases, water ionises more, so the value of Kw increases with temperature. See Table 1.2 for ionic product of water (Kw) at various temperatures.

Table 1.2: Ionic Product of Water (Kw) at Various Temperatures Temperature (°C) K_(w)(mol²·dm⁻⁶) pK_(w)= –log₁₀(K_(w)) 0 °C 1.14 × 10-¹⁵14.94 10 °C 2.93 × 10-¹⁵14.53 25 °C 1.00 × 10-¹⁴14.00 40 °C 2.92 × 10-¹⁴13.53 50 °C 5.47 × 10-¹⁴13.26 60 °C 9.61 × 10-¹⁴13.02 100 °C 5.13 × 10-¹³ 12.29

Note

· As temperature increases, K_(w) increases.

· This means that water ionises more at higher temperatures.

· However, the solution can still be neutral (i.e., [H+] = [OH-]) even though pH changes slightly.

· Kw links [H+] and [OH-]: if one rises, the other falls so that their product remains constant at given temperature.

· Kw is central when calculating pH in acidic or basic solutions.

· You are encouraged to use the link to watch the video on ionic product of water (Kw): https://youtu.be/Xfdj8JCCwI0 Relationship Between pH, pOH and pKw The relationship between pH, pOH, and pKw is fundamental to understanding acid-base chemistry. It is expressed mathematically as: pH + pOH = pKw At 25°C, the ionic product of water (Kw) is 1.0 × 10-¹⁴, so:

pKw = –log₁₀(Kw) = –log₁₀1.0 × 10-¹⁴= 14 ∴ pH + pOH = 14 As temperature increases, Kw increases and pKw decreases, hence the sum of pH and pOH becomes less than 14.

Summary of pH, pOH, and pKw at Different Temperatures:

Condition Kw (mol²·dm⁻⁶) pK_(w) pH + pOH At 25°C (neutral) 1.0 × 10-¹⁴14.00 14.00 At 50°C (neutral) 5.47 × 10-¹⁴13.26 13.26

Example 1

At 298 K, the pOH of a solution is 12.5. Calculate the pH of the solution. Give your answer to 1 decimal place.

Answer

Step 1: States the correct relation: pH + pOH = 14 pH = 14 − pOH Given: pOH = 12.5

Step 2: Substitutes correct values into the equation: pH = 14.00 – 12.5 = 1.5

Example 2

At 298 K, the pH of a solution is 10.5. Determine the concentration of hydroxide ions, [OH-], in the solution. Give your answer in mol/dm³ to 2 significant figures.

Answer

Step 1: Use the relation pH + pOH = 14 (at 298 K)

Step 2: Calculate the pOH pOH = 14 - 10.5 = 3.5

Step 3: Use the formula to find [OH-]:

[OH−] = 10⁻ᵖᴼᴴ= 10^(−3.5)Step 4: Calculate [OH-]:

[OH-] = 3.16 × 10⁻⁴mol/dm³ = 3.2 × 10⁻⁴mol/dm³

Example 3

A solution has a hydroxide ion concentration of [OH-] = 1.0 × 10⁻⁴mol/dm³.

Calculate the pOH of the solution.

a. Determine the pH of the solution.

b. State whether the solution is acidic, basic, or neutral.

Answer

a. Calculating pOH

Step 1: Use the formula pOH = –log₁₀[OH-]

Step 2: Substitute the value: pOH = –log₁₀(1.0 × 10⁻⁴) = 4.0 pOH = 4.0 Determining pH

Step 1: Use the relation pH + pOH = 14

Step 2: Substitute the value of pOH, pH = 14 – 4.0 = 10.0 pH = 10.0

b. Since pH = 10.0, which is greater than 7, the solution is basic Calculation of pH and pOH of Strong Acids and Bases Strong acids dissociate completely in water to release hydrogen ions (H+). This means that for monoprotic acids like HCl and HNO₃, the amount of hydrogen ions in the solution is the same as the amount of acid added, because each acid molecule gives one hydrogen ion.

General Steps

1. Write the dissociation equation—it goes to completion for strong species.

HCl_((aq)) → H++ Cl− H₂ S O₄(aq) → 2 H++ SO₄ ²⁻NaOH_((aq)) → Na++ OH−

2. Use stoichiometry to determine [H+] or [OH-].

e.g., HCl, HBr, HNO₃ [ H+] = [acid]ᵢₙᵢₜᵢₐₗ e . g . ,H₂ S O₄ 2[ H+] = [acid]ᵢₙᵢₜᵢₐₗ

3. Calculate pH or pOH using pH = –log [H+] or pOH = –log [OH-]

4. If you have pOH, convert it to pH using: pH = 14 – pOH

Example 4

What is the pH of 0.0025 mol/dm³ HCl?

Answer

Step 1: HCl is a strong acid, so it dissociates completely in water:

HCl → H+ + Cl- This means the concentration of hydrogen ions [H+] is equal to the concentration of HCl. So, [H+] = [HCl] = 0.0025 mol/dm³

Step 2: Use the pH formula, pH = -log₁₀[H+]

Step 3: Substitute the value, pH = -log₁₀(0.0025)

Step 4: Calculate the value pH = -log₁₀(2.5 × 10-³) pH = -(log₁₀(2.5) + log₁₀(10-³)) pH = -(0.398 - 3) pH = 2.60 pH of Very Dilute Strong Acids When the concentration of a strong acid is less than 1.0 × 10⁻⁶mol/dm³, we must consider the contribution of water to the hydrogen ion concentration.

Why include water’s contribution?

Water naturally ionizes slightly: H₂O ⇌ H+ + OH- At 25°C (298 K), this results in: [H+] = [OH-] = 1.0 × 10⁻⁷mol/dm³ So, when the acid concentration is below 1.0 × 10⁻⁶mol/dm³, the hydrogen ions from water are significant and must be added to those from the acid.

Example 5

Given: [H+] from acid = 1.0 × 10⁻⁷mol/dm³ Contribution from water = 1.0 × 10⁻⁷mol/dm³ Total [H+] = 1.0 × 10⁻⁷+ 1.0 × 10⁻⁷= 2.0 × 10⁻⁷mol/dm³ Now calculate the pH:

pH = -log₁₀(2.0 × 10⁻⁷) ≈ 6.70 For a diprotic strong acid like H₂ SO₄, the concentration of H+ ions is approximately equal to twice the initial concentration of the acid. The same pattern will also apply if the acid is triprotic (Phosphoric acid); it will be triple to initial concentration.

Calculating pOH for Strong Bases Steps to Calculate pOH for Strong Bases

1. Write the dissociation equation. Example: NaOH → Na+ + OH−

2. Determine [OH-] from concentration of the strong base. For monoprotic bases like NaOH or KOH, [OH−] = [NaOH]

3. Calculate pOH using: pOH = −log₁₀[ OH−]

4. Convert pOH to pH using the equation: pH + pOH = 14

Example 6

What is the pOH of a 0.05 mol/dm³ potassium hydroxide (KOH) solution?

Answer

Step 1: Identify the base and its dissociation.

Potassium hydroxide (KOH) is a strong base, so it dissociates completely in water:

KOH → K+ + OH- [OH-] = [KOH] [OH-] = 0.05 mol/dm³

Step 2: Use the pOH formula pOH = -log₁₀[OH-]

Step 3: Substitute the value pOH = -log₁₀(0.05)

Step 4: Calculate the value 0.05 = 5 × 10-² log₁₀(0.05) = log₁₀(5 × 10-²) = log₁₀(5) + log₁₀(10-²) = 0.699 - 2 = -1.301 pOH = -(-1.301) = 1.30 “This calculation is valid only when the base concentration is above the hydroxide level of pure water at 298 K (i.e., above 1.0×10⁻⁷M). In that case, water’s own OH- contribution is insignificant and can be ignored when calculating pOH.”

Example 7

What is the pH of a 1 × 10⁻⁸mol/dm³ sodium hydroxide (NaOH) solution?

Answer

Step 1: Contributors to [OH-]

a. From NaOH dissociation: 1.0 × 10⁻⁸mol/dm³

b. From water autoionisation: 1.0 × 10⁻⁷mol/dm³ (in pure water at 298 K)

Step 2: Add [OH-] together [ OH−]ₜₒₜₐₗ = 1.0 × 10⁻⁸+ 1.0 × 10⁻⁷= 1.1 × 10⁻⁷mol / dm³

Step 3: Calculate pOH pOH = −log¹⁰[OH−] = −log¹⁰(1.1 × 10⁻⁷) = 6.96

Step 3: Find pH using pH + pOH=14.00 pH = 14-pOH = 14-6.96 = 7.04

Activity 1.2 Exploring the pH Scale Materials needed Blank pH scale diagrams (0–14) – one per group, coloured markers or crayons, substance cards with names and pH values (e.g., lemon juice – pH 2, blood – pH 7.4, soap – pH 10), calculators for H+ concentration conversion Procedure

1. Explain how pH and hydrogen ion concentration are related.

2. Explain the formula in your own words.

3. On a large blank pH scale paper, label the regions:

a. Red for Acidic (pH 0–6)

b. Green for Neutral (pH 7)

c. Blue for Basic (pH 8–14)

4. Place each card at the correct position on the scale.

5. Write the approximate [H+] concentration below the pH points using:

[H+] = 10−pH

6. Pick one region of the pH scale to present: Acidic, Neutral, or Basic

7. Include

a. examples of substances in that region

b. their pH values

c. their [H+] concentration (pH 0 → [H+] = 1 M, pH 7 → [H+] = 1×10⁻⁷M, pH 14 → [H+] = 1×10-¹⁴M)

d. why pH matters (e.g., in food, body fluids, cleaning) Discuss

1. How can knowing pH help us in cooking?

2. Why is it important for farmers or doctors?

3. Why do we use a pH scale instead of saying the exact H+ concentration?

pK_(A)and pK_(B)Ionisation Constant Kₐfor Weak Acids Weak acids only partially break apart (dissociate) in water.

This means not all acid molecules produce hydrogen ions (H+). The process is reversible, so both the acid and ions stay in the solution. Because fewer H+ ions are formed, weak acids are less acidic than strong acids. The strength of a weak acid is measured using a value called the ionisation constant (Ka).

Ka is the dissociation constant for a weak acid. It measures how much a weak acid breaks apart (dissociates) in water.

Ka is not affected by concentration changes but is affected by temperature. A higher Ka means the acid dissociates more and is therefore a stronger weak acid.

Ka is used only for weak acids, not strong acids (which fully dissociate).

Table 1.3: Ka and pKa values of monoprotic weak acids at a temperature of 298K Weak Monoprotic Acid Kₐ pKₐ Acetic acid (CH₃COOH) 1.80 × 10⁻⁵4.74 Formic acid (HCOOH) 1.80 × 10⁻⁴3.75 Hydrofluoric acid (HF) 6.60 × 10⁻⁴3.17 Benzoic acid (C₆H₅COOH) 6.30 × 10⁻⁵4.2 Propanoic acid (CH₃CH₂COOH) 1.30 × 10⁻⁵4.87 Chloroacetic acid (ClCH₂COOH) 1.40 × 10⁻³2.86 Nitrous acid (HNO₂) 4.50 × 10⁻⁴3.35 Hydrazoic acid (HN₃) 1.90 × 10⁻⁵4.72 Hypochlorous acid (HOCl) 3.00 × 10⁻⁸7.4

Table 1.4: Ka and pKa values of polyprotic weak acids at a temperature of 298K Weak Polyprotic Acid Kₐ pKₐ Carbonic acid (H₂CO₃) – 1st dissociation 4.30 × 10⁻⁷6.37 Carbonic acid (H₂CO₃) – 2nd dissociation 5.60 × 10⁻¹¹10.25 Phosphoric acid (H₃PO₄) – 1st dissociation 7.10 × 10⁻³2.15 Phosphoric acid (H₃PO₄) – 2nd dissociation 6.30 × 10⁻⁸7.2 Phosphoric acid (H₃PO₄) – 3rd dissociation 4.50 × 10⁻¹³12.35 Sulfurous acid (H₂SO₃) – 1st dissociation 1.40 × 10⁻²1.85 Sulfurous acid (H₂SO₃) – 2nd dissociation 6.30 × 10⁻⁸7.2 Weak Polyprotic Acid Kₐ pKₐ Oxalic acid (H₂C₂O₄) – 1st dissociation 5.40 × 10⁻²1.27 Oxalic acid (H₂C₂O₄) – 2nd dissociation 5.40 × 10⁻⁵4.27 Deriving KₐExpression for a Weak Monoprotic Acid Let’s say the weak monoprotic acid is HA. It reacts with water like this:

HA (aq) ⇆ H+(aq)+ A−(aq) HA = weak acid H+ = hydrogen ion A- = conjugate base From the reaction, we apply the equilibrium law to get:

Kₐ = [H+]_(eq) [A−]_(eq)_ [HA]_(eq) This is the acid dissociation constant (Ka).

Use concentration assumptions Let the initial concentration of HA be ‘c’ mol/dm³.

Let ‘x’ mol/dm³ of it dissociate at equilibrium.

Then at equilibrium:

[HA] = c − x ;

[H+] = x ;

[A−] = x Substitute into the Ka Expression Kₐ = x . x_ c − x = x²_ c − x When dealing with weak acids, only a small fraction of the acid molecules dissociates into ions. This means:

x ≪ c or c ≫ x ⟹ c − x = c So, in the expression:

Kₐ = x²_ c x²= Kₐ × c x = √Kₐ × c [H+] = x ∴ [H+] = √Kₐ × c To determine the pH of a weak monoprotic acid, take the negative logarithm to the base ten of both sides of the equation log[H+] = − log( Kₐ × c)¹_ ₂ − log[H+] = −¹_ ₂ log Kₐ− ¹_ ₂ logc But − log[H+] = pH and − log Kₐ = pKₐ pH = ¹_ ₂ pKₐ− ¹_ ₂ logc

Example 8

Calculation: [H+], pH, and pKa of a weak acid Given:

Weak acid: Acetic acid (CH₃COOH) Initial concentration (c) = 0.10 mol/dm³ Ka = 1.8 × 10⁻⁵

Step 1: Write the dissociation equation CH₃COOH ⇌ H+ + CH₃COO-

Step 2: Set up the Ka expression Let ‘x’ be the amount that dissociates:

At equilibrium:

[H+] = x [CH₃COO-] = x [CH₃COOH] = 0.10 – X Kₐ = [H+]_(eq) [CH₃ COO−]_(eq)_______________ [CH₃ COOH]_(eq) = x²_ 0.10 − x Since x is small, assume 0.10 − x ≈ 0.10

Step 3: Substitute and solve for x Kₐ = x²_ 0.10 x²= 1.8 × 10⁻⁵× 0.10 = 1.8 × 10⁻⁶x = √1.8 × 10⁻⁶= 1.34 × 10⁻³So, [ H+] = 1.34 × 10⁻³mol / dm³

Step 4: Calculate pH pH = − log[H+] = − log(1.34 × 10⁻³) = 2.87

Step 5: Calculate pKa p Kₐ = − log(Kₐ) = − log(1.8 × 10⁻⁵) = 4.74 Ionisation Constant K_(b)for Weak Bases Weak bases only partially break apart (dissociate) in water. This means not all the base molecules produce hydroxide ions (OH-). The process is reversible, so both the base molecules and ions stay in the solution. Because fewer OH- ions are formed, weak bases are less basic than strong bases. The strength of a weak base is measured using a value called the ionisation constant (Kb).

Kb is the dissociation constant for a weak base. It measures how much a weak base break apart (dissociates) in water. Kb is not affected by concentration changes but is affected by temperature. A higher Kb means the acid dissociates more and is therefore a stronger weak base. Kb is used only for weak bases, not strong bases (which fully dissociate).

Table 1.5: Kb and pKb values of weak bases at a temperature of 298K Weak Base K_(b)(mol/dm³) pK_(b) Ammonia (NH₃) 1.80 × 10⁻⁵4.74 Methylamine (CH₃NH₂) 4.40 × 10⁻⁴3.36 Ethylamine (C₂H₅NH₂) 5.60 × 10⁻⁴3.25 Weak Base K_(b)(mol/dm³) pK_(b) Aniline (C₆H₅NH₂) 4.30 × 10⁻¹⁰9.37 Hydroxylamine (NH₂OH) 1.10 × 10⁻⁸7.96 Pyridine (C₅H₅N) 1.70 × 10⁻⁹8.77 Triethylamine ((C₂H₅)₃N) 6.30 × 10⁻⁵4.2 Dimethylamine ((CH₃)₂NH) 5.40 × 10⁻⁴3.27 Morpholine (O(CH₂CH₂)₂NH) 6.30 × 10⁻⁴3.2 Deriving K_(b)Expression for Weak Bases Let’s say the weak monoprotic or monoacidic base BOH, it reacts with water like this:

BOH (aq) ⇆ B+(aq)+ OH−(aq) BOH = weak base OH− = hydroxide ion B+ = conjugate acid From the reaction, we apply the equilibrium law to get:

K_(b) = [B+]_(eq) [OH−]_(eq)_ [BOH]_(eq) This is the base dissociation constant (Kb).

Use concentration assumptions Let the initial concentration of BOH be ‘c’ mol/dm³.

Let ‘x’ mol/dm³ of it dissociate at equilibrium.

Then at equilibrium:

[BOH] = c − x ;

[OH−] = x ;

[B+] = x Substitute into the Kb Expression K_(b) = x . x_ c − x = x²_ c − x When dealing with weak base, only a small fraction of the base molecules dissociates into ions. This means:

x ≪ c or c ≫ x ⟹ c − x = c So in the expression:

K_(b) = x²_ c x²= K_(b) × c x = √K_(b) × c [OH−] = x ∴ [OH−] = √K_(b) × c To determine the pOH of a weak base, take the negative logarithm to the base ten of both sides of the equation log[OH−] = − log( K_(b) × c)¹_ ₂ − log[OH−] = −¹_ ₂ log K_(b)− ¹_ ₂ logc But − log[OH−] = pOH and − log K_(b) = pK_(b) pOH = ¹_ ₂ pK_(b)− ¹_ ₂ logc

Example 9

Calculation: [OH-], pOH, pH, and pKb of a weak base

Example: Ammonia (NH₃)

Given:

Initial concentration of NH₃= 0.20 mol/dm³ Kb = 1.8 × 10⁻⁵

Step 1: Write the base dissociation equation NH₃+ H₂O ⇌ NH₄ + + OH-

Step 2: Set up the Kb expression Let x be the amount that dissociates:

At equilibrium:

[OH-] = x [NH₄ +] = x [NH₃] = 0.10 - x K_(b) = [NH⁴⁺]_(eq) [OH−]_(eq)_ [NH³]_(eq) = x²_ 0.20 − x Assume x is small:

K_(b) = x²_ 0.20

Step 3: Substitute Kb and solve for x 1.8 × 10⁻⁵= x²_ 0.20 x²= 1.8 × 10⁻⁵× 0.20 = 3.6 × 10⁻⁶x = √3.6 × 10⁻⁶= 1.90 × 10⁻³So, [OH-] = 1.90 × 10⁻³mol / dm³

Step 4: Calculate pOH pOH = − log[OH−] = − log(1.90 × 10⁻³) = 2.72

Step 5: Calculate pH pH = 14 − pOH = 14 − 2.72 = 11.28

Step 6: Calculate pKb pKb = -log(Kb) = -log(3.6 × 10⁻⁵) = 4.44

Activity 1.3 Understanding Partial Dissociation of Weak Acids and Bases Materials needed Two labelled beakers (A and B), Universal indicator solution or paper, Vinegar (weak acid), Hydrochloric acid (strong acid), distilled water Procedure

1. a. Discuss acids and bases in terms of Arrhenius and Brønsted-Lowy concepts.

b. Name everyday acids and bases that you encounter.

c. Discuss the difference between strong and weak acids or bases.

2. a. Label Beaker A as ‘Weak Acid (Vinegar)’ and Beaker B as ‘Strong Acid (HCl)’.

b. Pour equal volumes (e.g. 50 cm3) of vinegar into Beaker A and HCl into Beaker B.

c. Add a few drops of universal indicator to both solutions.

d. Observe and compare the colour change in each beaker.

e. Record the pH of each solution Use the table below to record your observations during the experiment comparing vinegar (a weak acid) and HCl (a strong acid).

Beaker Label

Type of Acid Colour with

Indicator Measured pH A Weak Acid (Vinegar) B Strong Acid (HCl)

i. What do the colours and pH values tell us about ion concentration?

ii. Repeat the experiment using weak and strong bases (e.g., Ammonia versus Sodium hydroxide).

3. a. What do you notice about the pH of vinegar compared to hydrochloric acid?

b. Why is the pH of vinegar higher than HCl even though both are acids?

c. What does this tell us about the number of hydrogen ions in each

solution?

d. How does this relate to the concept of partial break apart (dissociation)?

e. Would you expect a similar result with weak and strong bases?

Why or why not?

4. Consider a generic weak acid HA that dissociates as follows: HA ⇌ H+ + A-

a. Write the ionisation constant expression

b. Derive an expression for pH of weak acids in terms of C and x.

Assume the initial concentration of HA is ‘C’ mol/dm³.

Let ‘x’ be the concentration of H+ at equilibrium

Activity 1.4 Kₐ, K_(b), pKₐ, pK_(b)& pH Calculations Materials needed: Calculator Procedure

1. Use the formula pKa = -log (Ka) and pKb = -log (Kb) to calculate the missing values in the table below.

2. Use a calculator and write your steps.

Substance Kₐor K_(b)(given) pKₐor pK_(b)(your answer) Benzoic acid Kₐ= 6.3 × 10-5 Dimethylamine K_(b)= 5.4 × 10-4

3. Calculate the pH of a 0.100 moldm-3 acetic acid solution. Use the steps below:

a. Write the equation: CH₃COOH ⇌ CH₃COO- + H+

b. Set up Ka expression:

Kₐ = [H+]²_ [CH₃ COOH]

c. Assume [H+] = x, solve = x²____ 0.10

d. Find x (which is [H+]), then pH = -log(x)

4. Calculate the pH of 0.1 moldm-3 ammonia. Use Kb = x²____ 0.10

a. Find x = [OH-]

b. Calculate pOH = -log [OH-]

c. Use pH = 14 - pOH

Buffer Solutions and their Composition What is a Buffer Solution?

A buffer solution is a special mixture that resists changes in pH when small amounts of an acid or a base is added. It helps to keep the solution’s pH stable, which is important in many chemical and biological systems.

What is it made of?

A buffer solution usually contains

1. A weak acid and its salt (conjugate base) (acidic buffer), e.g., ethanoic acid (CH₃COOH) and sodium ethanoate (CH₃COONa). Or

2. A weak base and its salt (conjugate acid (basic buffer), e.g., ammonia (NH₃ ) and ammonium chloride (NH₄Cl) Why is it important?

Buffer solutions are used to

1. Keep the pH of our blood stable

2. Control pH in chemical reactions

3. Maintain pH in soaps, shampoos, and medicines Buffer Action Buffer action is how a buffer solution works to keep the pH steady, even when we add a little amount of acid or base.

Acidic Buffer (e.g., Vinegar and its Salt – CH₃COOH/CH₃COONa) An acidic buffer is a mix of a weak acid and its salt (which has a base part).

1. If you add acid (H+), the salt (CH₃COONa) soaks up the extra acid and turns it into the weak acid CH₃COOH. This stops the pH from dropping too much.

Reaction: H+ + CH₃COO- → CH₃COOH

2. If you add base (OH-), the weak acid reacts with the base to make water and salt. This stops the pH from rising too much.

Reaction: OH- + CH₃COOH → CH₃COO- + H₂O Basic Buffer (e.g., Ammonia and its salt – NH₃/NH₄Cl) A basic buffer is a mixture of a weak base and its salt (which has an acid part).

1. If you add acid (H+), the weak base reacts with the acid to form the salt (NH₄ +). This stops the pH from dropping. Reaction: H+ + NH₃→ NH₄ +

2. If you add base (OH-), the acid part (NH₄ +) reacts with the base to make NH₃ and water. This stops the pH from rising too much. Reaction: OH- + NH₄ + → NH₃+ H₂O Real-life Analogies to Understand Buffer Action

1. Body Temperature Analogy

When you are too hot, your body sweats to cool you down. When you are too cold, your body shivers to warm you up. It keeps your body temperature stable — just like a buffer keeps pH stable!

a. Sweating = the buffer removing excess acid

b. Shivering = the buffer adding acid when base is added

2. Cooking Analogy (Jollof Rice Example)

You accidentally add too much salt → You add a potato to soak it up or more rice to dilute it. It keeps the food flavour balanced — just like a buffer keeps the pH balanced!

a. Potato = absorbs extra salt → like the conjugate base absorbing acid

b. More rice or seasoning = balances taste → like the acid part of the buffer reacting with base Calculation of pH of a Buffer Solution Acidic Buffer For an acidic buffer (weak acid + salt), we use the Henderson-Hasselbalch equation:

pH = pKa + log [Salt (A-]_ [Acid (HA)] Where pKa is the negative log of Ka (acid dissociation constant) [Salt] is the concentration of the conjugate base (A-) [Acid] is the concentration of the weak acid (HA)

Example

Calculate the pH of a buffer made from:

0.10 mol dm-3 acetic acid (CH₃COOH) 0.10 mol dm-3 Sodium acetate (CH₃COONa) Ka of acetic acid = 1.8 × 10⁻⁵

Step 1: Find pKa pKa = -log (1.8 × 10⁻⁵) ≈ 4.74

Step 2: Use the formula pH = pKa + log [Salt (A-]_ [Acid (HA)] pH = 4.74 + log 0.10/0.10 = 4.74 Basic Buffer For a basic buffer (weak base + salt), use the formula:

pOH = pKb + log [Salt B H+]_ [Base B] Then: pH = 14 - pOH

Example 9

Calculate the pH of a buffer solution that contains:

0.20 mol dm-3 acetic acid (CH₃COOH) 0.10 mol dm-3 sodium acetate (CH₃COONa) Ka of acetic acid = 1.8 × 10⁻⁵Answer

Step 1: Calculate pKa pKa = -log (1.8 × 10⁻⁵) ≈ 4.74

Step 2: Use the Henderson-Hasselbalch equation pH = pKa + log [Salt (A-]_ [Acid (HA)] pH = 4.74 + log 0.10/0.20 pH = 4.74 + log(0.50) pH = 4.74 + (− 0.30) pH = 4.44

Example 10

Calculate the pH of a buffer solution that contains:

0.30 mol dm-3 ammonia (NH₃) 0.20 mol dm-3 ammonium chloride (NH₄Cl) Kb of ammonia = 1.8 × 10⁻⁵Answer

Step 1: Calculate pKb pKb = -log (1.8 × 10⁻⁵) ≈ 4.74

Step 2: Calculate pOH pOH = pKb + log [Salt ]_ [Base] = 4.74 + log ^(0.20)_ _(0.30) = 4.74 + log(0.667) = 4.74 + (− 0.18) = 4.56

Step 3: Calculate pH pH = 14 - pOH = 14 - 4.56 pH ≈ 9.44

Activity 1.5 Collaborative Learning

Topics

1. Buffer Solutions and Their Behaviour

2. pH Calculations Overview of the Activity

1. Organise yourself into groups of 3 or 4.

2. Each group member assumes a specific role- Researcher: Gathers key facts and relevant examples

Note-taker: Records important information and group ideas Timekeeper: Keeps the group on task and on time Presenter: Shares the group’s findings with the class

3. Topic Assignment and Research

a. Group A: Investigates buffer solutions, including their components, behaviour, and real-life applications.

b. Group B: Studies pH calculations using the Henderson-Hasselbalch equation, including how to calculate pH and component concentrations.

Groups use chemistry textbooks, printed guides, and teacher-approved resources for your research.

4. Collaboration Guidelines

a. Listen actively and respect one another’s ideas

b. Share tasks and support one another

c. Provide constructive feedback

5. Compile your findings and create a short, creative presentation. This can be a: poster, diagram, slide show or oral report.

6. Present your work to the class

7. Reflection

a. What is a buffer solution?

b. How does a buffer help maintain pH?

c. How do you calculate the pH of a buffer?

d. What did you enjoy or learn from your group experience?

Buffer Solution is made by combining

1. A weak acid and its conjugate base (salt) → Acidic Buffer

2. A weak base and its conjugate acid (salt) → Basic Buffer Acidic Buffer Preparation (e.g., CH₃COOH / CH₃COONa) Method A Mixing a Weak Acid and its Salt

1. Mix a measured amount of acetic acid (CH₃COOH) with sodium acetate (CH₃COONa).

2. Dissolve both in distilled water.

3. Adjust their concentrations based on the required pH.

4. Use the Henderson-Hasselbalch equation:

pH = pKa + log [Salt (A-]_ [Acid (HA)] Method B Reacting a Weak Acid with a Strong Base

1. Add NaOH (strong base) gradually to CH₃COOH (excess).

2. This neutralizes some of the CH₃COOH to form CH₃COONa.

3. The final solution contains both CH₃COOH and CH₃COONa — forming a buffer.

Basic Buffer Preparation (e.g., NH₃/ NH₄Cl)

Method A: Mixing a Weak Base and Its Salt

1. Mix a known concentration of ammonia (NH₃) with ammonium chloride (NH₄Cl).

2. Dissolve both in water to make the basic buffer.

3. Use the Henderson-Hasselbalch equation:

pOH = pK_(b) + log( [Salt]_ [Base]) Method B: Reacting a Weak Base with a Strong Acid

1. Add HCl (strong acid) slowly to NH₃(excess)

2. The reaction forms NH₄Cl, resulting in a buffer solution containing both NH₃and NH₄ +.

Applications of Buffer Solutions in Everyday Life

Buffer solutions are like helpful ‘pH bodyguards’. They stop things from becoming too acidic or too basic, keeping things balanced. Here are some fun and simple ways buffers help us every day:

1. In Our Body (Especially in Our Blood)

Our blood must stay at a steady pH (around 7.4) to keep us healthy. Buffers in the blood prevent it from becoming too acidic or too basic when we eat, run, or get sick. Imagine buffers as tiny superheroes protecting our blood!

2. In Toothpaste

When we eat sweet food, our mouth becomes more acidic, which can hurt our teeth. Toothpaste has buffers that help balance the pH in our mouth and protect our teeth from decay. Buffers help us to keep our smiles strong and healthy! (no mouth odour)

3. In Science Experiments

In school labs, we mix chemicals. The pH must stay steady so the results are correct. Buffers keep the pH from changing too much, so we get the right results. They make science safer and more accurate!

4. In Food and Drinks

Some drinks and foods need the right pH to taste good and stay fresh. Buffers help keep the pH steady so the food doesn’t spoil quickly or taste bad. For

example, soft drinks have buffers to balance their fizzy taste.

5. In Plants and Soil

Plants need healthy soil to grow. Buffers in the soil help keep the pH just right so the plants can absorb nutrients. Healthy soil = happy plants!

6. In Fish Tanks (Aquariums)

7. Fish can get sick if the water is too acidic or too basic. Buffers keep the water just right for the fish to live happily.

Activity 1.6 Preparing and Testing Buffer Solutions

Materials needed

1. Ethanoic acid solution (CH₃COOH) (weak acid)

2. Sodium ethanoate, CH₃COONa (salt of the weak acid)

3. Ammonium hydroxide, NH4OH (weak base)

4. Ammonium chloride, NH4Cl (salt of the weak base)

5. pH paper or universal indicator

6. Beakers, measuring cylinders, glass rods

7. Distilled water

8. Safety goggles and gloves

Step 1

a. Put on your goggles and gloves.

b. Mix a weak acid (0.100 mol dm-3) (like vinegar or ethanoic acid) with its salt (0.10 mol dm-3) (like sodium ethanoate) in a beaker.

c. Stir the solution gently using a glass rod.

This mixture creates a buffer solution that can resist pH changes.

Step 2: Measuring pH

a. Use pH paper or a universal indicator to measure the solution’s pH.

b. Record the value.

You are checking how acidic or basic the buffer is before testing it.

Step 3: Adding Acid and Base

a. Divide the buffer into 3 small beakers

b. Add a few drops of dilute HCl (acid) to the first one.

c. Add a few drops of dilute NaOH (base) to the second one.

d. Leave the third as a control (no change).

You are simulating what happens when an acid or base is added to a buffer.

Step 4: Testing Buffer Action

a. Measure the pH of all solutions in the three beakers again.

b. Compare the pH changes.

Buffers resist pH changes. You should see only small changes in pH when acid or base is added!

Repeat Steps 1-4, using a mixture of 0.100 mol dm-3 NH3(aq) and 0.1 mol dm-3 NH4Cl. This is a basic buffer.

Discussion Questions

1. Which sample had the biggest change in pH?

2. Did your buffer solution work well?

3. Why do you think some buffers work better than others?

Activity 1.7 Applications of Buffer Systems

This inquiry-based activity helps you explore real-world applications of buffer systems in biological and industrial contexts. Learners will work in groups, research a buffer system, and present their findings using guiding questions and a video resource.

Steps

1. Use the following video link to begin your investigation:

https://www.youtube.com/ watch?app=desktop&v=10Al4Z7W_zI The video explores how buffer systems work and their importance in biological and industrial environments.

2. Form small groups and choose a buffer system to investigate from one of these categories:

a. Biological buffers (e.g., blood, intracellular fluid, saliva)

b. Industrial buffers (e.g., food production, pharmaceuticals, agriculture)

3. Find answers to the following questions:

a. What are the components of your buffer system?

b. What is the optimal pH range maintained by the buffer?

c. What would happen if the buffer system failed?

d. How does this buffer system relate to the Henderson-Hasselbalch equation?

4. Group Presentations and Class Discussion

After all groups have presented, engage in a comparative discussion.

Focus on:

a. Common features of all buffer systems

b. How buffer composition matches the required pH range

c. The effects of buffer failure in different systems

d. Innovations in medicine and industry made possible through buffer chemistry

5. Design a Buffer System

Choose one of the following scenarios and propose a buffer system for it:

a. A sports drink that maintains pH for optimal electrolyte absorption

b. A soil additive that stabilises pH for specific crops

c. A pharmaceutical formulation that keeps a safe pH inside the body

Activity 1.8 Exploring Acids, Bases, pH, and Buffer Systems Materials needed pH paper, beakers, measuring cylinders, household substances (e.g., vinegar, baking soda solution, lemon juice, soap solution), weak acids and bases (e.g., acetic acid, ammonia), indicators, buffer salts (e.g., CH₃COONa, NH₄Cl), and distilled water.

Steps

1. Comparing Household Substances by pH

a. Collect common substances (e.g., lemon juice, baking soda, vinegar, soap solution).

b. Use pH paper to determine the pH of each and record.

c. Plot the results on a - pH scale.

d. What makes a substance acidic or basic, and identify neutral substances.

2. Understanding Acid/Base Strength and Dissociation

a. Classify the tested substances as strong acids, weak acids, neutral, weak bases, or strong bases based on pH values.

b. Write dissociation equations for:

i. Strong acid (e.g., HCl)

ii. Weak acid (e.g., CH₃COOH)

iii. Weak base (e.g., NH₃)

c. Compare extent of dissociation and relate it to pH values from Step 1.

3. Performing Ka, Kb, pKa, pKb and pH Calculations Given values of Ka(CH₃COOH) = 1.8 × 10⁻⁵and Kb (NH3) = 1.8 × 10⁻⁵, calculate:

a. pKa and pH of CH₃COOH

b. pKb and pH of NH3

4. Preparing and Testing Buffer Solutions.

a. Prepare buffer solutions using:

i. Weak acid and its conjugate base (e.g., CH₃COOH + CH₃COONa)

ii. Weak base and its conjugate acid (e.g., NH₃+ NH₄Cl)

b. Test the buffer’s ability to resist pH change by adding small amounts of HCl or NaOH.

5. Record and analyse results.

6. Exploring Real-Life Applications of Buffer Systems. Name buffer systems used in:

a. biological systems

b. industry

c. agriculture (soil pH control)

1. a. Define the term pH.

b. What is the mathematical relationship between pH and hydrogen ion concentration?

c. State the range of values on the pH scale and identify the type of solutions they represent.

i. pH < 7

ii. pH = 7

iii. pH > 7

2. A solution has a hydrogen ion concentration of [H+] = 3.2 × 10⁻⁴mol/dm³.

a. Calculate the pH and pOH of the solution.

b. State whether the solution is acidic, basic, or neutral.

c. Explain briefly the importance of this acidity level in everyday use.

d. How does temperature affect the ionic product of water (Kw)?

3. The pH of a solution is 5.2.

a. Calculate the concentration of hydrogen ions, [H+], in mol/dm³.

b. Comment on how the hydrogen ion concentration would change if the pH decreased to 4.2.

c. Give one example of a substance that could have this pH value.

4. A sample of a basic solution has a hydroxide ion concentration of [OH-] = 1.0 × 10-³ mol/dm³.

a. Calculate the pOH of the solution.

b. Determine the pH of the solution at 25°C.

c. Explain the significance of the equation pH + pOH = 14. When is it valid?

d. Describe one real-life use of a basic solution and explain why its pH is suitable for that use.

5. Lemon juice has a pH of 2.5, while blood has a pH of 7.4.

a. Calculate how many times greater the hydrogen ion concentration is in lemon juice compared to blood.

b. Describe how this large difference affects their properties and uses.

c. Explain why even a small change in blood pH can be harmful to the body.

6. You are designing a poster to educate your community on the importance of pH in everyday life.

a. Identify and describe three different areas of life where pH plays a key role.

b. For each area, state:

i. a common substance involved

ii. its pH value

iii. why its pH matters in that context

c. Discuss how incorrect pH levels can lead to problems in any one of the areas identified.

7. The table below shows the pH values of five household substances:

Substance pH Value Vinegar 3.0 Baking soda 9.0 Lemon juice 2.0 Milk 6.5 Ammonia solution 11.0

a. Arrange the substances in increasing order of acidity.

b. Which substance is closest to neutral pH?

c. Suggest a suitable universal indicator colour that might be observed for:

i. Lemon juice

ii. Baking soda

d. Why might ammonia solution be unsafe for use on skin?

8. A solution is prepared by mixing 3.60 × 10-³ mol NaOH with 5.95 × 10⁻⁴mol HCl, then diluted to 1.00 L.

a. Determine the pH of the resulting solution.

b. Explain why direct use of pH = –log[H+] would be incorrect before evaluating the reaction.

9. The pH of soil is measured at 8.42.

a. Calculate the [H+] and [OH-].

b. Suggest two practical ways farmers could adjust this soil pH if it were too high (basic).

10. a. Explain why using the pH scale is more practical than giving hydrogen ion concentrations in everyday life.

b. Describe how pH knowledge is useful in two of these areas: cooking, gardening, medicine, environmental water monitoring, or cleaning.

Explain why specific pH ranges matter in each.

11. Chemical runoff from nearby farmland caused the pH of a pond to drop from 7.8 to 6.4.

a. Calculate the change in hydrogen‐ion concentration between the two pH values.

b. Explain the potential effects of this pH change on fish and aquatic plant life.

c. Propose two sustainable farming practices that could prevent such acidification of water bodies.

Multiple Choice Questions

1. What does Ka represent in acid-base chemistry?

A. Acid boiling point B. Base dissociation constant C. Acid dissociation constant D. Activation energy

2. Which of the following correctly describes pKa?

A. The negative logarithm of the acid concentration B. The product of Ka and Kb C. The negative logarithm of Ka D. The inverse of pH

3. Which statement is TRUE about strong acids?

A. They have high pKa values B. They have low Ka values C. They ionize partially in solution D. They have low pKa values

4. The Kb of ammonia is 1.8 × 10⁻⁵. What is its pKb?

A. 5.0 B. 4.74 C. 6.22 D. 9.2

5. What is the relationship between Ka and the strength of a weak acid?

A. Higher Ka = weaker acid B. Lower Ka = stronger acid C. Higher Ka = stronger acid D. Ka has no effect on acid strength

6. Which pair of values would most likely represent a strong base?

A. Low Kb, high pKb B. High Kb, low pKb C. Low Kb, low pKb D. High Ka, high pKb

7. A solution has [H+] = 1 × 10⁻⁶mol/dm³. What is its pH?

A. 6 B. 7 C. 8 D. 10

8. The Ka of acetic acid is 1.8 × 10⁻⁵. Calculate its pKa.

A. 5.00 B. 4.74 C. 2.34 D. 3.76

9. If the pKa of a weak acid is 4.75, what can you infer about the pH of a buffer solution when [A-] = [HA]?

A. pH < 4.75 B. pH = 4.75 C. pH > 4.75 D. pH = 7

10. A weak base has a Kb of 3.2 × 10⁻⁶. What is its pKb?

A. 5.5 B. 6.5 C. 4.0 D. 3.0 Essay-type Questions

1. What is a buffer solution?

2. What are the two main components of an acidic buffer?

3. What is the function of a buffer solution?

4. How does an acidic buffer neutralise added acid?

5. How does an acidic buffer neutralise added base?

6. What is the Henderson-Hasselbalch equation?

7. Calculate the pH of a buffer made by mixing 0.2 M CH₃COOH and 0.1 M CH₃COONa. Ka of CH₃COOH = 1.8 × 10⁻⁵8. Explain why buffer solutions are important in biological systems.

9. Calculate the pH of a buffer prepared by mixing 0.3 M NH₃and 0.2 M NH₄Cl. Kb of NH₃= 1.8 × 10⁻⁵10. Design a buffer solution with a target pH of 5.0. Choose an appropriate weak acid and justify your choice.

11. What is the formula for calculating the pH of an acidic buffer solution?

12. Calculate the pH of a buffer solution containing 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵) and 0.1 M sodium acetate.

13. Calculate the Ka of a weak acid if a buffer made from it and its salt has a pH of 5.0 and [Salt]/[Acid] = 2.

14. You want to prepare a buffer solution with pH = 9.25 using ammonia (Kb = 1.8 × 10⁻⁵) and ammonium chloride. What should be the ratio [Base]/ [Salt]?

15. Describe how you would prepare 250 mL of a buffer solution with pH = 4.76 using acetic acid (Ka = 1.8 × 10⁻⁵) and sodium acetate.

16. Design a buffer system to maintain pH 7.4 for a biochemical experiment.

Choose appropriate components and justify your choice.

17. Explain the impact of dilution on the pH and buffer capacity of a solution made from a weak acid and its conjugate base.

In this section, you will learn what happens when salts dissolve in water. The two main ideas are hydration and hydrolysis. By the end of the section, you will be able to:

· describe what hydration is and what affects it.

· know how to tell if a salt solution will be acidic, basic, or neutral.

KEY IDEAS

· Hydration (of salts) is when water molecules surround and stick to the ions of a salt when it dissolves in water.

· Hydrolysis (of salts) is when a salt reacts with water to make the solution either acidic or basic.

Hydration of Salts

What is Hydration of Salts?

When you put a salt in water, it breaks apart into tiny, charged pieces called ions. These ions do not float around by themselves — water molecules quickly surround them. This process is called hydration.

When you stir table salt into water, you cannot see it anymore because the salt ions are now hydrated — water has surrounded them.

Example using Sodium chloride (NaCl) NaCl (s) → Na+ (aq) + Cl– (aq) This equation means that solid salt (NaCl) breaks apart into sodium ions (Na+) and chloride ions (Cl–) in water. These ions are now hydrated, meaning they are surrounded by water molecules.

Figure 2.1: Dissolution of NaCl in H2O

Hydrolysis of Salt

Hydrolysis of salts happens when a salt dissolves in water and the ions it forms react with water, changing the pH of the solution — making it either acidic or basic.

What Causes Hydrolysis?

Salts are made from acids and bases. If the salt comes from a strong acid and strong base (like NaCl), it stays neutral. If the salt comes from a weak acid or weak base, it can react with water to make the solution acidic or basic.

The hydrolysis of salt ions in water is an example of a Brønsted-Lowry acid-base reaction, where the ions act as either proton donors or proton acceptors when they interact with water molecules. The extent of hydrolysis depends on the strength of the parent acid or base. The weaker the parent acid or base, the greater the extent of hydrolysis of its conjugate ion.

Types of Salt Hydrolysis

Table 2.1: Resulting solution pH based on salt type Salt Type Parent Acid/Base Resulting Solution Strong acid + Strong base e.g., NaCl Neutral Weak acid + Strong base e.g., CH₃COONa Basic Strong acid + Weak base e.g., NH₄Cl Acidic Weak acid + Weak base e.g., NH₄CH₃COO CH₃COO NH₄ Depends on relative strengths (Kₐand K_(b) values) Cation Hydrolysis Cation hydrolysis happens when a positive ion (cation) from a salt reacts with water and releases H+ ions, making the solution acidic.

When Does Cation Hydrolysis Occur?

It usually happens when the salt comes from a weak base and a strong acid. The cation (positive ion) from the weak base reacts with water, forming an acidic

solution.

Example 1

Step-by-Step Cation Hydrolysis of Ammonium Chloride (NH₄Cl)

1. Dissolution of Salt

When NH₄Cl is added to water, it dissociates completely into its ions:

NH₄Cl (s) → NH₄ + (aq) + Cl– (aq)

a. NH₄ + is the cation from the weak base ammonia (NH₃).

b. Cl– is from the strong acid HCl and does not hydrolyse.

2. Identify the ion that hydrolyses Only the NH₄ + ion reacts with water. This is a cation hydrolysis process.

3. Reaction of NH₄

+ with Water NH₄ + (aq) + H₂O (l) ⇌ NH₃(aq) + H₃O+ (aq)

a. NH₄ + donates a proton (H+) to water.

b. This forms NH₃and H₃O+ (hydronium ion).

c. The H₃O+ ion makes the solution acidic.

Summary

1. NH₄Cl dissolves into NH₄ + and Cl–.

2. NH₄ + undergoes cation hydrolysis by reacting with water.

3. It produces H₃O+, which lowers the pH.

4. The resulting solution is acidic.

Example 2

Step-by-Step Cation Hydrolysis of Aluminium Chloride (AlCl₃)

1. Dissolution of salt When aluminium chloride (AlCl₃) dissolves in water, it dissociates into its ions:

AlCl₃(s) → Al³+ (aq) + 3Cl– (aq)

a. Al³+ is the cation that undergoes hydrolysis.

b. Cl– does not hydrolyse because it comes from a strong acid (HCl).

2. Identify the hydrolysing ion Al³+ is a small, highly charged cation with a high charge density. It strongly attracts water molecules.

3. Reaction of Al³+ with water (hydrolysis)

a. Al³+ (aq) + 6H₂O (l) ⇌ [Al(H₂O) ₆] ³+

b. The hydrated ion then loses a proton from one of its water molecules:

[Al(H₂O) ₆] ³+ ⇌ [Al(H₂O) ₅(OH)]²+ + H₃O+

c. The release of H₃O+ (hydronium ion) makes the solution acidic.

Summary

1. AlCl₃dissolves in water and releases Al³+ ions.

2. Al³+ attracts water molecules to form hydrated aluminium ions.

3. These hydrated ions donate protons to water, increasing H₃O+ concentration.

4. The solution becomes acidic due to this cation hydrolysis.

Anion Hydrolysis

Anion hydrolysis occurs when a negative ion (anion) from a salt reacts with water to form hydroxide ions (OH–), making the solution basic (alkaline).

When Does Anion Hydrolysis Happen?

Anion hydrolysis usually occurs when the salt comes from a strong base (like NaOH) and a weak acid (like acetic acid, CH₃COOH). In such cases, the anion (from the weak acid) reacts with water.

Example 3

Sodium Acetate (CH₃COONa)

When sodium acetate dissolves in water: CH₃COONa (s) → CH₃COO– (aq) + Na+ (aq)

1. Na+ does not hydrolyse (it comes from a strong base).

2. CH₃COO– (acetate ion) does hydrolyse: CH₃COO– (aq) + H₂O (l) ⇌ CH₃COOH (aq) + OH– (aq) This reaction:

a. produces OH–, increasing the pH.

b. makes the solution basic.

Summary

1. Anion hydrolysis makes solutions basic.

2. It happens when salts come from a strong base and weak acid.

3. The anion which is from a weak acid (like CH₃COO–, CN–, or CO₃²–) reacts with water, forming OH–.

Example 4

Step-by-Step Anion Hydrolysis of Sodium trioxocarbonate(IV) (Na₂CO₃)

1. Dissolution of the salt When sodium carbonate (Na₂CO₃) dissolves in water, it dissociates completely:

Na₂CO₃(s) → 2Na+ (aq) + CO₃²– (aq)

a. Na+ does not hydrolyse (it comes from a strong base).

b. CO₃²– is the conjugate base of a weak acid (H₂CO₃) and does hydrolyse.

2. Hydrolysis of CO₃²– ion CO₃²– (aq) + H₂O (l) ⇌ HCO₃

– (aq) + OH– (aq)

a. The carbonate ion reacts with water to form bicarbonate (HCO₃

–) and hydroxide (OH–).

b. OH– increases the pH, making the solution basic.

3. Further hydrolysis (optional) HCO₃

– (aq) + H₂O (l) ⇌ H₂CO₃(aq) + OH– (aq) Bicarbonate can also undergo further hydrolysis, but this reaction is weaker.

Summary

1. Na₂CO₃dissociates into Na+ and CO₃²–.

2. CO₃²– reacts with water to form OH–, increasing the pH. The solution becomes basic due to anion hydrolysis.

Whilst these examples can seem complex, the general principle of Table 2.1 holds for all of them.

Real-World Applications of Acidic, Basic and

Neutral Salt Solutions

Acidic Salt Solutions (pH < 7)

1. Fertilisers (e.g., Ammonium salts like NH₄Cl or NH₄NO₃): These release H+ ions in soil, lowering pH and aiding in the absorption of certain nutrients.

2. Industrial Metal Cleaning: Iron (III) chloride (FeCl₃) solutions are used to etch metals or clean machinery due to their mild acidity.

3. Electroplating and Dye Fixing: Acidic salts help in fixing dyes onto fabrics or plating metals in manufacturing.

Basic Salt Solutions (pH > 7)

1. Water Softening (e.g., Sodium carbonate Na₂CO₃): Removes hardness from water by precipitating calcium and magnesium ions.

2. Baking & Food Preparation (e.g., Sodium bicarbonate NaHCO₃): Baking soda is mildly basic and helps in leavening and neutralising acids in recipes.

3. Soap and Detergent Production: Basic salts are involved in the saponification process to make soap.

4. Antacid Formulations: Some basic salts (e.g., Magnesium hydroxide) are used in over-the-counter medicines to relieve heartburn.

Neutral Salt Solutions (pH ≈ 7)

1. Intravenous (IV) Fluids (e.g., NaCl - normal saline): Used in hospitals for hydration; being neutral prevents damage to tissues and blood cells.

2. Food Seasoning and Preservation: Table salt (NaCl) in solution is neutral and used in cooking, pickling, and curing meats.

3. Laboratory Reference Solutions: Neutral salts are used as controls or standards in chemical experiments due to their stable pH.

Factors Affecting the Extent of Hydrolysis

Several factors influence how much a salt undergoes hydrolysis in aqueous

solution. These include:

1. Strength of the Parent Acid or Base

The weaker the parent acid or base, the greater the extent of hydrolysis of its conjugate ion.

Example: Salts of weak acids like CH₃COO– or weak bases like NH₄ + hydrolyse more than those from strong acids/bases.

2. Charge on the ion Ions with higher charges, especially metal cations like Al³+ or Fe³+, tend to hydrolyse more strongly because they attract water molecules more intensely.

3. Size of the ion For ions with the same charge, smaller ions exhibit higher charge density, leading to stronger hydrolysis.

Example

a. Al³+ hydrolyses more than Na+ due to its smaller size and higher charge.

b. Among the ions Mg²+ and Ca²+, Mg²+ hydrolyses more than Ca²+ due to its smaller size and higher charge density.

4. Concentration of the Salt Solution

Dilute solutions often show a greater proportion of hydrolysis because water is in excess and equilibrium favours ion-water interaction.

5. Temperature An increase in temperature usually increases the extent of hydrolysis, especially if the hydrolysis reaction is endothermic.

Activity 2.1 Exploring Salt Hydrolysis

Materials needed Use the Reference Table on salt hydrolysis below Type of Salt (Source) pH Cation Type Anion Type Strong acid + strong base (neutral salt) e.g. MgCl₂, Na No hydrolysis (pH = 7) Strong base cations (e.g. Na+, K+, Ba²+) Strong acid anions (Cl–, Br–, I–, ClO₄

–) Strong acid + weak base (acidic salt) e.g. NH₄Br Cationic hydrolysis (pH < 7) Weak base cations (e.g.

NH₄ +, Cu²+, Zn²+) Strong acid anions (Cl–, Br–, I–, ClO₄

–) Weak acid + strong base (basic salt) e.g.

CH₃COOLi Anionic

hydrolysis (pH > 7) Strong base cations (Na+, K+, Ca²+) Weak acid anions (CH₃COO–, S²–, CO₃²–, PO₄³–) Steps

1. Identify salts that produce acidic aqueous solutions from the list below:

NH₄Cl, CH₃COONa, FeCl₃, NaCl, Na₂CO₃, KNO₃

2. Write balanced ionic dissociation equations (with state symbols) for NH₄Cl(aq) and FeCl₃(aq).

a. For each Step 2 cation, write the equation showing how it reacts with water (hydrolysis) to produce hydronium, H₃O+.

b. Using your equations, justify the acidity of the corresponding salt solutions.

3. From the salts listed in step 1, identify salts that produce basic aqueous solutions.

a. Write the balanced dissociation equations for the basic salt(s) identified in Step 3

b. Explain why these salts solutions are basic.

4. Identify salts that produce neutral aqueous solutions. Explain why aqueous

solution(s) of this /these salts is/are neutral.

Extension Task

1. Verify your findings by testing each of these salts in turn

2. Dissolve a spatula of each salt into 50 ml of water.

3. Using Universal indicator, pH paper or another suitable indicator test the resulting solutions for pH.

4. Compare this to your conclusions above.

Multiple choice Questions

1. Which of the following salts forms an acidic solution in water?

A. NaCl B. NH₄Cl

C. Na₂CO₃ D. KNO₃

2. What is the product of the hydrolysis of CH₃COO– in water?

A. CH₄ B. OH–

C. H₃O+ D. Na+

3. The hydration of an ion refers to:

A. Precipitation of salts B. Reaction of a salt with water C. Surrounding of ions by water molecules D. Evaporation of water

4. Which of the following salts is derived from a strong acid and strong base?

A. NH₄Cl B. Na₂CO₃

C. NaCl D. FeCl₃

5. Which ion contributes to the basicity of an aqueous solution?

A. NH₄ + B. CH₃COO– C. Fe³+ D. H₃O+

6. What type of solution does Na₂CO₃form in water?

A. Acidic B. Basic

C. Neutral D. Amphoteric

7. The hydrolysis of NH₄ + in water results in:

A. Release of OH–

B. Formation of NH₃and H₃O+

C. No reaction D. Formation of Na+

8. Which factor increases the extent of hydrolysis?

A. High salt concentration B. Use of strong acids C. Use of strong bases D. Dilute solution

9. Which salt will produce a neutral solution upon dissolution in water?

A. NH₄Cl B. NaCl

C. CH₃COONa D. FeCl₃

10. Which of the following is correct about hydration energy?

A. It is the energy needed to evaporate water B. It is the heat absorbed during hydrolysis C. It is the energy released when ions are surrounded by water molecules D. It is the heat required to ionise a salt

11. Which combination of ions will most likely produce a neutral salt?

A. Strong acid + strong base B. Weak acid + strong base C. Weak base + strong acid D. Weak acid + weak base

12. Hydration involves the interaction of:

A. Water molecules with covalent bonds B. Water molecules and ionic solutes C. Salt and oxygen D. Heat and acids

13. A salt solution has a pH of 5.0. This indicates the salt:

A. is neutral B. undergoes cation hydrolysis C. undergoes anion hydrolysis D. cannot be hydrolysed

14. Which of the following explains why FeCl₃produces an acidic solution?

A. Fe³+ is a weak acid B. Fe³+ reacts with water to release H+ C. FeCl₃is a basic salt D. Cl– reacts with water Structured Questions

1. Define hydration as it applies to ionic compounds.

2. What is meant by hydrolysis of salts?

3. State the difference between salts that produce acidic and basic solutions.

4. Explain why NaCl does not hydrolyse in water.

5. Write a balanced equation for the hydrolysis of NH₄ + in water.

6. Write the hydrolysis equation for CH₃COO– in water. State the nature (acidic/basic/neutral) of the solution.

7. Predict the pH of an aqueous solution of FeCl₃and justify your answer with an equation.

8. Explain why aqueous Na₂CO₃is basic using relevant ionic equations.

9. A student dissolves KNO₃in water. Predict the pH of the solution and explain your reasoning.

10. Using chemical equations, differentiate between hydrolysis of NH₄Cl and CH₃COONa.

11. Classify NH₄Cl, NaCl, CH₃COONa, and AlCl₃as forming acidic, basic, or neutral solutions. Explain.

12. Explain how the extent of hydrolysis is influenced by the strength of the parent acid or base and give examples.

13. Design an experiment to compare the acidity of NH₄Cl, NaCl, and Na₂CO₃.

Include expected results.

14. Discuss the environmental significance of salt hydrolysis.

15. Clarify the pH change of CH₃COONa in water with diagrams and equations.

In this section, you will learn about acid-base indicators, which are special chemicals that change colour when added to acids or bases. These colour changes help scientists know when a reaction is complete, especially during a process called titration. Indicators are usually weak acids or bases, and they show different colours depending on the pH of the solution. Common indicators include:

phenolphthalein – turns pink in bases and colourless in acids and methyl orange

– turns red in acids and yellow in bases. You will learn:

1. why indicators change colour at different pH values

2. which indicator to choose for different types of acid and base reactions

3. how to draw and read a titration curve – a graph that shows how the pH changes as one solution is added to another

4. important parts of the curve include the equivalence point (where the acid and base neutralise) and buffer region (where pH changes slowly) Through fun activities and practice problems, you will also see how indicators are used in real-life jobs, like: making medicine, checking clean water and testing products in factories. To understand Acid-Base Indicators and Titration Curves, refer to this section in the previous years

KEY IDEAS

· Analyte is the solution in the beaker or flask that is being tested.

· Endpoint is the point where the indicator changes colour, showing the titration is almost done.

· Equivalence point is the point in titration when the acid and base have completely reacted.

· Indicator is a chemical that changes colour in an acid or a base.

· Titration is a method where an acid is slowly added to a base (or vice versa) to find the exact point they react completely.

· Titration curve is a graph that shows how pH changes as more titrant is added.

Acid-Base Indicator

An acid-base indicator is a weak acid or base that changes colour when the pH of a solution changes (when it becomes more acidic or more basic). An acid- base indicator is a special chemical that changes colour when added to an acid or a base. It is made from a weak acid or base. It shows one colour in acid and a different colour in base.

How Does It Work?

The indicator exists in two forms: (unionised molecules in equilibrium with their ions)

1. One form (HIn), which represents the molecular form gives Colour 1

2. The other (In–) representing the ionic form gives Colour 2 These two forms are in equilibrium: HIn (Colour 1) ⇌ H+ + In– (Colour 2) When the amount of HIn and In– is the same, the colour looks like a mix of the two colours. The colour changes over a small pH range, depending on the indicator.

When Can We See the Colour Change?

Our eyes notice the colour change when the amount of HIn and In– changes a lot.

This usually happens around a pH range that is about 1 unit above or below the indicator’s pKa value.

Common Acid-Base Indicators and Their Colour

Changes

Table 2.2 shows some common indicators and how their colours change in acids and bases.

Table 2.2: Common Acid-Base Indicators

Indicator Colour in Acid Colour in Base

Litmus Red Blue

Methyl Orange Red Yellow

Phenolphthalein Colourless Pink

Bromothymol Blue Yellow Blue

Universal Indicator Red to Yellow (pH 1–6) Green to Purple (pH 7–14) Choice of Indicator in Acid-Base Titration

When performing a titration (a method to find out how much acid reacts with a base), we use an indicator to tell us when the reaction is finished. This point is called the endpoint or equivalence point. The equivalence point is the point at which stoichiometric amount added of titrant reacts completely with the analyte, leaving no excess. It represents the theoretical end point of the reaction.

End point in the titration is the experimental physical point where the indicator changes colour signalling the end of the titration But not all indicators work for all types of titrations. We must choose the one that changes colour at the right pH.

Simple Guide for Choosing Indicators

Table 2.3: Guide for choosing indicators Type of Titration pH at Equivalence Point Best Indicator Colour Change Strong Acid + Strong Base Around 7 (neutral) Any (e.g. Litmus, Phenolphthalein, Methyl orange) Varies depending on choice Strong Acid + Weak Base Less than 7 (acidic) Methyl orange Red → Yellow Type of Titration pH at Equivalence Point Best Indicator Colour Change Weak Acid + Strong Base Greater than 7 (basic) Phenolphthalein Colourless → Pink Weak Acid + Weak Base No sharp pH change No suitable indicator Use a pH meter instead

Note

· Choose an indicator that changes colour close to the equivalence point pH.

· For sharp colour changes, the indicator must have its working range within the steep part of the titration curve.

Titration Curves

A titration curve is a graph that shows how pH changes as one solution (like an acid or base) is slowly added to another during a titration. The x-axis shows the volume of the titrant added, and the y-axis shows the pH of the solution in the flask.

What Does the Curve Tell Us?

1. At first, the pH changes slowly.

2. Then there is a sudden jump in pH—this is the equivalence point.

3. After the jump, the pH levels off again.

Types of Titration Curves

Table 2.4: Types of titration curves and their equivalence point pH Titration Type Curve Shape Equivalence Point pH Strong Acid + Strong Base Steep jump around pH 7 About pH 7 Weak Acid + Strong Base Gentle rise, steep jump above pH 7 Around pH 8–9 Strong Acid + Weak Base Steep jump below pH 7 Around pH 4–6 Weak Acid + Weak Base Very gentle curve, no sharp jump No clear equivalence point Why Is It Useful?

1. Helps choose the best indicator (look for the steepest part).

2. Shows where the reaction is complete.

3. Used in chemistry labs, medicine, and water testing.

Titration Curve Diagrams

Refer to Figures 3.1, 3.2 & 3.3 for different titration curves.

1. Strong Acid vs Strong Base pH curve for the titration of 25.0 cm³of 0.10 mol dm⁻³HCl with 0.10 mol dm⁻³NaOH

Figure 3.1: Strong Acid vs Strong Base The following points can be deduced from the graph in Figure 3.1

1. initial pH = 1 (pH of 0.10 mol dm⁻³HCl)

2. pH changes only gradually until equivalence

3. very sharp jump in pH at equivalence: from pH 3 to pH 11

4. after the equivalence point the curve flattens out at a high value (pH of strong base).

5. pH at equivalence = 7

2. Weak Acid vs Strong Base pH curve for the titration of 25.0 cm³of 0.10 mol dm⁻³CH₃COOH with 0.10 mol dm⁻³NaOH

Figure 3.2: Weak Acid vs Strong Base The following points can be deduced from the graph in Figure 3.2

1. pH > 1 (pH of 0.10 mol dm–3 weak acid)

2. Addition of the strong base converts some of the weak acid to its conjugate base and creates a buffer. pH stays almost constant through the buffer region where both HA and A– are present. At half-equivalence point (12.5.0 cm3); pH = pKa

3. jumps in pH at equivalence from about pH 7.0 to 11.0, which is not as much of a jump as for a strong acid–strong base titration

4. after the equivalence point the curve flattens out at a high value (pH of strong base).

5. pH at equivalence is > 7

3. Strong Acid vs Weak Base pH curve for the titration of 25.0 cm³of 0.10 mol dm⁻³HCl with 0.10 mol dm⁻³NH₃

Figure 3.3: Strong Acid vs Weak Base The following points can be deduced from the graph in Figure 3.3

1. initial pH = 1 (pH of 0.10 mol dm⁻³HCl)

2. pH stays relatively constant through the buffer region to equivalence

3. jumps in pH at equivalence from about pH 3.0 to 7.0

4. after the equivalence point the curve flattens out at a low pH (pH of weak base).

5. pH at equivalence is < 7

4. Titration of a weak acid and a weak base pH curve for the titration of 25.0 cm³of 0.10 mol dm⁻³CH₃COOH with 0.10 mol dm⁻³NH₃

Figure 3.4: Weak Acid vs Weak Base The following points can be deduced from the graph in Figure 3.4

1. pH > 1 (pH of 0.10 mol dm⁻³weak acid)

2. addition of base causes the pH to rise steadily

3. changes in pH at the equivalence point is much less sharp than in the other titrations

4. after the equivalence point the curve flattens out at a low pH (pH of weak base).

5. pH at equivalence point is difficult to determine Drawing Titration Curves

1. Start with the pH of the first solution: Find the pH before adding anything (usually the acid or base being tested).

2. Add a little bit of the other solution (titrant): Each time you add, calculate the new pH.

3. Find the point where acid and base are equal (equivalence point): This is where the reaction is complete.

4. Add more titrant beyond that point: See how the pH keeps changing after the reaction ends.

5. Draw a graph: Put pH on the y-axis and volume added on the x-axis.

6. Join your points with a smooth line: The graph will have a curve, not sharp corners.

7. Mark the equivalence point: Show where the sharp rise or drop in pH happens.

8. If you see a slow change area, mark it as the buffer region: This is where the

solution resists pH change.

9. Add the indicator’s pH range on the graph: This helps you see if it changes colour at the right time.

Using the Curve to Pick the Right Indicator

1. Find the equivalence point on your graph (big jump in pH).

2. Look at the steep part of the curve near that point.

3. Choose an indicator that changes colour in that steep range.

4. Check if the indicator’s colour change fits

5. It should happen where the big pH change is.

Activity 3.1 Investigating Acid-Base Indicators

Materials needed

1. Common indicators: phenolphthalein, methyl orange, litmus

2. Labelled solutions: distilled water (A), orange juice (B), baking soda (C), vinegar (D), toothpaste solution (E), water (F), soap solution (G)

3. Droppers, test tubes, and white tiles or paper for clearer colour observation Steps

1. Put 10 cm3 of the solutions labelled A, B, C, D, E, F, and G.

2. Add two to three drops of phenolphthalein to solutions in the test tube.

3. Carefully observe the colour changes. Record the observations in the

table below.

Test Observation Deduction

A + phenolphthalein B + phenolphthalein C + phenolphthalein D + phenolphthalein E + phenolphthalein F + phenolphthalein G + phenolphthalein

4. Repeat steps 1,2 and 3 using methyl orange or litmus

Activity 3.2 Drawing a Strong Acid–Strong Base Titration Curve Instructions In this activity, you will draw the pH titration curve for a titration between 25.0 cm³ of 0.10 mol dm⁻³of a strong acid and 0.10 mol/dm³ strong base.

Use the titration data table below to plot the graph of the reaction Steps

1. Draw a horizontal axis labelled “Volume of Base Added (cm³)” from 0 to 50.

2. Draw a vertical axis labelled “pH” from 0 to 14.

3. Use the data table below to plot the pH at different volumes of base added.

4. Connect the points with a smooth curve.

5. Mark the equivalence point (25 cm³, pH = 7) with a dashed vertical line.

Selected Titration Data (Strong Acid with Strong Base)

Volume of Base Added (cm³) pH 0.0 1.00 5.0 1.30 10.0 1.60 15.0 2.30 20.0 3.50 24.0 6.10 24.9 6.90 25.0 7.00 25.1 7.10 26.0 10.00 30.0 12.00 40.0 13.00 50.0 13.50

Note

This is how the curve will look when accurately plotted:

6. Repeat steps 1 to 5. Then, using the titration data table for Titration A below, plot the graph for the titration of a weak acid with a strong base.

Titration A Volume of Base Added (cm³) pH 0.00 2.87 5.05 4.15 10.10 4.58 15.15 4.93 20.20 5.37 25.25 10.70 30.30 11.98 35.35 12.23 40.40 12.37 45.45 12.46

7. At what volume does the steepest change in pH occur?

8. Why is the pH at the equivalence point exactly 7.0?

9. Which indicator is most suitable for this titration?

10. What would happen if you used methyl orange for this titration?

Activity 3.3 Titration Technique

Materials needed

1. Glassware: Burette, Conical Flask (Erlenmeyer flask), Pipette (usually 25.0 cm³), Pipette Filler, Beaker, Funnel, retort stand and Clamp, White Tile

2. Solutions: a solution of known concentration (e.g., NaOH), a solution of unknown concentration (e.g., HCl), Distilled Water

3. Indicator: Phenolphthalein or Methyl Orange

4. Safety Equipment: Gloves, Lab Coat, Safety Goggles

Steps

Step 1: Preparing the Burette (Image 1)

A gloved hand carefully fills a clean burette with a titrant (usually a strong acid or base) using a small glass beaker.

Safety Tips

1. Always wear gloves and protective goggles when handling acids and bases.

2. Ensure the burette is clamped securely to a stand at eye level.

3. Fill below eye level to avoid splashes into the face.

4. Check for air bubbles in the nozzle and remove them before starting.

Step 2: Initial Setup (Image 2)

The full titration setup includes:

1. A vertical burette clamped to a stand.

2. An Erlenmeyer flask (conical flask) beneath the burette, containing a measured volume of analyte (unknown solution) and an indicator (e.g., phenolphthalein, which appears pink in basic solution).

Precision Tips

1. Record the initial burette reading (to the nearest 0.10cm3).

2. Ensure the tip of the burette is just above the flask but not submerged.

3. Use a white tile under the flask to make colour changes more visible.

Step 3: Performing the Titration (Image 3)

1. The titrant is added slowly and steadily from the burette into the flask while swirling the flask gently with the other hand.

2. As the titrant nears the endpoint, slow down the flow to dropwise.

Observation Tips

1. Watch for the first permanent colour change (e.g., pink to colourless if acid is added to base).

2. Avoid overshooting the endpoint by adding titrant too quickly.

3. Record the final burette reading immediately after the endpoint is reached.

4. Repeat the titration at least 2–3 times until two concordant results (within ±0.2cm³) are obtained.

Titration results are recorded in a table looking like this:

Burette Reading/cm³1 2 3

Final 22.80 46.80 22.50 Initial 0.00 24.40 0.00

Volume of acid used 22.80 22.40 22.50

Step 4: After the Endpoint (Image 4)

1. The flask now contains a neutralised solution, shown as clear and colourless.

2. Dispose of chemicals properly and rinse all glassware.

Step 5: Calculate the average titre using two concordant titre values to two decimal places.

Average titre = 22.40 + 22.50/2 = 22.45 cm³Step 6: Calculate the concentration of their unknown solution using their titration results and the formula C_(A) V_(A)_ C_(B) V_(B) = n_(A)_ n_(B) = mole ratio

Figure 3.5: Proper Titration Technique

Activity 3.4 Exploring Indicators, Titration Curves, and Titration

Technique Materials needed Phenolphthalein, methyl orange, bromothymol blue, hydrochloric acid (HCl), sodium hydroxide (NaOH), burette, pipette, conical flask, white tile, beakers, pH meter or pH paper, graph paper or digital plotting tool.

Steps

Step 1: Investigating Acid-Base Indicators

1. Test a range of acid-base indicators (e.g., litmus, phenolphthalein, methyl orange) using known acidic and basic substances (water, distilled water, vinegar, baking soda, hydrochloric acid solution, sodium hydroxide

solution)

2. Record the colour change at different pH levels and match to a pH scale.

3. Discuss how each indicator works and the concept of indicator transition ranges.

Step 2: Performing a Strong Acid–Strong Base Titration

Perform a titration of 25.0 cm³ of 0.10 mol/dm³ HCl with 0.10 mol/dm³ NaOH using phenolphthalein as the indicator. Accurately record volume readings and note the endpoint.

Step 3: Drawing and Interpreting the Titration Curve

1. Use the titration data to plot a titration curve (pH vs volume of base added). Label the initial pH, steep section, and equivalence point on the curve.

2. Interpret the graph and relate it to the theoretical understanding of acid- base neutralisation. Highlight where the indicator’s transition range fits on the curve.

1. a. Explain how acid-base indicators work in terms of pH and colour change.

b. Explain how an acid-base indicator works using the example of phenolphthalein.

2. a. Describe the colour changes of the following indicators across pH ranges:

i. Methyl orange

ii. Phenolphthalein

iii. Bromothymol blue

b. Identify the observable colour changes of methyl red in acidic and basic solutions.

3. a. Compare the suitability of phenolphthalein and methyl orange in a titration between:

i. Strong acid and strong base

ii. Weak acid and strong base

iii. Strong acid and weak base

b. Describe the difference between a strong acid–strong base titration and a weak acid – strong base titration.

4. Label the key parts (initial pH, steep rise, equivalence point) on a typical strong acid strong base titration curve.

5. a. A student titrates 25.0 cm³ of 0.10 mol/dm³ HCl with 0.10 mol dm-³ NaOH and observes the endpoint using phenolphthalein. Explain why this is an appropriate indicator choice for this titration.

b. In a titration between ethanoic acid (CH₃COOH) and NaOH: Justify the choice of phenolphthalein over methyl orange.

6. a. Analyse the differences in titration curves for the following combinations:

i. Strong acid vs. strong base

ii. Weak acid vs. strong base

iii. Strong acid vs. weak base

iv. Weak acid vs. weak base

b. Explain how the equivalence point of a titration helps in the selection of a suitable indicator.

7. a. Draw and annotate the titration curve for a weak acid–strong base titration. Indicate the buffer region and explain its significance.

b. Explain why a weak acid–weak base titration typically does not have a steep vertical portion on the curve, and why indicators are not ideal for it.

In this section, you will discover the fascinating world of redox reactions, where substances gain or lose electrons during chemical changes. You will learn what oxidation and reduction mean, how to identify substances that are gaining or losing electrons, and explore the real-life applications of redox reactions, including their roles in batteries, rusting, and metal extraction. Through simple explanations and hands-on experiments, you will understand how metals react, investigate oxidising and reducing agents, and practice writing half-equations in redox reactions. Additionally, you will discover how redox reactions generate electricity in batteries and fuel cells, learn about electrolytic cells and their applications like electroplating and purifying metals, and understand why metals corrode/rust(for iron) and how to prevent corrosion in everyday objects.

KEY IDEAS

· Corrosion a redox process in which metals react with substances in the environment, often oxygen and moisture, leading to rusting (in the case of iron) or deterioration.

· Displacement reaction: a reaction in which a more reactive metal displaces a less reactive metal from its compound.

· Electron transfer; the movement of electrons from one substance to another during a redox reaction.

· Oxidation is the loss of electrons by a substance during a chemical reaction; it may also involve the gain of oxygen or loss of hydrogen as well as increase in oxidation number.

· Oxidising agent: a substance that gains electrons and causes another substance to be oxidised.

· Reactivity series: a ranking of metals from most reactive to least reactive based on their ability to lose electrons and form positive ions.

· Redox reaction: a chemical reaction in which oxidation and reduction occur simultaneously.

· Reducing agent: a substance that loses electrons and causes another substance to be reduced.

· Reduction: the gain of electrons by a substance. It may also involve the loss of oxygen or gain of hydrogen as well as decrease in oxidation number.

Redox Reactions

Redox reactions cause changes in chemicals where some substances lose electrons (oxidation) and others gain electrons (reduction). These reactions happen all around us in Ghana — in our homes, farms, schools, and towns.

Redox reaction can be defined as a reaction in which electrons are transferred from one reactant to another, resulting in changes in oxidation states of the relevant species.

Real-Life Examples in Ghana

1. Gold Mining (Obuasi & Tarkwa): In gold mining areas, redox reactions help to remove gold from rocks.

2. Traditional Iron Work (Northern Ghana): Blacksmiths heat iron ore with charcoal. This causes a redox reaction that turns iron ore into real iron for tools and decorations.

3. Cocoa Fermentation: When cocoa beans are processed, redox reactions change their smell and colour, giving Ghanaian cocoa its world-famous taste.

4. Rusting Near the Coast (Elmina & Cape Coast): Metal gates, pipes, and buildings rust quickly in salty air. This is an oxidation reaction that damages metals and costs a lot of money to fix.

Definitions of Oxidation and Reduction

Oxidation and reduction can be defined in 4 broad ways. All of which could be applicable to any given situation. Choosing the best one to describe the processes will often depend upon the context.

In Terms of Oxygen

Oxidation: The gain of oxygen by a substance. For example, 4Fe + 3O₂→ 2Fe₂O₃ Iron (Fe) gained oxygen, so Fe is oxidised.

Reduction: The loss of oxygen from a substance.

For example, CuO + H₂→ Cu + H₂O Oxygen is removed from CuO (Copper oxide), hence, CuO is reduced In Terms of Hydrogen Oxidation: The loss of hydrogen from a substance. Example, in CH₄+ 2O₂→ CO₂+ 2H₂O Hydrogen is removed from methane (CH₄), CH₄is oxidised.

Reduction: The gain of hydrogen by a substance. In C₂H4 (g) + H2 (g) → C₂H6 (g), hydrogen is added to ethene (C₂H₄), C₂H₄is reduced.

In Terms of Electron Transfer

Oxidation: The loss of electrons. Example: Na → Na+ + e– (oxidation), Mnemonic: OIL — Oxidation Is Loss of electrons Reduction: The gain of electrons. Example, Cl₂+ 2e– → 2Cl– (reduction) Mnemonic: RIG — Reduction Is Gain of electrons In Terms of Oxidation Numbers Oxidation: An increase in oxidation number of relevant species.

Example: Fe²+ → Fe³+ (oxidation) The oxidation number of Fe²+ increased from +2 to +3 in Fe³⁺. Fe²+ is oxidised.

Reduction: A decrease in oxidation number of relevant species.

Example: MnO₄

– → Mn²+ (reduction) The oxidation number of Mn in MnO₄

– decreased from +7 to +2 in Mn²+. MnO₄

– is reduced.

Real-Life Examples of Redox Reactions

1. Combustion of Fuel (in Cars or Stoves)

Equation: CH₄+ 2O₂→ CO₂+ 2H₂O

Methane (CH₄) is oxidised (loses hydrogen). Oxygen (O₂) is reduced (gains hydrogen). This reaction produces energy (heat and light).

2. Displacement Reaction (Zinc and Copper Sulphate)

Equation: Zn + CuSO₄→ ZnSO₄+ Cu Zinc loses electrons → Oxidised.

Zn → Zn²⁺+ 2e- Copper ions (Cu²+) from CuSO₄gain electrons → Reduced.

Cu²⁺+ 2e-→ Cu

This is used in metal recovery and batteries.

3. Rusting of Iron (in Moist Air)

Equation: 4Fe + 3O₂+ 6H₂O → 4Fe (OH)₃ Iron (Fe) is oxidised to iron ions while oxygen (O₂) is reduced. This is a slow redox reaction that causes rust.

4. Respiration (in Living Cells)

Equation: C₆H₁₂O₆+ 6O₂→ 6CO₂+ 6H₂O + energy Glucose (C₆H₁₂O₆) is oxidised; Oxygen is reduced. This reaction gives energy to living things.

5. Electrolysis of Water (in Labs or Hydrogen Production) Equation: 2H₂O → 2H₂+ O₂ Water is split: H₂O → H₂(reduction at cathode) and H₂O → O₂(oxidation at anode).

H₂O is both oxidised and reduced. Oxidation number of hydrogen in H₂O is decreased from +1 to 0 (reduction) in H₂while that of oxygen is increased from -2 in H₂O to 0 in O₂(oxidation)

6. Corrosion of Copper (Green Patina)

Equation: 2Cu + O₂+ CO₂+ H₂O → Cu₂(OH)₂CO₃ Copper reacts with moist air and CO₂to form green copper carbonate. It is a slow oxidation process.

Activity 4.1 Oxidation and Reduction in Everyday Observations

Step 1

1. Observe these items or materials in Figures 4.1 and 4.2

2. What changes do you observe in these materials?

3. Why do you think the colour or texture has changed?

4. Have you seen anything like this in your community?

Figure 4.1: Images of fresh items Figure 4.2: Images showing redox processes

Step 2

Using the poster above define oxidation and reduction in terms of

1. Gain of oxygen / Loss of oxygen

2. Loss of hydrogen / Gain of hydrogen

3. Loss of electrons / Gain of electrons

4. Increase or decrease in oxidation number

Step 3

Using the reactions below:

1. Assign oxidation numbers

2. Identify the substances oxidised and reduced

3. Explain your reasoning in (1) and (2) 2Mg + O₂→ 2MgO 2Na + Cl₂→ 2NaCl Zn + Cu²+ → Zn²+ + Cu Fe + CuSO₄→ FeSO₄+ Cu H₂+ Cl₂→ 2HCl CH₃CH₂OH + [O] → CH₃COOH + H₂O 2H₂O₂→ 2H₂O + O₂ 4Fe + 3O₂→ 2Fe₂O₃ PbO₂+ 4H+ + 2Cl– → PbCl₂+ Cl₂+ 2H₂O 2Al + 3Cu²+ → 2Al³+ + 3Cu

The reactivity of metals refers to their tendency to lose electrons and form positive ions (cations). Some metals react quickly, while others react slowly or not at all.

A very reactive metal reacts easily with water, air, or acid. A less reactive metal reacts slowly or needs more heat to react.

The Reactivity Series

The Reactivity Series is a list that shows metals from the most reactive to the least reactive.

Here is a part of the list showing the most common metals:

Potassium (K) – most reactive Sodium (Na) Calcium (Ca) Magnesium (Mg) Aluminium (Al) Zinc (Zn) Iron (Fe) Tin (Sn) Lead (Pb) Copper (Cu) – not very reactive Gold (Au) – does not react easily Platinum (Pt) – Least Reactive Experimental Determination of Reactivity of Metals We can find out how reactive a metal is by doing simple experiments. The faster or more strongly a metal reacts, the more reactive it is.

Materials required Metals: Potassium, Sodium, Calcium, Magnesium, Zinc, Iron, Copper Cold water, dilute hydrochloric acid (HCl), test tubes and test tube rack, measuring cylinder, droppers or pipettes, Bunsen burner or source of steam, copper (II) sulphate solution, sandpaper (to clean metal surfaces), tongs and gloves (for safety) Procedure Experiment 1: Reaction with Water

1. Your teacher will demonstrate the reactions of potassium, sodium and calcium, watch these demonstrations and take notes.

2. Clean small pieces of each of the remaining metals using sandpaper.

3. Add about 10 cm³cold water into a test tube.

4. Add a clean piece of metal and observe what happens.

5. Record the speed of reaction (fizzing, movement, heat).

Expected Experimental Results

Experiment 1: Reaction of Metals with Cold Water

Metal Observation in Cold

Water Expected Result

Potassium (K) Reacts violently, sparks, lilac flame Very reactive Sodium (Na) Reacts vigorously, forms NaOH, hydrogen released Very reactive Calcium (Ca) Bubbles (effervescence), forms Ca (OH)₂ Reactive Magnesium (Mg) Very slow/no reaction Slightly reactive Zinc (Zn) No reaction Unreactive in cold water Iron (Fe) No reaction Unreactive in cold water Copper (Cu) No reaction Unreactive in cold water Gold (Au) No reaction Unreactive in cold water Experiment 2: Reaction with Dilute Acid

1. Pour dilute hydrochloric acid into a test tube.

2. Add a clean piece of metal to the acid. DO NOT REACT POTASSIUM, SODIUM OR CALCIUM WITH ACID.

3. Observe bubbling or temperature change.

4. Record how fast hydrogen gas is produced.

Expected Experimental Results

Metal Observation Expected Result

Magnesium (Mg) Rapid effervescence, produces H₂gas Highly reactive Zinc (Zn) Moderate effervescence Reactive Iron (Fe) Slow bubbling Less reactive Copper (Cu) No reaction Unreactive Gold (Au) No reaction Unreactive Experiment 3: Metal Displacement Reaction

1. Pour copper (II) sulphate solution into a test tube.

2. Add a piece of another metal (e.g., zinc or iron).

3. Leave it for a few minutes and observe any colour change or metal deposit.

4. Record whether displacement occurred.

Expected Experimental Results

Metal (Solid) Salt Solution Observation Expected

Result Magnesium

(Mg) Copper (II) sulphate (CuSO₄) Brown deposit of copper, blue solution fades Magnesium is more reactive than copper Zinc (Zn) Copper (II) sulphate (CuSO₄) Brown copper forms on zinc, blue fades Zinc displaces copper Iron (Fe) Copper (II) sulphate (CuSO₄) Reddish-brown copper forms on iron Iron is more reactive than copper Copper (Cu) Zinc sulphate (ZnSO₄) No visible reaction Copper is less reactive than zinc Copper (Cu) Iron (II) sulphate (FeSO₄) No visible reaction Copper cannot displace iron Zinc (Zn) Iron (II) sulphate (FeSO₄) Grey iron forms on zinc, solution fades Zinc displaces iron Experiment 4: Reaction with Oxygen

1. Clean the surface of each metal using sandpaper to remove any oxide layer. Your teacher will demonstrate the reactions of potassium, sodium and calcium. DO NOT ATTEMPT THESE YOURSELF.

2. Hold the metal with tongs and place it in a crucible or on a metal spoon.

3. Heat the metal strongly using a Bunsen burner or spirit lamp.

4. Observe the reaction: Look for signs like flame, colour change, or formation of oxide.

5. Allow the crucible to cool and note the colour and texture of the product formed.

6. Repeat for each metal and compare the intensity of the reaction.

Expected Experimental Results

Metal Observation (upon heating in air/ oxygen) Expected Result Potassium (K) Burns with a lilac flame Very reactive Sodium (Na) Burns with a bright yellow flame Very reactive Calcium (Ca) Burns with a red/orange flame Very reactive Magnesium (Mg) Burns with a bright white flame Highly reactive Zinc (Zn) Burns with bluish-green flame Moderately reactive Iron (Fe) Glows red-hot and forms black oxide Less reactive Copper (Cu) Forms black coating slowly Low reactivity Gold (Au) No reaction Unreactive

Activity 4.2 Investigating Metal Reactivity

Materials needed Test tubes and test-tube rack, beakers (100 mL and 250 mL), measuring cylinders (50 mL), metal samples: magnesium, zinc, iron, copper, salt solutions:

copper (II) sulphate, zinc sulphate, iron (II) sulphate, dilute hydrochloric acid (HCl, ~1 M), distilled water, sandpaper (for cleaning metal surfaces), dropper or pipette, thermometer Stopwatch or clock, stirring rod, gloves, goggles, and lab coats (for safety), tongs or forceps, waste container for used solutions.

This activity should be carried out in group.

Steps

1. Select a set of metals (e.g., magnesium, zinc, iron, copper), choose suitable reactions to investigate (e.g., with acids, water, or salt solutions), decide on measurements to record (e.g., fizzing rate, temperature change), and prepare a data table to organise the results.

2. Use the table below to record your observations and measurements during the metal reactivity experiment. Test each metal with a chosen solution (acid, water, or salt solution) and observe the results.

Metal Tested

Reaction Type

Rate of

Fizzing Temp.

Before (°C) Temp.

After (°C) Colour

Change Reaction

Observations

3. Discuss

a. what you observed and how it matched or differed from what you expected.

b. what may have affected the results (purity of metals, amount of acid, etc.).

4. Identify common patterns and create a class reactivity series. Rank the metals from most to least reactive based on your results.

5. Compare the group reactivity series with the standard reactivity series.

Oxidising Agent

An oxidising agent is a substance that gains electrons during a redox reaction and causes another substance to be oxidised (i.e., to lose electrons). The oxidising agent itself is reduced.

Example: Fe(s) + Cu²+(aq) → Fe²+(aq) + Cu(s) Here, Cu²+ is the oxidising agent because, it gains 2 electrons from Fe. It is reduced to Cu(s).

Characteristics of Oxidising Agents

Refer to Table 4.1 for the Characteristics of oxidising agents.

Table 4.1: Characteristics of oxidising agents Characteristic Explanation Electron acceptor Gains electrons from another substance during a redox reaction.

Causes oxidation Oxidises another substance by removing electrons from it.

Itself undergoes reduction While oxidising others, the oxidising agent is reduced (oxidation number decreases).

Often contains high oxidation states Begins in a high oxidation state and becomes reduced during the reaction.

Commonly non-metals or positive ions Examples include O₂, Cl₂, KMnO₄, K₂Cr₂O₇, H₂O₂, NO₃

–, Cu²+, Fe³+.

Strong oxidisers react readily They have a high tendency to gain electrons (e.g., halogens, KMnO₄).

Used in bleaching and disinfection Destroy organic materials and microbes (e.g., chlorine, ozone).

Reducing Agents

A reducing agent is a substance that loses electrons during a redox reaction and causes another substance to be reduced. It is itself oxidised in the process.

Reducing agents are often metals or species in low oxidation states.

Characteristics of Reducing Agents

Refer to Table 4.2 for the Characteristics of Reducing agents.

Table 4.2: Characteristics of Reducing agents Characteristic Explanation Electron donor Loses electrons to another substance during a redox reaction.

Causes reduction Reduces another substance by donating electrons to it.

Characteristic Explanation

Itself undergoes oxidation While reducing others, the reducing agent is oxidised (oxidation number increases).

Often contains low oxidation states Begins in a low oxidation state and becomes more positive after oxidation.

Commonly metals or negative ions Examples include Fe²+, Zn, H₂, CO, and SO₃²–.

Strong reducers react easily Alkali and alkaline earth metals (e.g., Na, K) are strong reducing agents.

Used in metal extraction and antioxidants Important in both industrial processes and biological systems.

Colour change may indicate reaction Can be observed when reducing agents react with coloured oxidising agents (e.g., KMnO₄becomes colourless).

Example of reaction: Fe(s) + Cu²+(aq) → Fe²+(aq) + Cu(s) Fe is the reducing agent because it loses electrons and is oxidised to Fe²+, reducing Cu²+ to Cu.

Tests for Oxidising and Reducing Agents

In Table 4.3 are common tests used to identify oxidising and reducing agents, along with observable changes and chemical explanations.

Table 4.3: Tests for Oxidising and Reducing Agents

Test Substance

Used To

Test For

Observation Explanation

Potassium iodide (KI) Oxidising agent

Solution turns brown or forms purple vapor Oxidising agent oxidises I– to I₂. Example:

Cl₂+ 2KI → 2KCl + I₂ Fe²+ ions Oxidising agent Pale green turns yellow- brown Fe²+ is oxidised to Fe³+.

Test Substance

Used To

Test For

Observation Explanation

Starch + Iodide Oxidising agent

Solution turns blue-black Oxidising agent releases I₂, which reacts with starch to form blue-black complex.

Potassium permanganate (KMnO₄, acidified) Reducing agent Purple

solution

becomes colourless Reducing agent reduces MnO₄

– to Mn²+.

MnO₄

– + 8H+ + 5e– → Mn²+ + 4H₂O Potassium dichromate (K₂Cr₂O₇, acidified) Reducing agent Orange

solution turns green Reducing agent reduces Cr₂O₇²– to Cr³+.

Cr₂O₇²– + 14H+ + 6e– → 2Cr³+ + 7H₂O Hydrogen peroxide (H₂O₂) Both Effervescence (O₂released) Decomposes; acts as reducing agent by giving off O₂.

Applications of Oxidising and Reducing Agents

Table 4.4 shows summary of Real-Life Applications of Oxidising and Reducing Agents Across Fields Such as Industry, Healthcare, and Water Treatment, and metallurgy.

Table 4.4: Applications of Oxidising and Reducing Agents

Application Area Agent Type Examples Explanation

Bleaching & Disinfection Oxidising agent Chlorine (Cl₂), Hydrogen peroxide (H₂O₂), Ozone (O₃) Oxidises stains, kills bacteria by breaking down cell walls.

Water Treatment Chlorine, Ozone Disinfects water by oxidising microbes and organic waste.

Rocket Propulsion

Potassium nitrate (KNO₃), Ammonium perchlorate Provides oxygen for combustion in fuels.

Application Area Agent Type Examples Explanation

Electrochemical Cells

Oxygen, Permanganate

(MnO₄

–) Accepts electrons at the cathode to drive the cell reaction.

Metallurgy Air (O₂), Oxygen-

enriched blast Oxidises impurities in ores such as FeS to Fe₂O₃.

Metal Extraction Reducing

agent Carbon (C), Carbon monoxide (CO), Aluminum (Al) Reduces metal oxides to obtain pure metals.

Photography (Developing) Hydroquinone,

Metol Reduces silver ions (Ag+) to metallic silver (Ag) on photographic paper or film.

Antioxidants in

Biology Ascorbic acid (Vitamin C), Glutathione Donate electrons to neutralise free radicals and protect cells.

Redox Titration Iron (II) sulfate (FeSO₄) Used to determine the concentration of oxidising agents.

Industrial Processes

Hydrogen gas (H₂), Carbon monoxide (CO) Used in refining metals and removing oxygen from compounds.

Activity 4.3 Understanding Oxidising and Reducing Agents

Using the following redox reaction write the definitions of oxidising and reducing agents using three different approaches: CuO(s) + H₂(g) → Cu(s) + H₂O(g)

1. a. Oxygen/Hydrogen Transfer Copper (II) oxide loses oxygen to form copper.

Hydrogen gains oxygen to form water.

Write your definitions:

Oxidising Agent:

Reducing Agent:

b. Electron Transfer Approach Cu²+ in CuO gains electrons (is reduced) to form Cu.

H₂loses electrons (is oxidised) to form H+ in H₂O.

Write your definitions:

Oxidising Agent:

Reducing Agent:

c. Oxidation Number Change Approach Cu changes from +2 to 0 (reduction).

H changes from 0 to +1 (oxidation).

Write your definitions:

Oxidising Agent:

Reducing Agent:

2. Using the following equations calculate the oxidation numbers and identify oxidising and reducing agents:

a. Zn + CuSO₄→ ZnSO₄+ Cu

b. MnO₄

– + Fe²+ → Mn²+ + Fe³+

3. Share your definitions and reaction analyses with your colleagues.

4. Using the equation: H₂O₂+ 2FeSO₄+ H₂SO₄→ Fe₂(SO₄) ₃+ 2H₂O, apply all three definitions to identify the oxidising and reducing agents.

5. Using the iron-oxygen reaction: 4Fe + 3O₂→ 2Fe₂O₃, identify:

a. the oxidising agent and justify your choice

b. the reducing agent and justify your choice

Oxidation Half-Reaction

An oxidation half-reaction shows the part of a redox (oxidation-reduction) reaction where a substance loses electrons. It focuses only on the oxidation process, not the entire redox equation.

Features of Oxidation Half-Reactions

Feature Description

Electron loss Electrons appear on the product side of the equation.

Oxidation number Increases during oxidation.

Charge balance Must be balanced for both mass and charge.

Occurs at The anode in electrochemical cells.

Examples of Oxidation Half-Reactions

Species Oxidised Oxidation Half-Reaction Explanation

Zinc (Zn) Zn(s) → Zn²+(aq) + 2e– Solid zinc loses 2 electrons to form Zn²+.

Iron (Fe) Fe²+(aq) → Fe³+(aq) + e– Iron (II) ion loses 1 electron to form Fe³+.

Hydrogen gas (H₂) H₂(g) → 2H+(aq) + 2e– Hydrogen gas is oxidised to hydrogen ions.

Carbon monoxide (CO) CO + H₂O → CO₂+ 2H+ + 2e– CO is oxidised in fuel cells.

Reduction Half-Reactions

A reduction half-reaction shows the part of a redox (oxidation-reduction) reaction where a substance gains electrons. It focuses only on the reduction process, not the full redox equation.

Key Features of Reduction Half-Reactions

Feature Description

Electron gain Electrons appear on the reactant side of the equation.

Oxidation number Decreases during reduction.

Charge balance Must be balanced for both mass and charge.

Occurs at The cathode in electrochemical cells.

Examples of Reduction Half-Reactions

Species Reduced Reduction Half-

Reaction Explanation

Copper (II) ion (Cu²+) Cu²+(aq) + 2e– → Cu(s) Cu²+ gains 2 electrons to form solid copper.

Iron (III) ion (Fe³+) Fe³+(aq) + e– → Fe²+(aq) Fe³+ gains 1 electron to form Fe²+.

Hydrogen ion (H+) 2H+(aq) + 2e– → H₂(g) Hydrogen ions gain electrons to form hydrogen gas.

Manganate (VII) ion (MnO₄

–) MnO₄

– + 8H+ + 5e– → Mn²+ + 4H₂O MnO₄

– is reduced to Mn²+ in acidic conditions.

Balancing Redox Reactions in Acidic Medium

Balancing redox (oxidation-reduction) reactions involves ensuring both mass and charge are conserved. In an acidic medium, this means using H+ ions and H₂O molecules to balance atoms and electrons to balance charge.

Systematic Steps

1. Split the reaction into two half-reactions Identify the species undergoing oxidation and reduction and write each half- reaction separately.

2. Balance all elements except H and O Start by balancing atoms such as metals or nonmetals that are not hydrogen or oxygen.

3. Balance oxygen atoms by adding H₂O Add one H₂O molecule for each missing oxygen atom.

4. Balance hydrogen atoms by adding H+ Add one H+ ion for each missing hydrogen atom.

5. Balance charge by adding electrons (e–) Add electrons to the more positive side to balance the charges.

6. Multiply half-reactions to equalise electrons Multiply one or both half-reactions so that the number of electrons lost equals those gained.

7. Add the half-reactions Combine both balanced half-reactions and cancel out electrons and any identical species on both sides (these are known as spectator ions).

8. Final check Ensure atoms and charges are balanced. Confirm the reaction is appropriate for an acidic medium (no OH– used).

Example 4.1

Unbalanced equation: Cr₂O₇²– + HNO₂→ Cr³+ + NO₃

– (Acidic Medium)

Step 1: Split into half-reactions Oxidation: HNO₂→ NO₃

– Reduction: Cr₂O₇²– → Cr³+

Step 2: Balance atoms except H, O Cr₂O₇²– → 2 Cr³+ HNO₂→ NO₃

–

Step 3: Balance O and H

Cr half: Cr₂O₇²– + 14 H+ → 2 Cr³+ + 7 H₂O N half: HNO₂+ H₂O → NO₃

– + 3 H+

Step 4: Balance charge with electrons Cr: 6 e– + 14 H+ + Cr₂O₇²– → 2 Cr³+ + 7 H₂O N: HNO₂+ H₂O → NO₃

– + 3 H+ + 2 e–

Step 5: Equalise electrons (multiply N half by 3) Nhalf × 3: 3 HNO₂+ 3 H₂O → 3 NO₃

– + 9 H+ + 6 e–

Step 6: Combine and cancel duplicates Final: Cr₂O₇²– + 3 HNO₂+ H+ → 2 Cr³+ + 3 NO₃

– + H₂O Cr₂O₇²– + 3 HNO₂+ 5H+ → 2 Cr³+ + 3 NO₃

– + 4H₂O Balancing Redox Reactions in Basic Medium Step-by-Step Method in Basic Solution

1. Divide the reaction into two halfreactions: oxidation and reduction.

2. Balance atoms other than H and O in each half-reaction.

3. Balance oxygen by adding H₂O.

4. Balance hydrogen by adding H+ as needed.

5. Neutralise H+ by adding OH– ions to both sides—this is the critical extra step in basic medium.

6. Combine H+ and OH– to form water, then simplify by removing excess H₂O.

7. Balance the charges by adding electrons (e–) to the more positive side of each halfreaction.

8. Multiply halfreactions so that electrons cancel when added together.

9. Add them, cancel identical species (H₂O, OH–, e–), and check that atoms and charges balance.

Example 4.2

Br– + MnO₄

– → MnO₂+ BrO₃

– in Basic Medium

1. Write halfreactions and balance atoms except H and O Oxidation: Br– → BrO₃

– Reduction: MnO₄

– → MnO₂ Balancing O and H (in acid medium first) Br- + 3H₂O → BrO₃

- + 6H+ MnO₄

- + 4H+ → MnO₂+ 2H₂O Add OH- to both sides of the equations to neutralise H+ Br- + 3H₂O + 6OH- → BrO₃

- + 6H+ +6OH- MnO₄

- + 4H+ + 4OH- → MnO₂+ 2H₂O + 4OH- Combine OH- and H+ to form water and simplify Br- + 3H₂O +6OH- → BrO₃

- + 6H₂O MnO₄

- + 4H₂O → MnO₂+ 2H₂O + 4OH- Simplify to net amounts Br- + 6OH- → BrO₃

- + 3H₂O MnO₄

- + 2H₂O → MnO₂+ 4OH- Balancing charges using electrons Br- + 6OH- → BrO₃

- + 3H₂O + 6e- MnO₄

- + 2H₂O + 3e- MnO₂+ 4OH- Equalise electrons by scaling halfreactions accordingly.

Br- + 6OH-→ BrO₃

- + 3H₂O + 6e- 2MnO₄

- + 4H₂O + 6e- → 2MnO₂+ 4OH- Cancel water and OH– duplicates, simplify.

Final balanced net ionic: H₂O + 2MnO₄ − + Br− → 2MnO₂+ BrO₃ −+ 2OH−

Example 4.3

Balance the reaction in basic medium: ClO– + Mn (OH)₂→ MnO₂+ Cl–

Step 1: Assign oxidation numbers ClO–: Cl is +1 Cl–: Cl is –1 → reduction Mn (OH)₂: Mn is +2 MnO₂: Mn is +4 → oxidation

Step 2: Write half-reactions Oxidation (Mn²+ to Mn⁴⁺in MnO₂): Mn (OH)₂→ MnO₂ Reduction (ClO– to Cl–): ClO− → Cl−

Step 3: Balance each half-reaction Oxidation Half-Reaction (Mn)

1. Balance Mn: Already balanced

2. Balance O: Mn (OH)₂has 2 OH (2 O, 2 H), MnO₂has 2 O Add 2 H+ to RHS to match 2 H in LHS Mn (OH)₂→ MnO₂+ 2H+

3. Neutralise H+: Add 2 OH– to both sides Mn (OH)₂+ 2OH- → MnO₂+ 2H+ +2OH-

4. Mn (OH)₂+ 2OH- → MnO₂+ 2H₂OBalance charge:

LHS: Mn (OH)₂is neutral 0+ -2 from (OH-) = -2 RHS: MnO₂(neutral) 0 + 2 OH– (–2) + 2 from (2H+) = 0 Add 2 e– to RHS Oxidation half-reaction: Mn (OH)₂+ 2OH- → MnO₂+ 2H₂O + 2e− Reduction Half-Reaction (Cl)

1. Balance Cl: already balanced

2. Balance O: ClO– has 1 O → add 1 H₂O to RHS

3. Balance H:2 H+ to RHS then 2OH- to both sides of the equation ClO- + 2H+ +2OH- → Cl- + H₂O +2OH-

4. Balance charge:

LHS: ClO– + 2 OH– + 2H+= -1 RHS: Cl– = –1, H₂O = 0, 2OH- = -2→ net = -3 Add 2 e– to LHS Reduction half-reaction: ClO- + 2H+ +2OH- +2e- → Cl- + H₂O +2OH- ClO- + H₂O- +2e- → Cl- +2OH-

Step 4: Add both half-reactions Mn (OH)₂+ 2OH- → MnO₂+ 2H₂O + 2e− ClO- + H₂O +2e- → Cl- +2OH- Add and cancel electrons:

Mn (OH)₂+ClO−+2OH−→MnO₂+2OH−+Cl−+H₂O Cancel 2OH– from both sides (spectator ions):

Balanced Redox Reaction in Basic Medium: Mn(OH)₂+ ClO− → MnO₂+ Cl− + H₂O

Activity 4.4 Understanding Redox Reactions

Figure 4.3: Images of rusting iron roofs (typical in Ghanaian buildings).

Figure 4.4: Image of rusting

Figure 4.5: Imagery related to gold mining (galamsey/artisanal sites)

1. Observe the images in Figures 4.3-4.5

a. Describe what you observed in these images.

b. What chemical changes might be happening in all situations?

2. Consider the reaction: 4Fe + 3O₂+ H₂O→ 2Fe₂O₃.H₂O

a. Break the reaction down into oxidation and reduction halves.

b. Which species is oxidised? Which is reduced? How can you tell?

3. a. Model galvanic action like water pipes or roofs in coastal towns Cu²+ + Fe⁰→ Cu⁰+ Fe²+

i. Assign oxidation states

ii. Identify which is oxidised/reduced

iii. Write corresponding halfequations

b. Represents gold recovery via the displacement method used in artisanal mining Au³+ + Zn⁰→ Au⁰+ Zn²+

i. Write and balance oxidation and reduction half-reactions

ii. Ensure electrons cancel and show the balanced overall equation

c. This redox pair can relate to water contamination issues near mining sites Fe³+ + SO₃²– → Fe²+ + SO₄²–

i. Balance halfreactions in an acidic medium

ii. Combine into a full balanced reaction

d. Balance halfreactions and write the overall balanced reactions

i. MnO₄

– + C₂O₄²– → Mn²+ + CO₂ (acidic medium)

ii. Pb (OH)₄²– + ClO– → PbO₂+ Cl– (basic medium) Show handling of OH– and H₂O to balance O, H, electrons.

iii. Fe (OH)₂→ Fe₃O₄+ H₂ (basic/anaerobic corrosion context)

4. Identify one redox process in your home or community.

Redox Titrations

Redox titrations are a way to measure how much of a substance is in a solution using a redox reaction. This is different from acid–base titrations, where the reaction involves hydrogen ions (H+) instead of electrons.

When Does the Reaction End?

1. The reaction ends when the right amount of each substance has been added and they have completely reacted.

2. This point is called the equivalence point – it is when the electrons lost and gained are equal.

How Do We Know It’s Done?

1. A colour change tells us the titration is complete.

2. Some reactions change colour on their own (self-indicating).

3. Others need a redox indicator to help us see the change.

Types of Redox Titrations

1. Permanganate titrations: utilising the strong oxidising power of MnO₄

–

2. Dichromate titrations: using Cr₂O₇²– as the oxidising agent

3. Iodometric/iodimetric titrations: based on the redox chemistry of iodine Summary Table Type Titrant Colour Change Indicator Common Use Permanganate KMnO₄ Purple → Colourless Self-indicating Fe²+, H₂O₂ Type Titrant Colour Change Indicator Common Use Dichromate K₂Cr₂O₇ Orange → Green/Blue Diphenylamine Fe²+, Alcohols Iodometric Na₂S₂O₃ Brown → Blue → colourless Starch Bleach, Cu²+, H₂O₂ Iodimetric I₂ Yellow-brown → colourless None/Starch Reducing agents (e.g.

SO₃²–) Standardisation of Potassium Tetraoxomanganate

(VII) (KMnO₄) Solutions Potassium permanganate (KMnO₄) is a strong oxidising agent, but it is not a primary standard because:

1. it is not 100% pure (may contain MnO₂)

2. it decomposes over time, especially in light or in neutral/alkaline solution Therefore, it must be standardised (accurately determined) before use in titrations.

Standardisation Procedure

To determine the exact concentration of KMnO₄by titration with a primary standard reducing agent, such as:

1. Ethanedioic acid (oxalic acid), H₂C₂O₄

2. Iron (II) sulfate, FeSO₄

3. Sodium oxalate, Na₂C₂O₄

Example: Standardising KMnO₄with FeSO₄

Apparatus & Chemicals

1. Burette, pipette, conical flask

2. KMnO₄solution (in burette)

3. Standard FeSO₄solution (in flask)

4. Dilute H₂SO₄(acidic medium)

5. White tile Reaction MnO₄ −+5Fe²⁺+8H+→Mn²⁺+5Fe³⁺+4H₂O Procedure

1. Rinse burette and fill with KMnO₄solution.

2. Pipette 25 cm³ of FeSO₄into conical flask.

3. Add about 20 cm³ of dilute H₂SO₄.

4. Titrate with KMnO₄until the first permanent pale pink colour appears.

5. Repeat for concordant titres.

Calculation Use:

1. Volume and known molarity of Fe²+

2. Balanced redox equation

3. Moles of Fe²+ → moles of KMnO₄

4. Calculate molarity of KMnO₄using:

Molarity = Moles of KMnO₄________________ Volume used (in dm³) Precautions

1. Use acidic medium only (dil. H₂SO₄) – never HCl (Cl– gets oxidised)

2. Protect KMnO₄solution from light and organic contamination

3. Use freshly prepared Fe²+ solution to avoid oxidation by air

Example 4.4: Standardising KMnO₄with FeSO₄

To find the exact concentration of a KMnO₄solution by titrating it against a 0.0200 mol/dm³ FeSO₄solution.

Titration Data

Volume of FeSO₄solution = 25.00 cm³ Molarity of FeSO₄= 0.0200 mol/dm³ Volume of KMnO₄used = 20.00 cm³ Balanced Redox Equation MnO₄ −+5Fe²⁺+8H+→Mn²⁺+5Fe³⁺+4H₂O From the equation, 1 mole of KMnO₄reacts with 5 moles of Fe²+ Step-by-Step Calculation

Step 1: Calculate moles of Fe²+ Moles of Fe²⁺= 25.00/1000 × 0.0200 = 5.00 × 10⁻⁴mol

Step 2: Use mole ratio to find moles of KMnO₄ From the equation:

Moles of KMnO₄= 1 __× 5.00 × 10⁻⁴= 1.00 × 10⁻⁴mol

Step 3: Calculate molarity of KMnO₄ Molarity of KMnO₄ = Moles___________ Volume in dm³= 1.00 × 10⁻⁴_________ 20.00 × 1000 = 0.00500 mol dm⁻³The exact concentration of the KMnO₄solution is 0.00500 mol/dm³ Standardisation of KMnO₄with Ethanedioic Acid Determining the exact concentration of potassium permanganate (KMnO₄)

solution using a primary standard solution of ethanedioic acid (oxalic acid, H₂C₂O₄).

Balanced Redox Equation

2MnO₄

– + 5H₂C₂O₄+ 6H+ → 2Mn²+ + 10CO₂+ 8H₂O Given Data Volume of ethanedioic acid solution = 25.00 cm³ Molarity of ethanedioic acid = 0.0200 mol/dm³ Volume of KMnO₄used = 20.00 cm³ Step-by-Step Calculation

1. Calculate the moles of H₂C₂O₄used:

Moles = (25.00/1000 ) × 0.0200 = 5.00 × 10⁻⁴mol

2. Use the mole ratio from the balanced equation:

Moles of KMnO₄= (2/5) × 5.00 × 10⁻⁴= 2.00 × 10⁻⁴mol

3. Calculate the concentration of KMnO₄ Molarity = moles / volume (in dm³) = 2.00 × 10⁻⁴_________ (20.00/1000 ) = 0.0100 mol/dm³ The exact concentration of the KMnO₄solution is 0.0100 mol/dm³.

Standardisation Using Sodium Thiosulphate

(Indirect Method) Standardisation of Sodium Thiosulphate – Indirect Iodometric Method Determining the exact concentration of sodium thiosulphate (Na₂S₂O₃) solution using the indirect iodometric method. This involves generating iodine from a primary standard (e.g., potassium dichromate) and titrating it with sodium thiosulphate.

Chemical Reactions Involved

Step 1: Iodine generation from potassium dichromate:

Cr₂O₇²– + 14H+ + 6I– → 2Cr³+ + 3I₂+ 7H₂O

Step 2: Titration of iodine with sodium thiosulphate:

I₂+ 2S₂O₃²– → 2I– + S₄O₆²– Apparatus and Reagents

1. Standard potassium dichromate solution (K₂Cr₂O₇)

2. Excess potassium iodide (KI)

3. Dilute sulphuric acid (H₂SO₄) or hydrochloric acid (HCl)

4. Sodium thiosulphate solution (Na₂S₂O₃)

5. Starch indicator (freshly prepared)

6. Burette, pipette, conical flask, white tile Procedure

1. Pipette 25 cm³ of standard K₂Cr₂O₇solution into a conical flask.

2. Add excess KI and acidify with dilute H₂SO₄.

3. Iodine (I₂) is liberated and turns the solution brown.

4. Titrate immediately with Na₂S₂O₃solution from the burette.

5. When the solution turns pale yellow, add a few drops of starch indicator.

6. Continue titration until the blue colour disappears (end point).

7. Repeat to obtain concordant titres.

Calculations

1. Calculate the moles of K₂Cr₂O₇used (primary standard).

2. Determine the moles of I₂produced using the balanced redox equation.

3. Use the stoichiometry to find moles of Na₂S₂O₃that reacted.

4. Calculate the concentration of Na₂S₂O₃: Molarity = moles / volume (in dm³)

Example 4.5

Calculation – Standardisation of Sodium Thiosulphate This example demonstrates how to calculate the exact concentration of sodium thiosulphate (Na₂S₂O₃) solution by titrating it with iodine generated from potassium dichromate (K₂Cr₂O₇).

Relevant Equations

Iodine generation: Cr₂O₇²– + 14H+ + 6I– → 2Cr³+ + 3I₂+ 7H₂O Titration with thiosulphate: I₂+ 2S₂O₃²– → 2I– + S₄O₆²– Given Data

1. Volume of K₂Cr₂O₇used = 25.00 cm³

2. Molarity of K₂Cr₂O₇= 0.0200 mol/dm³

3. Volume of Na₂S₂O₃used = 30.00 cm³ Step-by-Step Calculation

1. Calculate moles of K₂Cr₂O₇: Moles = (25.00 / 1000) × 0.0200 = 5.00 × 10⁻⁴ mol

2. Use mole ratio from the equation: 1 mol Cr₂O₇²– produces 3 mol I₂ So, moles of I₂= 5.00 × 10⁻⁴× 3 = 1.50 × 10–³ mol

3. Use I₂+ 2S₂O₃²– → 2I– + S₄O₆²–: 1 mol I₂reacts with 2 mol Na₂S₂O₃ Moles of Na₂S₂O₃= 1.50 × 10–³ × 2 = 3.00 × 10–³ mol

4. Calculate concentration of Na₂S₂O₃: Molarity = moles / volume in dm³ = 3.00 × 10–³ / (30.00 / 1000) = 0.100 mol/dm³ The exact concentration of the Na₂S₂O₃solution is 0.100 mol/dm³.

Standardisation of Sodium Thiosulphate Using

Potassium Iodate (KIO₃)

This guides learners through the process of determining the exact concentration of sodium thiosulphate (Na₂S₂O₃) solution using potassium iodate (KIO₃), a stable primary standard. Iodine is liberated by reaction with iodide ions in acidic medium and then titrated with sodium thiosulphate.

Key Chemical Reactions

1. Iodine generation: IO₃

– + 5I– + 6H+ → 3I₂+ 3H₂O

2. Iodine titration with sodium thiosulphate: I₂+ 2S₂O₃²– → 2I– + S₄O₆²–

Example Data

Volume of KIO₃solution = 25.00 cm³ Molarity of KIO₃= 0.0200 mol/dm³ Volume of Na₂S₂O₃used = 36.00 cm³ Step-by-Step Calculation

1. Calculate moles of KIO₃: Moles = (25.00 / 1000) × 0.0200 = 5.00 × 10⁻⁴ mol

2. Determine moles of I₂produced (from 1 mol KIO₃→ 3 mol I₂):

Moles of I₂= 5.00 × 10⁻⁴× 3 = 1.50 × 10–³ mol

3. Calculate moles of Na₂S₂O₃needed (1 mol I₂: 2 mol Na₂S₂O₃):

Moles of Na₂S₂O₃= 2 × 1.50 × 10–³ = 3.00 × 10–³ mol

4. Determine concentration of Na₂S₂O₃:

Molarity = moles / volume (in dm³) = 3.00 × 10–³ / (36.00 / 1000) = 0.0833 mol/dm³ The concentration of the sodium thiosulphate solution is 0.0833 mol/dm³.

Example 4.6: Redox Titration Example – Potassium Permanganate and Iron (II) Chemicals and Apparatus

1. Solution A: 0.05 mol/dm³ potassium tetraoxomanganate (VII) (KMnO₄)

2. Solution B: Iron (II) solution of unknown concentration

3. Dilute sulphuric acid (1M)

4. Burette (50 cm³), Pipette (25 cm³), Conical flask (250 cm³)

5. White tile, Retort stand and clamp Procedure

1. Set up the titration apparatus and fill the burette with solution A (KMnO₄).

2. Using a pipette, transfer exactly 25.0 cm³ of solution B into a clean conical flask.

3. Add approximately 10 cm³ of dilute sulphuric acid and swirl gently to mix.

4. Place the conical flask on a white tile under the burette.

5. Titrate against solution A until a permanent pale pink colour appears (endpoint).

6. Record your readings in a titration table.

7. Repeat the titration to obtain at least two concordant titres and calculate the average.

Balanced Redox Equation

MnO₄

– + 5Fe²+ + 8H+ → Mn²+ + 5Fe³+ + 4H₂O Analysis and Questions

1. Using your results and the equation above, calculate the concentration of Fe²+ in mol/dm³.

2. Solution B was prepared by dissolving 71.5 g of ammonium iron (II) sulphate hydrate, (NH₄)₂Fe(SO₄)₂·xH₂O, per dm³ of solution. Calculate the value of x, the number of water molecules of crystallisation in the salt.

3. A groundwater sample from a community near a mining site contains iron. The water appears clear when collected but turns reddish-brown after standing. Explain the chemical transformations occurring in the water and how this titration method accounts for these changes.

Answer

1. Concentration of Fe²+

Assume average titre of KMnO₄used = 23.50 cm³ (0.02350 dm³) Molarity of KMnO₄= 0.050 mol/dm³ Moles of KMnO₄= 0.050 × 0.02350 = 1.175 × 10–³ mol From the balanced equation:

1 mol MnO₄

– reacts with 5 mol Fe²+ Moles of Fe²+ = 5 × 1.175 × 10–³ = 5.875 × 10–³ mol Volume of Fe²+ solution used = 25.00 cm³ = 0.02500 dm³ Concentration of Fe²+ = 5.875 × 10–³ / 0.02500 = 0.235 mol/dm³

2. Value of x in (NH₄)₂Fe (SO₄)₂·xH₂O Mass of salt used per dm³ = 71.5 g Moles of Fe²+ = 0.235 mol (from above) Molar mass of anhydrous part = (2×18.04) + 55.85 + (2×96.06) = 284.06 g/mol Let x = number of water molecules (18.02 g/mol each) Molar mass = 284.06 + 18.02x 71.5 / (284.06 + 18.02x) = 0.235 => 284.06 + 18.02x = 71.5 / 0.235 ≈ 304.26 => 18.02x = 304.26 - 284.06 = 20.2 => x ≈ 1.12 → round to 1 (indicating monohydrate, but commonly x ≈ 6 for hexahydrate)

Note that actual known hydrate is x = 6

3. Iron Transformation in Groundwater

Fresh groundwater contains soluble Fe²+ ions, which are colourless.

Upon exposure to air (oxygen), Fe²+ is oxidised to Fe³+:

4Fe²+ + O₂+ 4H+ → 4Fe³+ + 2H₂O Fe³+ reacts with water to form reddish-brown precipitate of iron (III) hydroxide:

Fe³+ + 3H₂O → Fe (OH)₃+ 3H+ The titration method detects Fe²+ before oxidation or after converting Fe³+ back to Fe²+.

Activity 4.5 Redox Titration

1. Laboratory preparation Set up workstations with:

a. Equipment: burettes, pipettes, flasks, beakers, measuring cylinders, stands, clamps, thermometers, hot water baths.

b. Labelled Solutions:

i. Standard: KMnO₄(permanganate), Na₂S₂O₃(thiosulphate)

ii. Unknowns: Fe²+, C₂O₄²– (oxalate), I₂/KI

2. Form groups with roles:

a. Equipment Manager

b. Titrator (Operator)

c. Recorder

d. Analyst/Reporter

3. Procedure Distribution

a. Redox Titration Procedure Sheets

b. Conduct the titrations on rotational bases Titration 1: KMnO₄ vs Fe²+ Purpose: To determine the concentration of Fe²+ using standard KMnO₄.

You may use the link below to watch a video on titration of KMnO₄vs Fe²+ https://www.youtube.com/watch?v=ci4cHGLVZQY Materials

a. 0.020 mol/dm³ potassium permanganate (KMnO₄)

b. Iron (II) sulphate or Mohr’s salt (unknown concentration)

c. Dilute sulphuric acid (H₂SO₄)

d. Burette, pipette, conical flask, white tile, funnel, clamp stand Procedure

a. Rinse and fill the burette with KMnO₄solution. Record initial volume.

b. Pipette 25.0 cm³ of Fe²+ solution into a conical flask.

c. Add about 15 cm³ of dilute H₂SO₄.

d. Titrate with KMnO₄while swirling.

e. Stop when a faint pink colour persists for 30 seconds.

f. Record final volume and repeat for consistent titres.

Observations

a. KMnO₄is deep purple.

b. Fe²+ solution becomes faint pink at endpoint.

Safety

a. KMnO₄stains—avoid contact.

b. H₂SO₄is corrosive—use gloves and goggles.

Titration 2: KMnO₄ vs Ethanedioate (C₂O₄²–) Purpose: To determine the concentration of sodium ethanedioate using standard KMnO₄.

You may use the link below to watch a video on titration of KMnO₄ vs C₂O₄²– https://www.youtube.com/watch?v=GkroZP3oNEw Materials

a. 0.020 mol/dm³ KMnO₄

b. Sodium ethanedioate solution (unknown)

c. Dilute sulphuric acid

d. Water bath or hotplate, thermometer

e. Burette, pipette, conical flask Procedure

a. Pipette 25.0 cm³ of ethanedioate solution into a conical flask.

b. Add 20 cm³ of dilute H₂SO₄.

c. Warm the flask in a 60–70 °C water bath.

d. Fill the burette with KMnO₄and record initial volume.

e. Titrate while swirling; endpoint is faint pink persisting.

f. Repeat to get concordant titres.

Observations

a. Purple KMnO₄is decolourised.

b. Faint pink appears at endpoint in warm solution.

Safety

a. Use thermometer to monitor temperature.

b. Avoid skin contact with KMnO₄and H₂SO₄.

Titration 3: Iodine/Potassium Iodide vs Sodium Thiosulphate Purpose: To determine the iodine concentration using sodium thiosulphate. You may use the link below to watch a video on titration of Iodine/Potassium Iodide vs Sodium Thiosulphate https://www.youtube.com/watch?v=lkJyP6OYmAY&t=255s Materials

a. Iodine/potassium iodide solution (I₂/KI)

b. 0.1 mol/dm³ sodium thiosulphate (Na₂S₂O₃)

c. Fresh starch solution

d. Burette, pipette, conical flask Procedure

a. Pipette 25.0 cm³ of iodine solution into a conical flask.

b. Add 10 cm³ of potassium iodide (KI).

c. Titrate with sodium thiosulphate.

d. When brown fades to pale yellow, add 1–2 cm³ starch solution.

e. Continue titration until solution turns colourless.

f. Repeat for accuracy.

Observations

a. Brown iodine fades.

b. Blue-black colour appears with starch.

c. Solution becomes colourless at endpoint.

Safety

a. Iodine is an irritant—handle with care.

b. Use starch only near the endpoint. (when the solution turns pale yellow)

4. Problem-solving and analysis

a. Complete data tables: Record titration volumes and average results for all three reactions as shown in the titration table.

Titration table format Burette readings/ cm³1 2 3 Final reading 22.50 22.20 22.30 Initial reading 0.00 0.00 0.00 Titre 22.50 22.20 22.30 Average titre = 22.20 + 22.30/2 = 22.25cm3

b. Balanced redox equations: Review balanced equations and identify mole ratios Mn O₄ − + 5 Fe²⁺+ 8 H+ → Mn²⁺+ 5 Fe³⁺+ 4 H₂ O (mole ratio 1:5) 2Mn O₄ − + 5 C₂ O₄ − + 16 H+ → 2 Mn²⁺+ 10 CO₂ + 8 H₂ O (mole ratio 2:5) I₂ + 2 S₂ O₃ ²⁻→ 2 I− + S₄ O₆ ²⁻(mole ratio 1:2)

c. Calculations (with units and ratios): Calculate Unknown Concentrations C_(Mn)_(O)₄ − V_(Mn)_(O)₄ − _ C_(Fe)₂₊ V_(Fe)₂₊ = 1 _ _ C_(Mn)_(O)₄ − V_(Mn)_(O)₄ − _ C_(C)₂_(O)₄ − V_(C)₂_(O)₄ − = 2 _ _ C_(I)₂ V_(I)₂_ C_(S)₂_(O)₃ ₂₋ V_(S)₂_(O)₃ ₂₋ = 1 _ _

5. Identify potential experimental errors

a. Endpoint detection difficulties

b. Air bubbles in burette

c. Misreading meniscus

d. Temperature variations

6. Relate calculated concentrations to:

a. water quality standards

b. iron content in supplements

c. vitamin C and iodine analysis

Electrochemistry What is Electrochemistry?

Electrochemistry is the study of how electricity and chemical reactions are connected. It helps us understand:

1. how batteries work

2. why some metals rust /corrode

3. how we can use electricity to break or decompose substances apart (like water into hydrogen and oxygen) In electrochemistry, we look at how electrons move from one substance to another.

This movement can either:

1. Produce electricity (like in batteries), or

2. Use electricity to cause a chemical change (like in electrolysis).

Electrochemical Cell

An electrochemical cell is a device that uses chemical reactions to produce electricity or uses electricity to cause chemical changes. There are two main types of electrochemical cells:

1. Voltaic (or Galvanic) Cells – Generate electricity from chemical reactions (like in batteries).

2. Electrolytic Cells – Use electricity to make chemical reactions happen (like splitting water into gases).

Inside an electrochemical cell, electrons move from one part of the cell to another.

This movement of electrons creates electric current. The cell has two metal parts called electrodes, dipped in a liquid called an electrolyte.

Example: A Simple Voltaic Cell

If you connect a zinc rod (Zn) and a copper rod (Cu) in two solutions and join them with a wire and a salt bridge, electricity flows. This is a simple battery — a voltaic cell in action.

When a metal plate is dipped into a solution containing its own ions, an electrochemical process begins.

Example: Zinc metal in zinc ion solution (Zn in Zn²+) What happens?

1. Equilibrium is set up between the metal atoms and metal ions: Zn (s) ⇌ Zn²⁺(aq) + 2e−

2. Two things can occur depending on the metal and the ion concentration:

a. Oxidation: Some metal atoms lose electrons and go into the solution as positive ions.

b. Reduction: Some metal ions in solution gain electrons and stick to the metal plate as solid metal.

3. This causes a build-up of electrical charge on the metal plate, and a potential difference (voltage) is created between the metal and the solution.

Result

1. The metal becomes positively or negatively charged relative to the solution.

2. If connected in a complete circuit with another metal and its ions, an electrochemical cell is formed, and electricity can flow.

This process is the basis of all electrochemical cells – where metals “push” or “pull” electrons based on how easily they form ions.

Figure 4.6: Diagram of an electrochemical cell (Voltaic cell) Composition of an Electrochemical Cell Electrochemical Cell consists of the following main components

1. Electrodes: Electrodes are solid conductors, usually metals, where redox reactions take place.

Anode: Site of oxidation (loss of electrons). Electrons are released here. In voltaic cells, the anode is negative.

Cathode: Site of reduction (gain of electrons). Electrons are accepted here.

In voltaic cells, the cathode is positive.

2. Electrolytes: Electrolytes are ionic solutions that allow the movement of ions between electrodes and complete the internal circuit. They may be molten (melted) salts or aqueous solutions of salts

3. Salt or ion Bridge (or Porous Partition): A salt bridge allows ion flow between the two half-cells. It maintains electrical neutrality by letting ions balance charges.

4. External Circuit: The external circuit is usually a wire that connects the electrodes. Electrons flow from the anode to the cathode.

5. Voltmeter: The voltmeter is used to measure the potential difference (voltage) between the electrodes. It shows the cell’s electrical output.

Electrode Potential

This is the measure of the tendency of an electrode (metal or non-metal) in contact with its ions in solution to gain or lose electrons, resulting in the creation of an electric potential (voltage) between the electrode and the solution. In an electrochemical cell, each half-cell has its own electrode potential. When two half-cells are connected, electrons flow from the electrode with the lower (more negative) potential to the one with the higher (more positive) potential. This flow of electrons generates electricity.

Standard Hydrogen Electrode (SHE)

The Standard Hydrogen Electrode (SHE) is the reference electrode used to measure the standard electrode potentials (E°) of other half-cells.

Figure 4.7: A standard hydrogen electrode Pt(s) | H₂(g, 1 atm) | H+(aq, 1 M) Features of the SHE Electrode reaction: 2H+(aq) + 2e– ⇌ H₂(g) Electrode potential (E°): 0.00 V (by definition) Electrode material: Platinum (Pt) – inert and conducts electrons Gas used: Hydrogen gas (H₂) Pressure of gas: 1 atm or 100 kPa Concentration of H+: 1.0 mol/dm³ (1 M solution, e.g., HCl) Temperature: 298 K (25°C) – standard temperature Function of SHE

1. Acts as a reference half-cell in electrochemical measurements.

2. Used to measure standard electrode potentials of other half-cells.

Uses of the SHE

1. Measuring Standard Electrode Potentials: Connect the SHE to another half- cell to form a complete electrochemical cell and measure the voltage.

2. Comparing Oxidising/Reducing Powers

a. Positive E°: species is a better oxidising agent than H+.

b. Negative E°: species is a better reducing agent than H₂.

Limitations

1. Difficult to set up and maintain under laboratory conditions.

2. Platinum surface can be poisoned by impurities.

3. Not practical for routine lab use – replaced by secondary reference electrodes (e.g., calomel or Ag/AgCl).

Standard Electrode Potential (E°)

To compare different electrodes, we use standard electrode potentials measured under standard conditions. All values are compared against the Standard Hydrogen Electrode (SHE), which is assigned a potential of 0.00 V.

The Standard Electrode Potential, symbol E°, is the voltage of a half-cell under standard conditions, measured relative to the Standard Hydrogen Electrode (SHE), which is defined as 0.00 V.

Standard Conditions

1. Concentration: 1.0 mol/dm³ of ions

2. Temperature: 298 K (25°C)

3. Pressure: 1 atm (for gases)

4. Electrode: A clean metal or inert electrode (like platinum) if the substance is not a metal E° values are measured under standard conditions. A positive E° means the half- cell has a greater tendency to gain electrons (be reduced) than the hydrogen electrode. A negative E° means the half-cell is more likely to lose electrons (be oxidised). This is used to predict the direction of redox reactions and to calculate cell potentials.

Example

1. E°(Zn²+/Zn) = -0.76 V → zinc prefers to lose electrons (oxidation) when compared to the SHE.

2. E°(Cu²+/Cu) = +0.34 V → copper prefers to gain electrons (reduction) when compared to the SHE Applications

1. Predicting redox reaction feasibility or spontaneity

2. Designing electrochemical cells

3. Determining which species acts as oxidising or reducing agent

4. Used to predict the emf (electromotive force) of an electrochemical cell Factors That Affect Electrode Potential Values The main factors that affect electrode potential values are:

1. Concentration of ions Changes in the concentration of the metal ions in solution can shift the electrode potential. According to the Nernst equation, increasing the concentration of the ions involved in the half-cell increases the electrode potential for reduction reactions and decreases it for oxidation reactions.

The Nernst Equation

The Nernst Equation is used to calculate the electrode potential of a half- cell or full electrochemical cell under non-standard conditions. It accounts for the effect of ion concentration or gas pressure on the cell potential. The two forms are:

a. Base-10 Logarithmic Form (Common in Textbooks) Standard Form (at 25°C or 298 K) E = E° − 0.0591/n log([Red]_ [Ox] )

b. Full Nernst Equation E = E° − RT___ nF In([Red]_ [Ox] ) Where E = electrode potential (V) E° = standard electrode potential (V) R = gas constant = 8.314 J mol–¹ K–¹ T = temperature in Kelvin (K) n = number of electrons transferred F = Faraday constant = 96500 C mol–¹ [Red], [Ox] = concentrations or partial pressures of reduced and oxidised species

Example 4.7

Cell reaction: Fe³+ + e– ⇌ Fe²+ Standard electrode potential, E° = +0.77 V Temperature = 25°C = 298 K [Fe²+] = 0.10 mol/dm³, [Fe³+] = 1.00 mol/dm³ n = 1 (number of electrons transferred)

a. Using the Base-10 Logarithmic Form E = E° − 0.0591/n log([Red]_ [Ox] ) E = 0.77 − 0.0591/1 log(^(0.10)_ _(1.00)) = 0.77 − 0.0591(− 1) = 0.77 + 0.0591 = 0.8291 V

b. Using the Natural Logarithmic Form E = E° − RT___ nF In([Red]_ [Ox] ) E = 0.77 − 8.314 × 298/1 × 96500 In(0.10_ 1 ) = 0.77 − 0.0257 × (− 2.3026) = 0.77 + 0.0591 = 0.8291 V Conclusion Both forms of the Nernst equation give the same result (0.8291 V) at 25°C.

· Use the simplified log version at 25°C for quick calculations.

· Use the natural log version (ln) for accuracy and variable temperature calculations.

2. Temperature: Electrode potentials vary with temperature.

The standard electrode potential is measured at 25°C (298 K). Deviating from this temperature can affect reaction kinetics and equilibrium, thereby altering the potential.

3. Pressure (for Gaseous Electrodes): Affects electrode potential when gases are involved (e.g. in hydrogen or oxygen electrodes). Changes in the partial pressure of a gas can shift the position of equilibrium and thus affect the potential.

4. Nature of the Electrode Metal or Material: Different materials have different tendencies to lose or gain electrons. The intrinsic properties of the metal (e.g., ionisation energy, electronegativity, lattice energy) influence its electrode potential.

5. Presence of Complexing Agents: Complexation can lower or raise the free ion concentration, thus changing the electrode potential. Complexes can stabilise certain oxidation states, shifting the equilibrium and affecting the potential (e.g., Ag+ + 2NH₃→ [Ag (NH₃)₂]+).

6. Type of Solvent: The solvent can influence ion solubility and reaction kinetics. Solvent properties like dielectric constant and solvation energy affect the electrode reaction and therefore the potential.

7. Surface Condition of the Electrode: Oxidation, corrosion, or impurities on the electrode surface can alter potential. Surface films may inhibit electron transfer or change the effective surface area.

Activity 4.6 Understanding Electrode Potential

Step 1: Team up with your peers to explore and carry out this task. In your group appoint a recorder, researcher, presenter, and coordinator.

You may use the video link below to carry out Activity 4.6.

https://chatgpt.com/c/68931de5-8d9c-832a-9226-f87bcc491972

Step 2: In your groups discuss

a. What happens when a metal is placed in a solution of its own ions?

b. Why does a potential difference arise?

c. Use drawings to explain how a potential is developed at the metal-solution interface.

d. What determines the direction and magnitude of electron movement?

Step 3: Research and discuss how the following factors affect electrode potential:

Factor Research Questions

(a) Nature of electrode How does the metal’s reactivity influence potential?

(b) Temperature of electrolyte What happens to the potential when temperature increases or decreases?

(c) Concentration of ions How does ion concentration shift the equilibrium and potential?

(d) Pressure of gas (for gaseous electrodes) How does gas pressure (e.g., H₂(g) in SHE) affect potential values?

Step 4: Complete the chart below as a group. Indicate whether each factor increases, decreases, or has no effect on electrode potential, and explain why.

Factor Effect on E°

(↑ / ↓ / ↔) Explanation Real-World

Example

Nature of the electrode Temperature of electrolyte Concentration of electrolyte Pressure of gas species

Step 5: Group presentation of charts

Step 6: Whole-class discussion, clarifying misconceptions Activity4.7 Understanding Electrode Potentials and Voltaic Cells

1. Observe Figure 4.8.

a. Can you measure the voltage of just one end of it? Why or why not?

b. What do you need to measure a voltage or potential difference?

c. Does voltage exist at a single point, or is it always between two points?

d. How do you think this relates to measuring the potential of a single electrode in electrochemistry?

Figure 4.8: Dry cell batteries wired to a voltmeter

2. Observe Figure 4.9. Describe SHE in their own words.

Figure 4.9: The standard hydrogen electrode (SHE)

3. Observe Figure 4.10.

Figure 4.10: A typical voltaic cell

a. Identify the anode, cathode, salt bridge, voltmeter and electrolytes in the diagram.

b. State the standard conditions.

c. Write the redox equation for this cell

4. a. Explain how SHE is used as a reference.

b. Discuss examples like SHE || Cu²+/Cu and interpret the meaning of positive or negative E° values.

5. Representing Voltaic Cells from Half-Cells: Consider the half cells

a. Metals in contact with ions: Mg/Mg²+,

b. Gas - ions: Cl₂/Cl– Draw, label and describe each cell, noting oxidation and reduction.

6. Writing Cell Notation

Example: Zn | Zn²+ || Cu²+ | Cu

a. Write cell notation for Voltaic Cells drawn in step 5 above.

b. Identify anode and cathode

Note

Symbols: single line (phase), double line (salt bridge), left = anode and right = cathode.

7. Write the following cell notations

a. Zn²+/Zn and Ag+/Ag

b. Fe³+/Fe²+ and SHE

c. Cu²+/Cu and SHE

Activity 4.8 Video Explanation of a Voltaic Cell

Materials needed: Laptop or tablet

1. You may use the link below to watch a video on how a Voltaic cell works.

https://chatgpt.com/c/6894d49e-ea04-832b-a653-62b7d2eb6607 As you watch the video, identify:

a. What materials make up the anode and cathode?

b. How do electrons move through the external circuit?

c. Which species is oxidised, and which is reduced?

d. How does the salt bridge help maintain neutrality?

e. Which electrode becomes negatively charged and why?

2. Pause the video at key moments:

a. What process occurred at each electrode?

b. In which direction are the electrons moving? How do we know?”

c. Ask: “What is the salt bridge doing here?”

3. a. What might happen if one electrode material changes?

b. How would the voltage change with different ion concentrations?

c. What do you predict will happen if the salt bridge is removed?

d. How would the reaction cease if one reactant runs out?

Calculations of standard electrode potentials The standard electrode potential (E°) of a half-cell is the potential difference between it and the Standard Hydrogen Electrode (SHE), when both are under standard conditions:

For a voltaic cell:

E°_(cell)= E°_(cathode)− E°_(anode) Cathode → where reduction occurs (gain of electrons) Anode → where oxidation occurs (loss of electrons) Steps in Calculating E°cell

1. Write the two half-equations for the redox reaction.

2. Identify which is the oxidation and which is the reduction.

3. Assign the correct E° values from (reduction potentials only).

4. Apply the formula: E°_(cell)= E°_(cathode)− E°_(anode)

5. Interpret the sign:

a. If E°cell > 0, reaction is spontaneous under standard conditions.

b. If E°cell < 0, reaction is non-spontaneous under standard conditions.

Example 4.8

A cell is made from a magnesium electrode in Mg²+ (aq) and a copper electrode in Cu²+ (aq), both at 1 mol dm–³.

Given:

E°(Mg²+/Mg) = −2.37 V E°(Cu²+/Cu) = +0.34 V Calculate E°cell and state whether the reaction is spontaneous.

Answer

Step 1 – Write the half-equations:

Reduction (Cathode): Cu²+ + 2e– → Cu E° = +0.34 V Oxidation (Anode): Mg → Mg²+ + 2e– E° = −2.37 V

Step 2 – Apply formula:

E°cell = (+0.34) − (−2.37) E°cell = 0.34 + 2.37 = +2.71 V

Step 3 – Interpretation:

E°cell = +2.71 V → Positive, so the reaction is spontaneous.

Overall cell reaction: Mg(s) + Cu²+(aq) → Mg²+(aq) + Cu(s) Common Variations in Questions

1. Given E°cell and one half-cell value, find the other half-cell potential.

2. Predict spontaneity of a reaction.

3. Rank metals or ions by oxidising/reducing strength using E° values.

4. Use E° values to predict displacement reactions.

Predicting the Feasibility of a Reaction Using Standard

Cell Potential and Gibbs Free Energy

1. Relationship Between E°_(cell) and ΔG° The link between standard cell potential and Gibbs free energy is given by the thermodynamic equation: ΔG° = −nF E°_(cell) Where:

ΔG° = Standard Gibbs free energy change (J or kJ mol–¹) n = Number of moles of electrons transferred in the balanced redox equation F = Faraday constant = 96,500 C mol–¹ E°cell = Standard cell potential (V)

2. Prediction Rules

a. If E°_(cell)> 0 → ΔG° < 0 → Reaction is feasible/spontaneous under standard conditions.

b. If E°cell = 0 → ΔG° = 0 → Reaction is at equilibrium under standard conditions.

c. If E°cell < 0 → ΔG° > 0 → Reaction is non-feasible/non-spontaneous under standard conditions.

3. Step-by-Step Procedure

a. Write the balanced redox reaction and determine n (electrons transferred).

b. Look up or calculate E°_(cell): E°_(cell)= E°_(cathode)− E°_(anode)

c. Calculate ΔG° using: ΔG° = −nF E°_(cell)

d. Interpret the sign of ΔG° to predict feasibility.

Example 4.9

Given E°(Zn²+/Zn) = −0.76 V E°(Cu²+/Cu) = +0.34 V Predict whether the reaction: Zn₍ₛ₎+ Cu²+ (aq) → Zn²+ (aq) + Cu₍ₛ₎is feasible under standard conditions.

Answer

Step 1 – Identify electrodes Cathode (reduction): Cu²+ + 2e– → Cu E° = +0.34 V Anode (oxidation): Zn → Zn²+ + 2e– E° = −0.76 V

Step 2 – Calculate E°_(cell) E°_(cell)= (+0.34) − (−0.76) = 1.10 V

Step 3 – Calculate ΔG° n = 2 F = 96,500 C mol–¹ ΔG° = −(2) (96,500) (1.10) = −212,300 J mol–¹ = −212.3 kJ mol–¹

Step 4 – Interpretation ΔG° is negative → reaction is spontaneous under standard conditions.

Experiment to Determine the Emf of a Voltaic Cell Aim: To construct a Daniell cell and determine its electromotive force (EMF) under standard conditions.

Materials needed

1. Zinc strip (clean, polished)

2. Copper strip (clean, polished)

3. 1.0 mol dm–³ zinc sulphate solution (ZnSO₄)

4. 1.0 mol dm–³ copper sulphate solution (CuSO₄)

5. Salt bridge (e.g., KNO₃or NH₄NO₃in agar gel)

6. Digital voltmeter (high resistance, ≥10 MΩ)

7. Connecting wires and crocodile clips

8. Two 250 cm³ beakers Procedure

1. Pour 1.0 mol dm–³ ZnSO₄ into one beaker and 1.0 mol dm–³ CuSO₄ into another.

2. Place the zinc strip in the ZnSO₄solution and the copper strip in the CuSO₄

solution.

3. Connect the zinc electrode to the negative terminal of the voltmeter and the copper electrode to the positive terminal using wires and clips.

4. Insert the ends of the salt bridge into each beaker to allow ion flow and maintain electrical neutrality.

5. Observe and record the voltage displayed on the voltmeter as soon as the circuit is complete.

Expected Reaction in the Daniell Cell

Anode (oxidation): Zn(s) → Zn²+(aq) + 2e– Cathode (reduction): Cu²+(aq) + 2e– → Cu(s) Overall reaction: Zn(s) + Cu²+(aq) → Zn²+(aq) + Cu(s) Expected Result Standard EMF of Daniell cell:

E°_(cell)= E°_(cathode)- E°_(anode) E°_(cell)= (+0.34) − (−0.76) = +1.10 V Measured EMF under ideal conditions ≈ 1.10 V.

Precautions

1. Ensure electrodes are clean and free from oxides before use.

2. Avoid drawing current for long periods to prevent polarisation.

3. Keep the salt bridge saturated and ensure no air bubbles block ion flow.

4. Use solutions at 1.0 mol dm–³ and maintain temperature at 25 °C for standard conditions.

Applications of Principles of Electrode Potentials

Application Principle Example

Predicting Feasibility of

Redox Reactions

A redox reaction is feasible under standard conditions if E°_(cell)= E°_(cathode)− E°_(anode)> 0.

Zn displaces Cu²+ because E°(Cu²+/Cu) = +0.34 V > E°(Zn²+/Zn) = −0.76 V.

Predicting Direction of

Electron Flow

Electrons flow from electrode with lower E° (more negative, stronger reducing agent) to one with higher E°.

In a Daniell cell, electrons flow from Zn (−0.76 V) to Cu (+0.34 V).

Electrochemical Series and

Reactivity Elements arranged in order of increasing E° help predict metal reactivity and displacement reactions.

Mg (−2.37 V) is more reactive than Zn (−0.76 V).

Calculating EMF

of Cells E°_(cell)= E°_(cathode)− E°_(anode) gives the voltage of a cell.

For Zn–Cu cell: E°_(cell)= 0.34 − (−0.76) = +1.10 V.

Determining Gibbs Free Energy

Change ΔG° = −nF E°_(cell)links electrical work to thermodynamic feasibility.

For E°_(cell)= +1.10 V, n = 2, ΔG° = −212.3 kJ mol–¹.

Industrial Electrolysis

E° values predict which electrode reactions occur and the potential required.

Electroplating: E° values help choose conditions for desired metal deposition.

Application Principle Example

Corrosion Prevention

More reactive metals (negative E°) act as sacrificial anodes to protect less reactive metals.

Zn or Mg blocks protect steel hulls from rusting.

Electroplating and

Metal Refining

E° values predict deposition order during electrolysis.

Silver plating: Ag+ reduced at cathode, silver anode dissolves.

Designing Batteries and Fuel

Cells Half-cells with large positive E°_(cell)produce high-voltage cells.

Li-ion batteries use Li/Li+ (very negative E°) with strong oxidising cathode.

Analytical Chemistry

Redox titrations rely on predictable E° values to detect equivalence points.

Permanganate titration of Fe²+ uses known E° values for endpoint detection.

Activity 4.9 Understanding Standard Electrode Potentials and

Connection with Gibbs Free Energy (ΔG°)

Materials Mini E° table: Mini Electrochemical Series (Standard Reduction Potentials) Half-reaction (reduction) E° / V MnO₄

– + 8H+ + 5e– → Mn²+ + 4H₂O +1.51 Cl₂+ 2e– → 2Cl– +1.36 Ag+ + e– → Ag +0.80 Fe³+ + e– → Fe²+ +0.77 Cu²+ + 2e– → Cu +0.34 2H+ + 2e– → H₂(g) 0.00 Pb²+ + 2e– → Pb −0.13 Fe²+ + 2e– → Fe −0.44 Zn²+ + 2e– → Zn −0.76 Al³+ + 3e– → Al −1.66 All tabulated E° values are for reduction

1. Understanding Standard Electrode Potentials

E° values refer to reduction under standard conditions (1 M, 25 °C, 1 atm) e.g.

Zn²+ + 2e– → Zn E° = –0.76 V Cu²+ + 2e– → Cu E° = +0.34 V More positive E° → stronger oxidising agent (more likely to be reduced).

More negative E° → stronger reducing agent (more likely to be oxidised).

2. Calculating Standard Cell Potentials (E°_(cell)) Using the Mini Electrochemical Series above:

a. Identify the two half-reactions and their E° values

b. Label both as reduction

c. Determine which is to be oxidised (more negative E°) vs. reduced (more positive E°)

d. Reverse the oxidation reaction’s E° sign

e. Apply: E°cell = E°cathode − E°anode

f. ΔG°= −nF E°cell where n= moles of electrons transferred.

F = 96500 C mol⁻¹Interpretation

a. E°cell >0 ⇒ spontaneous under standard conditions

b. E°cell <0 ⇒ non-spontaneous under standard conditions.

c. A negative ΔG° confirms spontaneity

3. Calculate E°cell and ΔG°for a Zn/Cu cell.

a. Consider the simple Cell Cu²⁺+2e−→Cu E°= +0.34 V Zn²⁺+2e−→Zn E°= −0.76 V

i. Identify anode and cathode.

Cu²+/Cu is more positive ⇒ cathode (reduction);

Zn²+/Zn becomes oxidation.

Reverse oxidation: Zn→Zn²⁺+2e− (E° = +0.76 V after sign change).

ii. Calculate E°cell.

E°cell = E°cathode − E°anode E°cell = 0.34 − (−0.76) = +1.10 V

iii. Determine ΔG° for this reaction.

ΔG°= −nF E°cell n = 2 (two electrons) ΔG∘ = −2(96500) (1.10) = −2.123×10⁵J ≈ −212 kJ Negative ΔG°, positive E°cell ⇒ feasible/spontaneous.

b. Consider the simple Cell Fe³⁺+e− → Fe²⁺E° = +0.77V Al³⁺+3e− → Al E° = –1.66V

i. Identify which is oxidised and reduced.

ii. Calculate E°cell and ΔG°.

Activity 4.10 Build and Explore Voltaic Cells

Goal: Assemble simple cells and discover which pairings give higher emf.

Materials needed

1. Metals (strips/nails): Zn, Cu, Fe, Pb, graphite (pencil lead)

2. Electrolytes: 0.5 – 1.0 M solutions (e.g., CuSO₄, ZnSO₄, FeSO₄, NaCl)

3. Salt bridge: filter paper soaked in saturated KNO₃/NaNO₃(or salt-soaked paper towel)

4. Beakers/label tape; multimeter (DC, 0–2 V range); crocodile clips;

stopwatch Mini E° (reduction) reference (25 °C, 1 M, 1 atm) Ag+ + e– → Ag +0.80 V Fe³+ + e– → Fe²+ +0.77 V Cu²+ + 2e– → Cu +0.34 V 2H+ + 2e– → H₂ 0.00 V Pb²+ + 2e– → Pb −0.13 V Fe²+ + 2e– → Fe −0.44 V Zn²+ + 2e– → Zn −0.76 V Procedure

1. In small groups, set up two half-cells e.g., Zn/ZnSO₄and Cu/CuSO₄ Fe/FeSO₄and Cu/CuSO₄ Pb/Pb(NO₃)₂and Cu/CuSO₄ Cu/CuSO₄and Ag/AgNO₃, connected by a salt bridge.

Which electrode do you expect to be the anode/cathode? Why?

2. Connect to the multimeter or voltmeter (red to predicted cathode).

a. Record emf and polarity.

b. Draw cell diagrams

3. Using standard reduction potentials from the mini electrochemical series, calculate the theoretical E°cell.

a. Compare measured versus calculated values

b. Discuss possible sources of error: non-standard concentrations, temperature changes, contact resistance, electrode polarisation

4. Electrode Reactions Exploration using Worksheet. Research and complete the following for each battery type:

a. Lead-acid battery Anode half-reaction (oxidation):

Cathode half-reaction (reduction):

Overall balanced equation:

b. Alkaline cell Anode half-reaction (oxidation):

Cathode half-reaction (reduction):

Overall balanced equation:

c. Lithium-ion battery Anode half-reaction (oxidation):

Cathode half-reaction (reduction):

Overall balanced equation:

d. Nickel-cadmium battery (NiCd) Anode half-reaction (oxidation):

Cathode half-reaction (reduction):

Overall balanced equation:

e. Fuel cells (H₂–O₂) Anode half-reaction (oxidation):

Cathode half-reaction (reduction):

Overall balanced equation:

5. Discuss why certain battery types are used for specific applications.

Activity 4.11 Storage & Fuel Cell Technologies

1. Watch the following videos How batteries work - Adam Jacobson | TED-Ed How a Fuel Cell Electric Vehicle Works on Vimeo https://www.youtube.com/watch?v=Rwy6WozMbt8 https://www.youtube.com/watch?v=SRt1KOP9a2g

2. For each video, pause at key moments to discuss and

a. List the components you can recognise. What does each do in one phrase?

b. Write the anode and cathode.

c. State features that make fuel cells better suited than batteries for long-haul vehicles.

d. the component of fuel cell systems that improves their lifespan to match that of diesel engines.

Fill the table as you watch the video System Key compo- nents spot- ted Evidence of anode vs cathode action Energy transforma- tions Advantages over die- sel-powered Challenges/ risks H₂fuel-cell EV Hydrogen-rich fuel system (on-board reformer, etc.)

Battery electric (reference)

3. Concept Mapping for Hydrogen Fuel Cells: Use the diagram in Figure 4.11 as visual reference for Hydrogen Fuel Cells. Develop four main branches from the central topic and add key subcomponents to each.

a. Fuel Cell Components

b. Electrochemical Reactions

i. Anode (oxidation)

ii. Cathode (reduction) Overall

c. Energy Transformations

d. Advantages over Batteries & Diesel

Figure 4.11: Hydrogen fuel cell

Electrolysis What is Electrolysis?

Electrolysis is a process that uses electricity to break down chemical compounds.

How Does it Work?

Electricity passes through a solution or liquid of the substance, called an electrolyte.

The electricity causes chemical changes, breaking down the substance. This process stores energy in the form of new chemical bonds.

Key Points

1. Electrolysis needs electricity to work.

2. It is the opposite of a battery, which makes electricity from chemicals.

3. Electrolysis helps create new substances and store energy.

The word “electrolysis” comes from Greek words “electron” (electricity) and “lysis” (breaking down).

Electrolytic Cell

What is an Electrolytic Cell?

An electrolytic cell is a setup that uses electrical energy from an external source (like a battery or power supply) to force a chemical reaction that would not happen on its own. It is the opposite of a galvanic or voltaic cell (which produces electricity from a spontaneous reaction). In an electrolytic cell, electricity is supplied to drive a non-spontaneous redox reaction.

Basic Components

1. Power Source – a battery or DC supply to push electrons into the system.

2. Electrodes – two conductors dipped in the electrolyte:

a. Cathode (–): where reduction happens (gain of electrons).

b. Anode (+): where oxidation happens (loss of electrons).

Note

In electrolytic cells, the cathode is negative, and the anode is positive because of the external power supply.

3. Electrolyte – a molten ionic compound or ionic solution that conducts electricity (provides free-moving ions).

Figure 4.12: An Electrolytic Cell

How it Works

1. The external power supply pushes electrons into the cathode (negative electrode). Positive ions (cations) in the electrolyte move to the cathode to gain electrons → reduction occurs.

2. The power supply pulls electrons from the anode (positive electrode).

Negative ions (anions) in the electrolyte move to the anode to give up electrons → oxidation occurs.

3. The overall reaction is a decomposition (breaking down) of the electrolyte or compound.

Factors That Affect Selective Discharge of Species During

Electrolysis

1. Position in the Electrochemical Series (discharge/reduction potential) Cathode (−): Ions with a more positive E° (easier to reduce) discharge first.

Example: In aqueous NaCl, H+ (from water) is reduced to H₂rather than Na+ to Na metal.

Anode (+): Species that are easier to oxidise discharge first.

Example: With enough Cl–, Cl₂forms; otherwise, OH– from water is oxidised to O₂.

Electrochemical Series for Cations

K+ Na+ Ca²⁺Mg²⁺Al³⁺Zn²⁺Fe²⁺Sn²⁺Pb²⁺H+ Cu²⁺Hg²⁺Ag+ Au+

Ease of discharge increases Electrochemical Series for Anions F- SO4²⁻NO3- Cl- Br- I- OH- Ease of discharge increases The electrochemical series is an arrangement of ions in order of their relative ability to accept or release electrons. Cations lower in the series readily accepts electrons at the cathode and get reduced than those higher in the series Anions lower in the series give out electrons more easily and are therefore discharged in preference to those higher in the series

2. Concentration of ions High concentration favours that ion (shifts the Nernst potential).

This applies mainly to anions in aqueous solutions in which the halide ion concentration is higher than OH- concentration. In the case of SO₄ ²⁻and NO₃

- ions, OH- ions are still preferentially discharged even if their concentrations are higher. They are extremely difficult to discharge in aqueous solutions because they have high oxidation potential and will not easily give up electrons.

Example: Concentrated brine (NaCl): Cl- is discharged to form Cl₂at the anode instead of OH- ions from H₂O.

In Dilute brine: O_(2(g))is formed at the anode instead (from OH–).

3. Nature of the electrodes (inert vs active/corrodible) Inert anodes (Pt/graphite): They are electrodes that conduct electrons but do not undergo oxidation or reduction. The electrolyte supplies the anion for oxidation (e.g., Cl₂or O₂) Active anodes: These are electrodes that actively participate in the electrochemical cell reaction (oxidation or reduction). They can dissolve and supply metal ions.

Example: Cu anode in CuSO₄: the anode (Cu₍ₛ₎) dissolves to Cu²+; at the cathode, Cu²+ plates out (forms Cu₍ₛ₎) This is the basis of copper refining.

No O_(2(g))forms.

Electrolysis of Molten Compounds in Inert

Electrodes When an ionic compound is melted (molten), its ions are free to move and carry a charge.

1. Molten lead (II) bromide, PbBr₂₍ₗ₎

a. Cathode: Pb²⁺+ 2e− → Pb₍ₗ₎

b. Anode: 2Br− → Br_(2(g))+ 2e−

c. Overall: PbBr₂₍ₗ₎→ Pb₍ₗ₎+ Br_(2(g))

2. Molten sodium chloride, NaCl₍ₗ₎

a. Cathode: Na_((aq)) + + e− → Na₍ₗ₎

b. Anode: 2Cl_((aq)) − → Cl_(2(g))+ 2e−

c. Overall: 2NaCl₍ₗ₎→ 2Na₍ₗ₎+ Cl_(2(g))

3. Molten alumina (Al₂O₃)

a. Cathode: Al_((aq))³+ + 3e– → Al₍ₗ₎

b. Anode: 2O_((aq))²– → O_(2(g))+ 4e–

c. Overall: 2Al₂O₃₍ₛ₎→ 4Al₍ₗ₎+ 3O_(2(g)) Electrolysis of Aqueous Solutions using Inert Electrodes Assume inert electrodes (Pt/graphite) and room temperature. For each electrolyte:

(i) half-reactions, (ii) observations, (iii) products, (iv) change to electrolyte.

Dilute NaCl (aq)

Item Details

Half-reactions Cathode (–): 2H₂O + 2e– → H₂(_(g))+ 2OH– Anode (+): 4OH– → O_(2(g))+ 2H₂O + 4e– (water/OH– oxidised) Observations Cathode: Colourless bubbles (H₂).

Anode: Colourless bubbles (O₂), relights glowing splint.

Products Cathode: Hydrogen gas (H₂).

Anode: Oxygen gas (O₂).

Item Details

Electrolyte change

Solution becomes alkaline (OH– builds up) → NaOH_((aq)) forms; [Cl–] largely unchanged.

Concentrated NaCl (aq) (brine)

Item Details

Half-reactions Cathode (–): 2H₂O + 2e– → H_(2(g))+ 2OH– Anode (+): 2Cl– → Cl_(2(g))+ 2e– (high [Cl–] favours formation of Cl₂) Observations Cathode: Effervescence (H₂).

Anode: Greenish-yellow gas with pungent smell (Cl₂).

Products Cathode: Hydrogen gas (H₂).

Anode: Chlorine gas (Cl₂).

Electrolyte change

Solution becomes alkaline (OH–) → NaOH(aq) forms;

[Cl–] decreases as Cl₂leaves; some dissolved Cl₂may form HOCl/ClO–.

CuSO₄ (aq) Item Details

Half-reactions Cathode (–): Cu²⁺+ 2e– → Cu₍ₛ₎ Anode (+): 2H₂O → O_(2(g))+ 4H+ + 4e– (SO₄ ²⁻not discharged) Observations Cathode: Reddish-brown copper deposit.

Anode: Bubbles of O₂; blue colour of solution fades over time.

Products Cathode: Copper metal (Cu).

Anode: Oxygen gas (O₂).

Electrolyte change [Cu²⁺] falls (blue fades); [H+] rises → solution becomes more acidic (effectively forming H₂SO₄with SO₄ ²⁻).

Dilute H₂SO₄ (aq) Item Details

Half-reactions Cathode (–): 2H+ + 2e– → H_(2(g)) Anode (+): 2H₂O → O_(2(g))+ 4H+ + 4e– Observations Cathode: Bubbles of H₂.

Anode: Bubbles of O₂. Gas volume ratio ~2:1 (H₂:O₂).

Products Cathode: Hydrogen gas (H₂).

Anode: Oxygen gas (O₂).

Electrolyte change Overall acid concentration ~unchanged (H+ consumed at cathode ≈ H+ produced at anode).

AgNO₃ (aq) Item Details

Half-reactions Cathode (–): Ag+ + e– → Ag₍ₛ₎ Anode (+): 2H₂O → O_(2(g))+ 4H+ + 4e– (NO₃

– not discharged) Observations Cathode: Grey/silver crystals deposit.

Anode: Bubbles of O₂.

Products Cathode: Silver metal (Ag).

Anode: Oxygen gas (O₂).

Electrolyte change [Ag+] decreases; [H+] increases → solution becomes acidic (effectively some HNO₃formed).

Concentrated KI (aq)

Item Details

Half-reactions Cathode (–): 2H₂O + 2e– → H₂(g) + 2OH– (K+ not reduced) Anode (+): 2I– → I₂+ 2e– Item Details Observations Cathode: Bubbles of H₂.

Anode: Brown colour near anode (I₂); with starch, blue- black colour.

Products Cathode: Hydrogen gas (H₂).

Anode: Iodine (I₂).

Electrolyte change

Solution becomes alkaline (OH–) near cathode; [I–] falls; I₂ may form I₃

–/hypoiodite in alkaline medium.

Electrolysis with Active Electrodes

1. Copper (II) sulphate solution with copper electrodes (active anode) Half-reactions Cathode (−): Cu²⁺+ 2e− → Cu₍ₛ₎ (copper plates out) Anode (+): Cu₍ₛ₎→ Cu²⁺+ 2e− (copper anode dissolves) Observations Cathode: Reddish-brown Cu coating forms, gets thicker.

Anode: Copper electrode thins; flakes.

Solution: Blue colour ~constant (if anode and cathode areas/currents are balanced).

Products at each electrode Cathode: Copper metal.

Anode: Copper (II) ions (from dissolving anode).

Change to the electrolyte [Cu²⁺] is maintained ~constant (Cu²+ used at cathode is replenished at anode); SO₄ ²⁻unchanged; pH ~unchanged.

2. Concentrated sodium chloride solution with a mercury cathode Half-reactions Cathode (Hg, −): Na+ + e−+ Hg → Na (Hg) (sodium amalgam) Anode (+): 2Cl− → Cl₂(g) + 2e− Observations Anode: Greenish-yellow Cl₂gas (pungent) bubbles off.

Cathode: Silvery amalgam film forms on Hg surface.

Smell: Sharp, irritating odour of Chlorine near the anode.

Products at each electrode Cathode: Na (Hg) (sodium amalgam, intermediate).

Anode: Chlorine gas (Cl₂).

Change to the electrolyte [Cl−] Cl− decreases (Cl₂leaves).

[Na+] Na+ effectively becomes NaOH(aq) (solution turns more alkaline).

Safety note Mercury is toxic—industrial process, not for school labs.

3. Electrolysis of water using nickel-based electrodes (Ni works in alkaline media, e.g., 20–30% KOH/NaOH; Ni surfaces cycle between Ni (OH)₂at the anode.)

Half-reactions (alkaline) Cathode (−): 2H₂O + 2e− → H₂(g) + 2OH− Anode (+): 4OH− → O₂(g) + 2H₂O + 4e− Observations Cathode: H₂gas bubbles.

Anode: O₂gas bubbles (about half the cathode volume for equal time).

Electrodes: Anode may darken (Ni (OH)₂film). Slight warming of electrolyte.

Products at each electrode Cathode: Hydrogen gas (H₂).

Anode: Oxygen gas (O₂).

Gas volume ratio: H₂:O₂≈2:1 Change to the electrolyte Net water is consumed → alkaline solution becomes slightly more concentrated over time; pH stays high (OH– made at cathode ≈ used at anode), composition (KOH/NaOH) essentially unchanged aside from dilution effects.

4. Electrolysis of copper (II) sulphate solution (CuSO₄(aq)) using graphite (inert) electrodes Half-reactions Cathode (–): Cu²⁺(aq) + 2e− → Cu(s) Anode (+): 2H₂O(l) → O₂(g) + 4H+(aq) + 4e− Reason: SO₄ ²⁻is very hard to oxidise; water is oxidised to oxygen.

Overall (net) ionic equation Multiply the cathode reaction by 2 and add:

2Cu²⁺(aq) + 2H₂O(l) → 2Cu(s) + O₂(g) + 4H+(aq) In the presence of SO₄ ²⁻, the rise in H+ effectively means H₂SO₄is formed in solution.

Observations Cathode (–): Reddish-brown copper metal plates onto the electrode; cathode mass increases.

Anode (+): Colourless gas (O₂) bubbles; a glowing splint relights. Graphite remains largely intact (may slowly erode at high current).

Solution: The blue colour fades with time (as [Cu²⁺] falls).

Products at each electrode Cathode: Copper metal (Cu) (solid coating).

Anode: Oxygen gas (O₂).

Change to the electrolyte [Cu²⁺] decreases → blue colour fades.

[H+] increases → solution becomes more acidic (effectively H₂SO₄(aq) forms with the sulphate present).

SO₄ ²⁻essentially unchanged (spectator).

Activity 4.12 Components of an Electrolytic Cell

Materials needed Blank cell template or access to one digital simulation or short video, and a printed reference diagram

1. Use the link below to watch the video on electrolysis or use the diagram in Figure 4.13.

https://chatgpt.com/c/68ad0880-8cdc-8325-a4f3-a96f75ea370e

Figure 4.13: Electrolytic cell parts

2. Draw a labelled electrolytic cell diagram. The details of the diagram should include:

Anode, cathode, electrolyte, power supply/source, direction of electron flow and direction of current flow

3. Explain the function(s) of the various parts of the diagram.

a. the roles of electrodes,

b. describes electron/ion flow, another discusses the power source, etc.

Activity 4.13 Modelling an Electrolytic Cell

Materials

1. Desk signs: Anode Bank (+), Cathode Bank (−), Power Supply (Manager)

2. Electron tokens (paper chips)

3. Role cards: Cations (+), Anions (−), Tellers, Manager

4. Tape arrows on floor: Wire path (external circuit) + Electrolyte zone (between banks)

5. A4 recording sheets (diagram + notes) Roles

1. Bank Manager (Power Supply): pushes tokens (electrons) from the Cathode → Anode through the wire.

2. Tellers: receive or give out tokens at each bank.

3. Cations (+): move to Cathode Bank.

4. Anions (−): move to Anode Bank.

Procedure

Step 1

Create a visual setup: Draw or position “Anode Bank” (left, positive) and “Cathode Bank” (right, negative), include the Power Supply “Manager” in the middle, and arrowed paths for electron flow (Anode → Power Supply → Cathode).

Step 2: Introduce Aqueous Ions & Discharge Rules

1. Add water into the mix: Introduce H+ and OH– ions from water.

2. Discharge rules for ions:

a. Cathode (–): If metal ions are less reactive than H+, they get reduced;

otherwise, H+ → H₂forms.

b. Anode (+): Halides (Cl–, Br–, I–) are discharged first. If none, OH– → O₂is released.

3. Use priority slips: Give each ion a “priority card” (e.g., high to low) and role-play their turn to get discharged.

Step 3: Trace movements: As a class, recap the paths each ion and electron took. Discuss the Manager’s role (Power Supply):

Step 4: Diagram & Record In small groups, sketch a simple diagram with:

1. Anode / Cathode “banks”

2. Electron flow arrows

3. Ion paths (H+, OH–, etc.) and discharge outcomes

Activity 4.14 Selective Discharge and Products

Step 1: Write down factors that affect ions that are discharged during electrolysis.

Step 2: Discus the following factors

a. Position in reactivity series

b. Ion concentration

c. Electrode type

d. Applied voltage

Step 3: Model by example comparisons (demonstration)

a. Molten salt (e.g., NaCl): Na+ discharged at cathode, Cl– at anode.

b. Dilute aqueous NaCl: H₂forms at cathode if H+ discharge is more favourable; Cl₂at anode unless OH– (from water) is favoured.

Step 4: Predict the outcomes of Electrolysis with inert versus active electrodes

a. Molten NaCl

b. Inert electrodes in dilute NaCl solution

c. Inert electrodes in concentrated NaCl solution

d. Active copper electrode in CuSO₄solution Write half-equations and state the key factor accounting for the discharge.

Step 5: Share your findings with your class.

Activity 4.15 Electrolysis Circus Inquiry

Materials needed

1. Electrodes: Inert (graphite rods, platinum wires, or carbon rods)

2. Power source: 6–12 V DC supply or battery pack

3. Electrolysis cell/container: Beaker (100 – 250 mL)

4. Connecting wires with crocodile clips

5. Electrolytes (examples)

a. Copper (II) sulphate solution (CuSO₄, 0.1 – 0.5 M)

b. Sodium chloride solution (NaCl, dilute and concentrated)

c. Potassium iodide solution (KI, 0.1 – 0.5 M)

d. Dilute sulphuric acid (H₂SO₄, 0.5 M)

6. Measuring cylinders and droppers

7. Safety equipment: Goggles, gloves, lab coats

8. Indicators/test reagents

a. Blue and red litmus paper

b. Glowing splint (for O₂test)

c. Burning splint (for H₂test) Safety Precautions

1. Always wear goggles and gloves.

2. Do not touch electrodes or solutions while current flows.

3. Handle acids and copper salts with care; wash spills immediately.

4. Work with low voltage DC only (not mains electricity).

Steps

1. In small groups investigate and state the ions present, possible products, half- equations of the following:

a. Electrolysis of CuSO₄solution using Cu electrodes

b. Electrolysis of NaCl solution (dilute vs. concentrated) using C electrodes.

c. Electrolysis of KI solution using C electrodes.

d. Electrolysis of dilute H₂SO₄using C electrodes.

2. Circus Experiment Stations

Group Work Instructions

a. Roles within each group (e.g., setup technician, recorder, product tester, safety officer).

b. Rotate roles between trials so everyone participates.

c. Record all observations in a results table (electrolyte, cathode product, anode product, evidence/test used).

d. Compare group results and discuss any differences.

3. Experimental Setup

a. Place 100 mL of electrolyte in the beaker.

b. Insert two inert electrodes into the solution, keeping them apart but not touching.

c. Connect the electrodes to the DC power supply using wires and crocodile clips. Mark the anode (+) and cathode (−) clearly.

d. Prepare test papers/reagents for identifying gases and products.

e. Switch on the power supply (6–9 V recommended).

4. Observe bubbling at the electrodes (gas evolution).

5. Collect gases (if possible) using an inverted test tube or test near the electrode.

6. Test products

a. At cathode (−): Test gas with a burning splint (H₂“pop” test).

b. At anode (+): Test with glowing splint (O₂relights), starch (iodine turns blue-black), or chlorine test (bleaches litmus).

7. Record observations (colour changes, gas evolution, electrode deposits).

8. Repeat the experiment for different electrolytes:

a. CuSO₄→ copper deposit at cathode, O₂at anode.

b. NaCl (dilute) → H₂at cathode, O₂at anode.

c. NaCl (concentrated) → H₂at cathode, Cl₂at anode.

d. KI → H₂at cathode, I₂at anode (starch test).

e. Dilute H₂SO₄→ H₂at cathode, O₂at anode.

Discussions

1. What factors determine which ion is discharged at the electrodes?

2. How do results differ between dilute and concentrated NaCl solutions?

3. Why is copper deposited in the electrolysis of CuSO₄?

4. How does this link to industrial applications (e.g., electroplating, chlor- alkali process)?

Electroplating means covering one metal with a thin layer of another metal using electricity. The object to be coated is made the cathode (–), and the coating metal is the anode (+). Both are placed in a solution that has ions of the coating metal.

When electricity passes, metal ions move from the solution and stick to the object (cathode). The coating metal (anode) dissolves to replace the ions in solution.

Practical Applications of Electrolysis

1. Extraction of Metals: Used to extract reactive metals (e.g., aluminium, sodium, magnesium) from their ores.

2. Electroplating: Example: Silver plating spoons; chrome plating car parts.

3. Manufacture of Chemicals

a. Production of important substances: Chlorine, hydrogen, sodium hydroxide (by electrolysis of brine).

b. Oxygen and hydrogen (by electrolysis of water).

4. Purification of Metals

5. Medical and Everyday Uses

a. Electrolysis used for hair removal (cosmetic).

b. Rechargeable batteries involve reversible electrolysis processes.

Experiment Copper Electroplating

Materials

1. Copper sulphate solution

2. Copper strip (anode, +)

3. Metal object to coat (key, coin, etc.)

4. DC power supply (3 – 6 V battery or adapter)

5. Wires with clips

6. Beaker or plastic cup

7. Sandpaper or steel wool

8. Soap water (for grease)

9. Mass balance

10. Vinegar (for cleaning) Procedure

Step 1: Clean the object

1. Rub with sandpaper to make it shiny.

2. Wash in soap water, rinse.

3. Dip quickly in vinegar, rinse again. (Clean objects plate better!)

Step 2: Set up the cell

1. Pour copper sulphate solution into the beaker.

2. Determine the mass of the object to be coated using a mass balance.

3. Connect the object to the negative (–) terminal → cathode.

4. Determine the mass of the copper strip using a mass balance.

5. Connect the copper strip to the positive (+) terminal → anode.

6. Put both into the solution, but don’t let them touch.

Step 3: Electroplate

1. Switch on power (3 – 4 V).

2. Let it run for 15 – 30 minutes.

3. Watch: copper slowly coats the object.

Step 4: Finish

1. Turn off power.

2. Take out the object, rinse in water, and dry.

3. Determine the mass of the object.

4. Look at the new copper coating!

5. Determine the mass of the copper strip using a mass balance.

6. Compare the mass gained by the coated object to the mass lost by the copper strip.

What you should see A thin, reddish-brown copper layer form on the object. The copper strip gets thinner as copper moves from it. If plating is uneven, it means the object was not cleaned well or current was too high/low.

Extraction of Aluminium

Aluminium is the most common metal in the Earth’s crust. It is never found unreacted because it reacts easily with oxygen. We cannot use carbon to extract it (like iron) because aluminium holds oxygen more strongly (aluminium is higher in the electrochemical series than hydrogen). So, we use electrolysis.

Step 1: From Bauxite to Alumina (Bayer Process)

1. Crush bauxite ore.

2. Mix with hot sodium hydroxide → aluminium dissolves.

3. Filter out red mud (iron oxides, silica).

4. Cool the solution → aluminium hydroxide forms.

5. Heat aluminium hydroxide strongly → pure alumina (Al₂O₃).

Step 2: Setting up the Electrolytic Cell (Hall – Héroult Process)

1. A steel tank lined with carbon acts as the cathode (–).

2. Carbon anodes (+) hang from above.

3. Alumina is dissolved in molten cryolite (Na₃AlF₆) to lower melting point (to ~960 °C).

4. Direct current (DC) is applied:

a. Carbon lining = negative terminal.

b. Carbon anodes = positive terminal.

Step 3: What Happens During Electrolysis

1. Strong current passes through the cell.

2. At cathode (–): Al³+ + 3e– → Al (molten aluminium collects at the bottom).

3. At anode (+): O²– + C → CO₂(carbon is used up).

4. Aluminium is tapped out regularly.

5. Anodes must be replaced often.

6. More alumina is added from time to time.

Figure 4.14: Electrolysis Process of Aluminium from Alumina Extraction (Refining) of Gold by Electrolysis Gold is often found mixed with other metals (like silver and copper). For jewellery and electronics, we need very pure gold. This is done using the Wohlwill process.

Step 1: Preparation

1. Impure gold (95–98% pure) is melted and shaped into anodes (+).

2. Thin pure gold sheets are prepared as cathodes (–).

3. The electrolyte is a solution of gold chloride in hydrochloric acid.

4. The tank is kept warm (about 60–70 °C).

Step 2: Setup

1. Impure gold anodes are hung in the solution.

2. Pure gold cathodes are placed in the same solution.

3. Anodes are connected to the positive terminal, cathodes to the negative terminal of a DC supply.

Step 3: Electrolysis Process

1. At anode (+): Au → Au³+ + 3e– (impure gold dissolves).

2. At cathode (–): Au³+ + 3e– → Au (pure gold is deposited).

Result

1. Pure gold coats the cathode.

2. Impurities fall off as anode slime or stay in solution.

3. Valuable metals (like silver and platinum) can later be recovered from the slime.

Faraday’s Laws of Electrolysis

Introduction to Faraday’s Laws of Electrolysis

What Are Faraday’s Laws?

These laws, formulated by Michael Faraday in 1833, establish the quantitative relationship between electric charge and chemical changes at electrodes during electrolysis. They are foundational in electrochemistry.

Faraday’s First Law

Statement: The mass of a substance produced (deposited or liberated) at an electrode during electrolysis is directly proportional to the total electric charge passed through the electrolyte.

Mathematical Expression: m ∝ Q ⇒ m = Z × Q m: mass of substance (g) Q: electric charge passed (Coulombs, Q = I × t) Z: electrochemical equivalent (g/C) Interpretation More current or longer duration → more substance deposited, assuming the efficiency of the process remains the same.

Figure 4.15: Visual overview of Faraday’s First Law of Electrolysis Faraday’s Second Law Statement: When the same amount of charge passes through different electrolytes, the masses of substances liberated at the electrodes are directly proportional to their chemical equivalent weights (equivalent weights = molar mass ÷ valence).

Mathematical Expression, m₁:m₂= E₁:E₂ m₁, m₂: Masses of produced substances E₁, E₂: Equivalent weights of the substances Implication Given equal charge, substances with higher equivalent weights yield more mass in proportion.

Figure 4.16: Visual overview of Faraday’s second Law of Electrolysis Combined Mathematical Form Both laws can be combined into a single equation:

m = ((M × I × t)_ (n × F) Where:

m = mass of substance produced (g) M = molar mass of the substance (g/mol) I = current (amperes) t = time (seconds) n = number of electrons transferred per ion F = Faraday constant (96,500 C/mol) Given I = 1.0 A, t = 1.0 h = 3600 s; M(Cu) = 63.55 g mol⁻¹; F = 96,500 C mol⁻¹ Reaction at the cathode: Cu²⁺+ 2 e− → Cu(s) from the balanced equation, 2 mol of e− deposited 1mol of Cu n = It__ zF = 1.0 × (1 × 60 × 60)______________ 2 × 96500 = 0.01865 mol m(Cu) = n × M = 0.01865 mol × 63.55 gmol⁻¹= 1.19 g

Activity 4.16 Metal Extraction

Images of Bauxite ore → aluminium (left) and Alluvial gold ore → refined gold (right)

1. State the differences you can see between the raw ore and the final shiny metal in the image above.

2. a. How aluminium is obtained

i. Bauxite → purified to alumina

ii. Alumina is dissolved in molten cryolite

iii. At the cathode: aluminium metal forms

iv. At the anode: oxygen gas is released

b. Why is electricity needed to separate aluminium from its ore?

3. a. How gold is refined

i. Impure gold is made into an anode (+).

ii. Pure gold sheet is the cathode (–).

iii. Electrolyte = chloroauric acid.

iv. Pure gold slowly collects on the cathode

b. How does Faraday’s First Law explain why more electricity produces more gold?

4. a. List both negative and positive impacts of mining in Ghana.

i. Problems (Negatives)

ii. Benefits (Positives)

b. How can we balance the benefits of mining with its problems?

5. Discuss how metal extraction supports Ghana

a. How does mining help Ghana grow?

b. What responsibilities come with this growth?

Activity 4.17 Exploring Faraday’s Laws of Electrolysis

1. a. Calculate the mass of copper deposited when a current of 2.0 A passes through copper (II) sulphate solution for 30 minutes.”

Discuss the steps and answer as a class.

b. How long must a current of 3 A be passed through molten aluminium oxide (Al₂O₃) to produce 2.7 g of aluminium? (M(Al) = 27 g/mol, n = 3)

2. a. Think about what Faraday’s Laws say

i. The more electricity you use, the more substance is deposited.

ii. Different metals deposit in amounts based on their charge and mass.

b. Where are these laws used in Ghana. Use these prompts

i. Electroplating

ii. Purification

iii. Extraction

iv. Batteries

3. Write down two benefits of using electrolysis in Ghana and two challenges

4. Share your findings with colleagues.

Corrosion in Ghana

Ghana’s warm climate, long coastline, and growing industries make metals wear out quickly. In coastal towns, the salty sea breeze causes corrosion. In the north, metals face wet and dry seasons that speed up damage. Everyday items like iron roofing sheets rust fast if they are not protected. Even historic places such as Cape Coast and Elmina Castles need constant repair because of the sea air. Understanding corrosion science is important for Ghana’s future. It helps us protect metals by learning how corrosion works and how to prevent it.

What is Corrosion?

Corrosion is when a metal changes into a compound (like an oxide, hydroxide, or salt) because of chemical reactions with its surroundings.

It is an electrochemical process:

1. The metal loses electrons (oxidation).

2. Other substances in the environment gain electrons (reduction).

The Reactions

General Process

1. Anodic (Oxidation): M → Mⁿ+ + ne– (The metal atom becomes a metal ion).

2. Cathodic (Reduction):

a. In acidic solutions: 2H+ + 2e– → H₂

b. In neutral/alkaline solutions: O₂+ 2H₂O + 4e– → 4OH–

Example: Zinc in Acid

Anode: Zn → Zn²+ + 2e– Cathode: 2H+ + 2e– → H₂ Overall: Zn + 2H+ → Zn²+ + H₂ In short: Corrosion is when metals ‘rust away’ because of electrochemical reactions with the environment. Learning how it works helps Ghana protect its buildings, industries, and heritage.

The Electrochemical Nature of Corrosion

Why Electrochemical?

Corrosion happens like a tiny battery on the surface of the metal

1. Some parts of the metal act like the anode (they lose electrons).

2. Other parts act like the cathode (they gain electrons).

3. Water and oxygen around the metal help carry the electrons and ions.

The Process Step by Step

1. At the Anode (Oxidation)

The metal loses electrons and turns into ions.

Example: Fe → Fe²⁺+ 2e−

2. At the Cathode (Reduction)

Electrons released at the anode travel to another part of the metal.

In acidic water: 2H+ + 2e− → H₂(hydrogen gas forms) In neutral/alkaline water: O₂+ 2H₂O + 4e− → 4OH−

3. Overall Effect

The metal surface changes into compounds like iron oxides (rust), zinc salts, or hydroxides. The bond between the metal and the oxide layer becomes weaker and flaky, causing it to come off with much less force than if it were pure metal.

Everyday Example

An iron nail left in salty water rusts quickly. One part of the nail becomes the anode (metal dissolves). Another part becomes the cathode (oxygen reacts with electrons). Together, the nail behaves like a little battery until it is eaten away.

Why It Matters in Ghana

1. Roofing sheets rust in the salty sea breeze.

2. Old bridges, cars, and even historic castles need constant maintenance.

Understanding corrosion helps us protect and save money by painting, coating, or galvanising metals.

Factors Affecting the Rate of Corrosion

The rate of corrosion depends on both the environment and the properties of the metal. The Table 4.3 below summarises key factors, their effects, and everyday examples learners can relate to.

Table 4.3: Factors that affect the rate of corrosion Factor Effect on Corrosion Rate Example Moisture (Water Availability) More moisture speeds up corrosion.

Iron rusts faster in rainy seasons.

Factor Effect on Corrosion Rate Example

Oxygen Supply Higher oxygen speeds up cathodic reactions.

Steel near the sea corrodes quickly.

Electrolytes (Salts/Acids) Salts and acids increase conductivity, making corrosion faster.

Metals corrode faster in seawater.

Temperature Higher temperature increases reaction speed.

Faster rusting in hot, humid coastal Ghana.

Metal Purity & Composition Impurities form anode–cathode sites; alloys resist corrosion.

Stainless steel resists rust better than iron.

pH of the Environment Acidic conditions increase corrosion.

Acid rain corrodes buildings and cars.

Contact with

Dissimilar Metals

The more reactive metal corrodes faster (galvanic corrosion).

Zinc corrodes when in contact with copper.

Surface Condition

Scratches or rough surfaces corrode faster.

Painted iron resists rust longer than bare iron.

Rusting Rusting is a specific form of corrosion that affects iron and its alloys. It produces iron (III) oxide (Fe₂O₃ . nH₂O ), commonly known as rust.

Rusting is defined as an electrochemical process in which iron reacts with oxygen and water to form hydrated iron (III) oxide (Fe₂O₃.nH₂O). Refer to Figure 4.17 for the rusting of iron.

Figure 4.17: Rusting of iron Experiment: Conditions Needed for Rusting Aim: Find out which conditions (water, oxygen/air, salt) are needed for iron to rust.

Materials needed

1. 5 clean iron nails (same size)

2. 5 test tubes + rack (or 5 small clear jars with lids)

3. Tap water

4. Boiled (hot) water (recently boiled to remove dissolved air)

5. Cooking oil (a few mL)

6. Calcium chloride or silica gel (drying agent) in a little gauze or paper sachet

7. Salt (NaCl) to make salt water (about 1 teaspoon in 50 mL)

8. Labels/marker, measuring cylinder/spoon, tongs

9. A few drops of vinegar/lemon juice (acid rain test)

10. Paper towels, gloves, eye protection Safety

1. Wear goggles and gloves.

2. Hot water can burn—use tongs and let it cool a bit before handling.

3. Do not touch silica gel/calcium chloride with bare hands.

4. Dispose of liquids down the sink with plenty of water; dry agents go to trash (or as per school rules).

Set-ups (label the tubes) Tube A – Water + Air (control for rusting)

1. Add enough tap water to cover the nail halfway.

2. Insert clean nail so part is in water and part in air.

3. Leave open or loosely capped.

Tube B – No Oxygen (water present)

1. Fill with freshly boiled, hot water (nearly to the top).

2. Add a clean nail, so that it is fully submerged

3. Add a thin layer of oil to float on top (keeps air out).

Tube C – No Water (air present)

1. Put a small sachet of calcium chloride/silica gel in the tube (keeps air dry).

2. Add a dry nail (do not add water). Cap tightly.

Tube D – Salt Water (water + air + salt)

1. Make salt solution (1 tsp salt in 50 mL water).

2. Add enough to cover a clean nail halfway. Leave open/loosely capped.

Tube E – Water + Air and acid (acidic environment)

1. Add enough tap water to cover the nail halfway.

2. Add a few drops of vinegar or lemon juice to acidify the conditions.

3. Insert clean nail so part is in water and part in air.

4. Leave open or loosely capped.

Procedure

1. Clean each nail with a paper towel (remove grease/rust).

2. Set up Tubes A–E as above and label them.

3. Leave the rack undisturbed for 2–5 days (check daily).

4. Observe and record: colour changes, brown rust, cloudiness, bubbles.

5. Take photos or draw what you see each day.

Results table Tube Conditions Day 1 Day 2 Day 3 Day 4/5 Rust? (Yes/ No/Rate) A Water + Air B Water, no oxygen (oil layer) C Dry air, no water D Salt water + air E Water and air and acid Conclusion prompts

1. Which tubes rusted? Which did not?

2. What does this show about the conditions needed for rusting?

3. Which factor made rusting faster? (salt, acid)

4. How can we prevent rusting in real life? (paint, oil, galvanising, stainless steel) Explanations

1. Tube A (water + air): Rust forms at the waterline.

Both oxygen and water are present → Fe → Fe²+ + 2e– (anode); O₂ + H₂O + e– → OH– (cathode) → rust (Fe₂O₃·xH₂O).

Tube B (no oxygen): No rust.

Water but no oxygen (oil stops air).

2. Tube C (no water): No rust.

Oxygen present but no water → reaction cannot proceed.

3. Tube D (salt water): Fastest rusting.

Salt increases conductivity→ speeds the electrochemical reactions.

4. Optional E (acidic): Faster rusting than A.

Acid provides H+, which speeds oxidation of iron.

Experiment to Investigate the Conditions That Affect

Rate of Rusting

Materials

1. Iron nails of uniform size

2. Test tubes with stoppers

3. Test tube rack

4. Sodium chloride solution (various concentrations)

5. Distilled water

6. Dilute acids (HCl)

7. Tap water

8. Thermometer

9. Water bath

10. Iron filings and iron nails (different surface areas)

11. Labels Procedure

1. Effect of Electrolytes

Place identical iron nails in separate test tubes containing

a. Distilled water

b. 1% NaCl solution

c. 5% NaCl solution

d. 10% NaCl solution

e. Observe for 3-5 days and record the extent of rusting.

2. Effect of pH Place identical iron nails in separate test tubes containing

a. Neutral water (pH 7)

b. Slightly acidic solution (pH 5)

c. Acidic solution (pH 3)

d. Observe for 3-5 days and record the extent of rusting.

3. Effect of Temperature

Prepare three identical setups with iron nails in salt water.

a. Keep one at room temperature,

b. One in a refrigerator and

c. One in a warm water bath.

d. Observe for 3-5 days and record the extent of rusting.

4. Effect of Surface Area

a. Place an iron nail in one test tube

b. Place an equivalent mass of iron filings in another test tube.

c. Add equal volumes of salt water to both.

d. Observe for 3-5 days and record the extent of rusting.

Expected Results

1. Electrolytes: Higher salt concentration leads to faster rusting

2. pH: Lower pH (more acidic) accelerates rusting

3. Temperature: Higher temperature increases the rate of rusting

4. Surface Area: Larger surface area (iron filings) results in faster rusting Conclusion The rate of rusting increases with

1. higher concentration of electrolytes

2. lower pH (more acidic conditions)

3. higher temperature

4. greater surface area of the metal Experiment to Investigate Ways of Preventing Rusting Materials

1. Clean iron nails

2. Test tubes with stoppers

3. Test tube rack

4. Paint

5. Oil

6. Zinc strips

7. Magnesium strips

8. Copper strips

9. Grease

10. Galvanised nails

11. Saltwater solution (electrolyte)

12. Labels Procedure

1. Physical barriers

a. Coat different iron nails with paint, oil and grease

b. Leave one nail uncoated as a control.

c. Place each in separate test tubes with salt water.

2. Sacrificial protection

a. Connect clean iron nails to zinc strip, magnesium strip and copper strip

b. Place each pair in separate test tubes with salt water.

c. Include an unconnected iron nail as a control.

3. Galvanisation comparison Place a galvanised nail and a regular iron nail in separate test tubes with salt water.

4. Observe all setups for 1 week and record observations.

Expected results

1. Physical Barriers: Nails coated with paint, oil, or grease show little to no rusting

2. Sacrificial Protection: Nails connected to zinc or magnesium show little to no rusting (these metals are more reactive and sacrifice themselves). Nails connected to copper show accelerated rusting (galvanic corrosion)

3. Galvanisation: Galvanised nail shows resistance to rusting Conclusion Various methods can effectively prevent rusting by either creating a barrier to oxygen and water or by providing sacrificial protection through more reactive metals.

Preventing Rusting

Rusting happens when iron + oxygen + water react. We can stop or slow down rusting in two main ways:

1. Non-Redox Methods (Blocking air or water) These methods do not involve chemical reactions; they just cover the iron to stop oxygen and water touching it.

Painting → e.g., painted gates, cars.

Oiling or Greasing → e.g., bicycle chains.

Plastic Coating → e.g., wire coverings.

Tin Plating → food cans are coated with tin to stop rust.

2. Redox Methods (Sacrificial protection using another metal) These methods use another metal that reacts more easily than iron. That metal is “sacrificed” (it corrodes first) to protect the iron.

a. Galvanising → coating iron with zinc. Zinc rusts first, protecting the iron underneath.

b. Sacrificial Anode → attaching blocks of magnesium or zinc to ships, underground pipes, or oil rigs. These metals corrode instead of the iron.

c. Alloying → making stainless steel (iron + chromium + nickel).

Chromium forms a thin, protective oxide layer that stops rusting.

Everyday Examples (Ghana)

1. Painted roofing sheets → non-redox protection.

2. Galvanised roofing sheets → redox protection (zinc protects the iron).

3. Stainless steel cutlery → alloy protection.

Impact of Rusting and Corrosion in Everyday Life

1. On Buildings and Homes

a. Roofing sheets: Rust causes holes → leakages during rain.

b. Pipelines: Corrode and burst → water wastage and high repair costs.

c. Bridges and structures: Rust weakens iron beams → can cause accidents.

2. On Transport

a. Cars, buses, and bicycles rust if not painted or maintained → dangerous and costly repairs.

b. Ships and boats corrode quickly in salty sea water → need regular maintenance.

c. Airplanes must be protected from corrosion to avoid mechanical failures.

3. On Environment and Heritage

a. Historic buildings like Cape Coast and Elmina Castles require constant maintenance because of salty coastal air.

b. Corroded tanks and containers can leak harmful chemicals into soil and water.

4. On the Economy

a. Billions are spent worldwide each year repairing or replacing corroded equipment.

b. In Ghana, money that could improve schools and hospitals is also used to fix rust damage (e.g., bridges, vehicles, smelters).

5. On Daily Life

a. Rust makes tools blunt and unsafe.

b. Household items like nails, cutlery, and fences lose strength.

c. Corroded cooking utensils can contaminate food.

Activity 4.18 Understanding Rusting and Corrosion

1. Use the video link below to watch corrosion of metals https://chatgpt.com/c/68b0bef7-ad60-8324-96c7-43d8c372569b

2. Observe the two metal objects below.

a. What happened to this one (the brown metal)?

b. Why do you think the other one is not brown?

3. a. Write your own definition of “rusting” and “corrosion.”

b. Compare your definitions with a partner and improve it.

4. What differences do you notice in how these metals have changed?

5. a. What differences do you notice in colour and texture?

b. What might those colours tell us about the products formed

c. Write the reaction steps of rusting process (anodic and cathodic processes).

d. Write the overall reaction.

e. Compare your answers.

6. Compare galvanic and non-galvanic methods for protecting iron from rust. What are the advantages and disadvantages of each method?

7. What do the words ‘rust’ and ‘decay’ have in common?

a. Use the link below to watch the video on rusting https://www.youtube.com/watch?v=h1A0yAcavu0

i. What changes do you see on the metal surface?

ii. What role do water and air play in this transformation?

b. Use the link below to watch the video on Fruit Decay https://www.youtube.com/ watch?v=S12zZhdOckc&list=PL0FB8D619E5C752AE

i. How does the fruit change?

ii. What do you think causes those changes?

Activity 4.19 Rusting of Metals

1. Mystery Rust Stations

Aim: Work out which conditions made some nails rust and others not.

Materials Pre-made station cups with nails (teacher prepares dry air; water only; salty water; oil-covered water; vinegar; painted nail; galvanized nail if available), labels A–F.

Procedure

a. Move around Stations A–F. Look only (no touching).

b. Observe nail colour, flakes, liquid, bubbles, smell.

c. State the condition you think is inside (air / water / salt / oil / acid / paint).

d. Write one why sentence for each station: “We think A is salt water because ___.”

2. Rust Race (3–5 days) Goal: Make a setup that rusts an identical nail the fastest.

Rules (one-variable only): 1 nail • 1 cup • 50 ml liquid • add only ONE extra (1 tsp salt or 1 tsp sugar or 1 tsp vinegar).

Do this

a. Choose the one change you’ll test (salt? acid?). Label cup with team + variable.

b. Set up the nail in the liquid.

c. Each day, score rust 0–3. Note colour/flakes/cloudiness.

d. Final day: Share results, which set up generated the most rust? How did you ensure that this race was fair?

Safety Do not touch rusty nails with bare hands; use paper towel; wash hands.

Rust Score Key

a. 0 = shiny, no change

b. 1 = slight colour change

c. 2 = obvious brown spots or flakes

d. 3 = heavy rust / lots of flakes

3. Crime Scene Investigation: The Rusty Disaster

Task: Your class has been called to investigate a crime scene where a metal structure failed because of corrosion.

a. Look at the photos and objects from the “crime scene” (e.g., rusted bolt, broken gate hinge, roof sheet).

b. Write down the clues you see rust colour, flakes, cracks, peeled paint, or green coating.

c. Read the short witness cards

d. In groups, decide the main cause of the failure

e. Fill in a simple Claim-Evidence-Reasoning sheet:

i. Claim: What caused the failure.

ii. Evidence: What clues you saw.

iii. Reasoning: Why that evidence shows corrosion.

f. Share your group’s conclusion with the class.

4. Anti-Rust Challenge checks (2–3 days) Goal: Create the best rust-prevention method with limited materials.

Do this

a. Coat half of a nail (e.g., oil, wax, paint, tape, plastic wrap). Leave the other half bare.

b. Put all nails in the same liquid (class agrees). Label and predict which works best and why (barrier keeps out water/oxygen).

c. Each day, score rust 0–3 on each half; take photos/sketches.

d. Make a poster: method, before/after, scores, and why it worked (or not).

Rust Score Key

a. 0 = shiny, no change

b. 1 = slight colour change

c. 2 = obvious brown spots or flakes

d. 3 = heavy rust / lots of flakes

5. One-Variable Rust Test + Graph Task: Work in groups to test how one factor (salt, temperature, covering, or oxygen) affects rusting.

Steps

1. Choose one variable to test

a. Salt level (0, ½, 1, 2 tsp in water)

b. Temperature (sunlight, shade, fridge)

c. Oxygen (cup half-full vs filled to the top and sealed with oil)

d. Covering (painted vs unpainted nail)

2. Set up your experiment

a. Place identical nails in cups of water with your chosen change.

b. Label each cup clearly.

3. Daily check (3–5 days)

a. Look for colour change, flakes, or cloudiness.

b. Score rust on a scale: 0 (no rust) → 3 (lots of rust).

c. Record results in a table.

4. Make a graph

a. Plot rust level (0–3) on the Y-axis.

b. Plot your variable change (salt level, temp, etc.) on the X-axis.

5. Share findings

a. Show your graph to the class.

b. Complete the sentence: “Rust was fastest when ___ because ___.”

Safety Do not touch rusty nails with bare hands; use paper towels and wash hands after.

Rust Score Key

a. 0 = shiny, no change

b. 1 = slight colour change

c. 2 = obvious brown spots or flakes

d. 3 = heavy rust / lots of flakes

Activity 4.20 Rust Investigations & Anti-Rust Methods

1. Gallery Walk

a. Post your experiment results and rust prevention method.

b. Rotate around the room.

c. Add a sticky note with a question or comment to each group’s poster.

2. Rust Prevention Advertisement

a. Create a 60-second commercial for your best prevention method.

b. Show why it works and convince others to use it.

3. Real-Life Impact Stations

a. Visit stations with photos of rust on bridges, cars, and pipes.

b. Discuss with your group:

i. How does rust affect safety?

ii. How does it cost money?

4. Concept Mapping Challenge: Work in teams to draw a big map linking

a. Rusting process

b. Prevention methods

c. Real-life impacts

Part A Multiple Choice Questions

1. Which of the following best defines oxidation?

A. Loss of protons B. Gain of electrons C. Loss of electrons D. Gain of neutrons

2. Which of the following metals is the most reactive?

A. Copper B. Zinc

C. Iron D. Potassium

3. In the reaction: Zn + CuSO₄→ ZnSO₄+ Cu, which substance is reduced?

A. Zinc B. Copper

C. Sulphate D. Oxygen

4. Which observation best indicates a metal is reacting with acid?

A. Effervescence B. Formation of a precipitate C. Colour change D. Condensation

5. What is the reducing agent in the reaction: Fe₂O₃+ 3CO → 2Fe + 3CO₂?

A. Fe₂O₃ B. CO

C. Fe D. CO₂

6. Which of the following metals will not displace hydrogen from dilute acid?

A. Zinc B. Magnesium

C. Copper D. Iron

7. A student places iron in a solution of copper (II) sulphate. What is observed?

A. Gas evolved B. Copper is deposited C. A white precipitate form D. Nothing happens

8. Which metal will displace iron from iron (II) sulphate?

A. Copper B. Gold

C. Magnesium D. Silver

9. Which of the following is not an oxidation reaction?

A. Rusting of iron B. Combustion of methane C. Electrolysis of water D. Freezing of water

10. What is the role of electrons in redox reactions?

A. Gained in oxidation only B. Lost in both oxidation and reduction C. Transferred from reducing to oxidising agents D. Used up in neutralisation

11. Which of the following reactions is NOT a redox reaction involving changes in oxidation states of elements?

A. Zn + Cu²+ → Zn²+ + Cu B. HCl + NaOH → NaCl + H₂O C. 2Mg + O₂→ 2MgO D. 4Fe + 3O₂→ 2Fe₂O₃

12. Which of the following reactions does NOT involve oxidation or reduction?

A. Ag+ + Br– → AgBr B. 2H₂S + SO₂→ 3S + 2H₂O C. 2[Ag (NH₃)₂]+ + Cu → Cu[(NH₃)₄]²+ + 2Ag D. 2Al + 2OH– + 6H₂O → 2Al (OH)₄

– + 3H₂

13. In the reaction below, which substances act as oxidising agents?

2MnO₂+ 4KOH + O₂+ Cl₂→ 2KMnO₄+ 2KCl + 2H₂O A. KMnO₄only B. MnO₂only C. MnO₂and O₂ D. O₂and Cl₂ Essay Questions

1. a. List the metals above in order of decreasing reactivity.

b. Define oxidation, in electronic terms, using one example from above.

c. Define reduction, in terms of oxidation number, using one example from above.

d. State and explain which is the strongest reducing agent in the examples above.

e. State and explain which is the strongest oxidising agent in the examples above.

f. Deduce whether a gold coin will react with aqueous magnesium nitrate.

2. How can you use displacement reactions to test which metals are more reactive than others?

3. You have magnesium, zinc, and copper. How would you carry out an experiment to find out which one reacts the fastest with dilute hydrochloric acid?

4. Why do some metals like potassium and sodium react with water, but others like copper do not?

5. Redox reactions happen in many places around us. Give three real-life examples of redox reactions and explain their importance.

6. What is the chemical equation for rusting? Explain how both oxidation and reduction happen during the rusting of iron.

7. Plan a simple classroom experiment that shows both oxidation and reduction happening at the same time.

8. Some metals react with oxygen, some with water, and others with acids.

Compare their reactions and use your observations to arrange the metals in order of reactivity.

9. Imagine putting a piece of iron into a solution of silver nitrate. Explain what will happen.

10. A science laboratory in Kumasi received five unlabelled bottles, each containing a clear solution. Preliminary tests were conducted, and the following observations were recorded:

Solution Test with KI Test with KMnO₄ Test with K₂Cr₂O₇ pH A Brown colour formed Purple colour remains Orange colour remains 3.5 B No change Decolourised Green colour formed 6.5 C Brown colour formed Decolourised Green colour formed 2.0 D No change Purple colour remains Orange colour remains 7.0 E Brown colour formed Purple colour remains Green colour formed 4.0 The laboratory supervisor believes these solutions might be: hydrogen peroxide (H₂O₂), sodium chloride (NaCl), iron (II) sulphate (FeSO₄), sodium sulphite (Na₂SO₃), and copper (II) sulphate (CuSO₄).

a. Based on the test results, identify each solution (A through E), justifying your answers with specific reference to the chemical behaviour of oxidising and reducing agents.

b. Design a flowchart showing how you could systematically identify these five solutions using only two chemical tests. Explain your reasoning.

11. a. Write the oxidation and reduction half-equations for the following reaction 2Mg + O₂→ 2MgO

b. Assign oxidation numbers to all elements in this reaction and identify the species oxidised and reduced 2Fe₂O₃+ 3C → 4Fe + 3CO₂

12. a. Balance the redox equation in an acidic medium MnO₄

– + Fe²+ → Mn²+ + Fe³+

b. Balance the equation in basic medium Cr (OH)₃+ ClO– → CrO₄²– + Cl– Redox titrations

13. a. Design an experiment to compare the reactivity of three metals: Mg, Zn, and Cu, using dilute HCl. Describe how to determine which is most reactive.

b. A student mixed iron nails with copper (II) sulphate solution and observed a brown deposit. Explain the redox changes using half- equations.

14. Evaluate the environmental implications of redox reactions in mining (e.g., gold or bauxite extraction). Suggest one way to reduce negative impacts.

15. a. Define the term standardisation in titration.

b. Write the balanced redox equation for the reaction between KMnO₄ and Fe²+ in acidic solution.

c. What colour change is observed at the endpoint when KMnO₄is used as the titrant?

d. Identify the oxidising and reducing agents in this redox titration.

16. a. If 25.0 cm³ of Fe²+ solution requires 22.60 cm³ of 0.0200 mol/dm³ KMnO₄, calculate the number of moles of KMnO₄used.

b. Using your answer above, calculate the number of moles of Fe²+ that reacted.

c. From the moles and volume of Fe²+, calculate its concentration in mol/dm³.

d. Describe two safety precautions specific to KMnO₄during titration.

17. a. A student obtained titre values of 21.30 cm³, 22.60 cm³, and 22.50 cm³. Identify the concordant titres and justify the selection.

b. Compare the advantages of using KMnO₄as a self-indicator versus using an external indicator like starch.

c. A sample of Fe²+ was exposed to air for some time. Explain how this affects the standardisation titration results.

d. Explain the role of dilute sulphuric acid in the titration and why hydrochloric acid is not preferred.

18. a. A solution was made by dissolving 55.6 g of (NH₄)₂Fe(SO₄)₂·6H₂O per dm³. Calculate the theoretical concentration of Fe²+ in mol/dm³ and compare it with an experimentally determined value. Discuss possible sources of error.

b. A student standardised KMnO₄and obtained a concentration higher than expected. Propose and evaluate at least two experimental errors that may have caused this deviation.

c. Design an inquiry-based experiment to compare the rate of reaction between KMnO₄and Fe²+ under different acid concentrations.

Describe how you would collect and interpret the data.

19. a. State the chemical formula for ethanedioic acid and its role in redox titration.

b. What is the chemical formula of the ethanedioate ion?

c. Write the balanced redox equation for the reaction between KMnO₄ and C₂O₄²– in an acidic medium.

d. What colour change is observed at the endpoint of this titration?

e. Identify the oxidising agent and the reducing agent in the reaction.

20. a. If 25.0 cm³ of ethanedioate solution requires 20.00 cm³ of 0.0200 mol/dm³ KMnO₄to reach the endpoint, calculate the moles of KMnO₄used.

b. Use the balanced equation to calculate the moles of ethanedioate ions that reacted.

c. Calculate the concentration of the ethanedioate solution in mol/dm³.

d. Explain why the titration mixture must be heated to 60–70°C before starting the titration.

21. a. Describe the effect of not heating the titration mixture on the accuracy of the results.

b. A student consistently observed a delayed colour change at the endpoint. Suggest possible reasons for this.

c. Compare the endpoint visibility of KMnO₄vs ethanedioate titrations and KMnO₄vs Fe²+ titrations.

d. Discuss the importance of an acidic medium in the KMnO₄and ethanedioate titration.

22. a. A student dissolved 6.30 g of ethanedioic acid dihydrate (H₂C₂O₄·2H₂O, molar mass = 126 g/mol) in 250 cm³ of solution.

Calculate the theoretical concentration in mol/dm³ and compare with a value obtained from titration.

b. Suppose the KMnO₄solution was not properly standardised.

Describe how this would affect the accuracy of the ethanedioate concentration obtained.

c. Design an investigation to determine the effect of temperature on the rate of the redox reaction between KMnO₄and ethanedioate. Include how you would ensure reliability and safety.

23. a. What is the role of sodium thiosulphate in redox titration?

b. Write the ionic equation for the reaction between iodine and sodium thiosulphate.

c. What is the function of potassium iodide (KI) in the titration process?

d. What indicator is commonly used in this titration, and what colour change does it show at the endpoint?

e. Identify the oxidising and reducing agents in the titration.

24. a. A student titrated 25.0 cm³ of iodine solution with 0.100 mol/dm³ sodium thiosulphate. The average titre was 23.50 cm³. Calculate the moles of Na₂S₂O₃used.

b. From the balanced equation, determine the moles of iodine that reacted.

c. Calculate the concentration of iodine in the 25.0 cm³ sample.

d. Explain why potassium iodide is added in excess during the titration.

25. a. Compare the visibility and sharpness of the endpoint in this titration versus that of a KMnO₄titration.

b. A student added starch at the beginning of the titration instead of near the endpoint. What could go wrong and why?

c. During the titration, the solution in the conical flask was left to stand for too long before the titration. What could happen, and how would it affect the results?

d. Why is it important to carry out the titration away from direct sunlight?

26. a. A solution containing iodine was standardised using sodium thiosulphate. After analysis, the concentration of iodine was found to be lower than expected. Suggest possible experimental errors that may explain this observation.

b. Design an experiment to investigate the effect of light exposure on the stability of iodine solution. Include a control and a method of measuring changes in iodine concentration.

c. Suppose you want to analyse iodine content in a seaweed extract.

Outline the titration procedure and modifications needed for sample preparation and reliability.

Part B Multiple Choice Questions

1. In a voltaic cell, which type of reaction occurs at the anode?

A. Reduction B. Oxidation

C. Neutralisation D. Precipitation

2. The function of the salt bridge in a voltaic cell is to:

A. Provide a pathway for electrons to flow B. Maintain electrical neutrality by allowing ion movement C. Increase the voltage of the cell D. Reduce resistance in the external circuit

3. Which of the following is a correct example of a redox reaction producing electricity?

A. Zn + CuSO₄→ ZnSO₄+ Cu B. HCl + NaOH → NaCl + H₂O C. AgNO₃+ NaCl → AgCl + NaNO₃ D. CaCO₃→ CaO + CO₂

4. In a Daniell cell, the zinc electrode is in ZnSO₄solution and the copper electrode is in CuSO₄solution. Which statement is correct?

A. Electrons flow from copper to zinc B. Zinc is reduced C. Copper is oxidised D. Zinc is the anode

5. In a voltaic cell, electrons flow:

A. From the cathode to the anode through the salt bridge B. From the anode to the cathode through the external circuit C. From the cathode to the anode through the external circuit D. From the anode to the cathode through the salt bridge

6. Which of the following best explains why a redox reaction can produce electricity?

A. The transfer of electrons between reactants is directed through an external circuit B. The reaction occurs very slowly C. The reaction involves gases that move freely D. The reaction increases the temperature of the solution

7. A voltaic cell consists of a magnesium electrode in Mg²+ solution and a silver electrode in Ag+ solution. Which electrode is the anode and why?

A. Silver, because Ag is more reactive than Mg B. Silver, because Ag+ ions are more easily reduced C. Magnesium, because Mg is more reactive and more easily oxidised D. Magnesium, because Mg²+ ions are more easily reduced Essay Type Questions

1. Describe the basic components of a voltaic cell and explain the role of each in generating electricity.

2. Explain, with the aid of a diagram, how the Daniell cell works, including:

redox reactions, electron flow, and purpose of salt bridge.

3. Compare and contrast the operation of a voltaic cell and an electrolytic cell.

4. A magnesium–copper voltaic cell is set up under standard conditions.

Write the half-reactions, calculate E°cell, state spontaneity.

5. Given the following half-cell reduction potentials:

E°(Ag+/Ag) = +0.80 V1 E°(Zn²+/Zn) = −0.76 V1 Calculate the standard cell potential (E°cell) for the cell:

Zn(s) | Zn²+(aq) || Ag+(aq) | Ag(s)

6. Given the following reduction potentials:

E°(Cu²+/Cu) = +0.34 V E°(Pb²+/Pb) = −0.13 V

a. Write the balanced overall cell reaction for the voltaic cell.

b. Calculate E°cell.

c. Predict whether the reaction is spontaneous under standard conditions.

7. A galvanic cell is constructed from a magnesium half-cell and a tin half- cell.

E°(Mg²+/Mg) = −2.37 V E°(Sn²+/Sn) = −0.14 V

a) Identify the anode and cathode

b) Calculate the standard cell potential

c) Determine the value of ΔG° for the cell reaction at 298 K.

(F = 96,500 C mol–¹; n = 2)

8. Consider the following half-reactions and their E° values:

Cl₂(g) + 2e– → 2Cl–(aq) E° = +1.36 V Br₂(l) + 2e– → 2Br–(aq) E° = +1.09 V

a) Which species is the stronger oxidising agent?

b) Write the overall cell reaction when these two half-cells are connected.

c) Calculate E°cell.

9. You are given the following data E°(Fe³+/Fe²+) = +0.77 V E°(I₂/I–) = +0.54 V

a) Calculate E°cell for the reaction1 2Fe³+(aq) + 2I–(aq) → 2Fe²+(aq) + I₂(s)

b) Calculate ΔG° for the reaction (n = 2; F = 96,500 C mol–¹).

c) Comment on the spontaneity of the reaction.

d) Explain how changing the concentration of Fe³+ ions might affect the cell potential.

10. (a) Define standard electrode potential.

(b) State the condition under which a redox reaction is feasible based on E°cell.

(c) List two industrial applications of electrode potential principles.

(d) State the electrode potential sign convention for a reduction reaction.

11. (a) Given E°(Zn²+/Zn) = −0.76 V and E°(Cu²+/Cu) = +0.34 V, calculate the standard cell potential for a Daniell cell and predict if the reaction is spontaneous.

(b) Arrange the following metals in order of increasing reactivity based on their E° values: Fe²+/Fe (−0.44 V), Al³+/Al (−1.66 V), Cu²+/Cu (+0.34 V).

(c) State two ways in which electrode potentials are used in corrosion prevention.

12. A fuel cell operates with the following half-reactions O₂(g) + 4H+(aq) + 4e– → 2H₂O(l) E° = +1.23 V H₂(g) → 2H+(aq) + 2e– E° = 0.00 V

(a) Identify the anode and cathode.

(b) Calculate E°cell.

(c) State whether the reaction is feasible under standard conditions.

(d) Explain how the concept of sacrificial protection works using electrode potentials.

13. In an experiment, a student measures the EMF of a Zn–Ag cell under non- standard conditions and obtains a value lower than the standard EMF.

(a) Suggest two factors that could cause this decrease.

(b) Using the Nernst equation, explain how ion concentrations affect the EMF.

(c) Discuss the role of electrode potentials in designing rechargeable batteries, giving examples of suitable electrode materials and their E° values.

14. (a) Define electrode potential.

(b) State two standard conditions for measuring standard electrode potential.

(c) State two applications of standard electrode potentials in chemistry.

15. (a) Arrange the following metals in order of increasing reactivity using their standard electrode potentials: Mg, Cu, Zn.

(b) Using standard electrode potentials, predict whether a solution of silver nitrate will oxidise copper metal.

(c) Explain why potassium cannot be obtained by the electrolysis of aqueous potassium chloride solution, using electrode potentials.

16. (a) A galvanic cell is constructed from Zn/Zn²+ and Cu/Cu²+ half-cells.

(i) Draw a labelled diagram of the cell.

(ii) Write the half-cell reactions and the overall cell reaction.

(iii) Calculate E°cell given: E°(Zn²+/Zn) = –0.76 V, E°(Cu²+/Cu) = +0.34 V.

(iv) State whether the reaction is spontaneous.

(b) Explain, with the aid of standard electrode potentials, why hydrogen is liberated when zinc reacts with dilute hydrochloric acid but copper is not.

17. Given the standard electrode potentials E°(Fe³+/Fe²+) = +0.77 V, E°(Zn²+/Zn) = –0.76 V, E°(Cu²+/Cu) = +0.34 V,

(a) Predict whether zinc metal will reduce Fe³+ to Fe²+ in solution.

(b) Justify your answer with a suitable calculation of E°cell.

(c) Suggest one industrial process that uses this principle.

18. An unknown metal M has a standard electrode potential of –0.44 V for M²+/M.

(a) Predict whether M will displace copper from a CuSO₄solution.

(b) Predict whether M will liberate hydrogen gas from dilute H₂SO₄.

(c) Support your predictions with E°cell calculations.

Part C Electrolytic Cells and Applications

1. (a) Define an electrolytic cell.

(b) State the role of the anode and cathode in an electrolytic cell.

(c) Identify one everyday application of electrolysis.

(d) List two examples of electrolytes commonly used in electrolytic cells.

2. (a) Explain why a direct current (DC) power supply is used in electrolytic cells instead of alternating current (AC).

(b) Describe what happens to cations and anions during electrolysis.

(c) Give two examples of how electrolysis is applied in industry.

(d) Illustrate with a simple diagram the movement of ions in an electrolytic cell.

3. (a) Compare the processes at the electrodes of an electrolytic cell with those in a voltaic cell.

(b) Predict the products at the electrodes when aqueous sodium chloride

solution is electrolysed. Explain your reasoning.

(c) Analyse the importance of electroplating in protecting metals used in daily life.

(d) Justify the use of graphite electrodes in the electrolysis of molten sodium chloride instead of metal electrodes.

4. Evaluate the environmental impact of using electrolysis in industries such as aluminium extraction or the chlor-alkali process.

5. Design a simple classroom demonstration to show the electrolysis of water and explain how this connects to sustainable fuel production.

6. Synthesise information from industrial and household uses of electrolysis to propose new innovative applications in modern technology.

7. Critically assess the economic advantages and disadvantages of using electrolysis for large-scale metal extraction.

Part D Faraday’s Laws of Electrolysis

1. a. State Faraday’s First Law of Electrolysis.

b. Write the formula that combines current, time, molar mass, and charge to calculate the mass deposited at an electrode.

c. What is the value of the Faraday constant (F)?

d. Which scientist formulated the laws of electrolysis?

e. At which electrode is copper deposited when a solution of copper

(II) sulphate is electrolysed with copper electrodes?

2. a. Explain why more copper is deposited when electrolysis runs for a longer time.

b. A current of 2.0 A passes through silver nitrate solution for 30 minutes. Calculate the mass of silver deposited.

(M_Ag = 108 g mol–¹, n=1, F=96,500 C mol–¹)

3. a. Using Faraday’s Second Law, explain why passing the same charge through solutions of silver nitrate and copper sulfate produces more silver than copper.

b. Why does aluminium extraction require a large amount of electricity?

c. Which electrode process in the Hall–Héroult cell releases oxygen gas?

4. A current of 3.0 A is passed through molten sodium chloride for 10 minutes. Calculate the mass of sodium formed. (M_Na = 23 g mol–¹, n=1)

5. Compare the processes of electroplating copper onto jewellery and electrorefining impure gold using Faraday’s Laws.

6. In electrolysis of copper (II) sulfate using inert electrodes, explain why the solution becomes less blue with time.

7. Describe how Faraday’s First Law can be used to determine the value of Avogadro’s number experimentally.

8. A plating industry requires 5.0 g of silver to coat a set of ornaments.

Calculate the quantity of electricity needed.

9. Ghana’s VALCO smelter consumes massive amounts of electricity for aluminium production. Use Faraday’s First Law to explain why electricity cost is the largest factor in aluminium price.

10. Mining in Ghana provides both gold and bauxite (aluminium ore). Discuss how Faraday’s Second Law helps explain differences in refining costs between gold and aluminium.

11. Design an experiment using Faraday’s First Law to calculate the charge on an electron. What measurements would be needed?

12. Evaluate the environmental impacts of large-scale aluminium and gold electrolysis plants in Ghana, connecting them to Faraday’s Laws.

13. A gold refinery passes a current of 500 A for 24 hours through an impure gold anode. Estimate the mass of gold refined and discuss how this contributes to Ghana’s export revenue.

Part E

1. a. Define corrosion and distinguish it from rusting.

b. State two conditions necessary for the rusting of iron.

c. Write a balanced chemical equation for the rusting process in the presence of water and oxygen.

2. a. List three methods of preventing rusting and state how each works.

b. Why does a zinc coating (galvanisation) protect iron better than a paint coating?

c. Explain briefly why aluminium appears resistant to corrosion even though it is very reactive.

3. A bridge built near the sea rusted severely within five years, while a similar bridge inland remained rust-free

a. Suggest reasons for the difference

b. Design a simple experiment to compare the effect of salt water and fresh water on the rusting of nails.

4. A class tested rusting using identical iron nails:

Condition Rust score after 72 h (0–3) Dry air 0 Tap water 1 Salt water 3 Boiled water + oil layer 0

a. Identify the independent variable and state two controlled variables.

b. Rank the conditions from slowest to fastest rusting and give a brief reason.

c. Explain why boiled water plus oil prevented rusting.

d. Predict what happens if a copper strip touches the iron nail in salt water. State which metal corrodes; explain using anode/cathode ideas.

5. a. Explain how corrosion affects transportation, infrastructure, and the economy.

b. Suggest and evaluate two large-scale strategies to minimise corrosion damage.

c. If you were a materials engineer, which corrosion-prevention method would you recommend for pipelines in Ghana and why?