OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2017: Question 22

✅ Required Apparatus & Setup

• Zinc half-cell: Zn rod dipped in a 1 mol dm⁻³ Zn²⁺(aq) solution.
• Iron half-cell: Platinum (Pt) electrode dipped in a mixture of 1 mol dm⁻³ Fe²⁺(aq) and 1 mol dm⁻³ Fe³⁺(aq) ions.
• Circuit: High-resistance voltmeter connected via external wires to both electrodes.
• Salt bridge: Filter paper soaked in a saturated ionic solution (e.g., KNO₃) connecting the two solutions.

💡 Standard Conditions Required

• Temperature: 298 K (or 25 °C)
• Concentration: 1 mol dm⁻³ for all aqueous ions.
• Pressure: 100 kPa (or 1 atm) — not strictly applicable here as no gases are involved, but good practice to state.

🧠 Exam Technique & Drawing Tips

When drawing half-cells containing two aqueous ions of the same element (like Fe²⁺ and Fe³⁺), a platinum electrode is mandatory because neither ion is a solid metal to act as an electrical contact. Ensure your salt bridge dips properly into both solutions and the voltmeter reads the potential difference.

❌ Common Student Errors

• Omitting platinum (Pt) for the iron half-cell.
• Writing incorrect concentrations (e.g., writing 1 mol(e⁻) instead of 1 mol dm⁻³ ).
• Failing to label the salt bridge or voltmeter clearly.

📐 Step-by-Step Calculation

1. Identify half-equations from Table 22.1:
   System 1: Zn²⁺ + 2e⁻ ⇌ Zn (E° = -0.76 V) -> Oxidation happens here (more negative)
   System 3: Fe³⁺ + e⁻ ⇌ Fe²⁺ (E° = +0.77 V) -> Reduction happens here (more positive)
2. Apply formula:
   E°(cell) = E°(right) - E°(left)
   E°(cell) = 0.77 - (-0.76)
3. Calculate final value:
   E°(cell) = +1.53 V (or 1.53 V)

❌ Common Calculation Traps

Be careful with negative signs when subtracting standard electrode potentials. Subtracting -0.76 is the same as adding 0.76. The mark scheme ignores signs in the final answer box, but including the positive sign is good chemical practice.

✅ Correct Answers

• Strongest reducing agent: Zn (or Zinc)
• Strongest oxidising agent: MnO₄⁻ (Permanganate ion)

💡 Core Chemical Principles

• The most negative E° value indicates the equilibrium that lies furthest to the left, meaning the reduced species (Zn) loses electrons most easily and is therefore the strongest reducing agent.
• The most positive E° value indicates the equilibrium that lies furthest to the right, meaning the oxidised species (MnO₄⁻) gains electrons most easily and is therefore the strongest oxidising agent.

📐 Step-by-Step Construction (Systems 2 and 4)

1. Write out the two half-equations:
   System 2: SO₄²⁻(aq) + 2H⁺(aq) + 2e⁻ ⇌ SO₃²⁻(aq) + H₂O(l) (E° = +0.17 V)
   System 4: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇌ Mn²⁺(aq) + 4H₂O(l) (E° = +1.51 V)
2. Reverse the more negative half-equation (System 2) to act as oxidation (electrons on the right):
   SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻
3. Multiply to balance electrons:
   Multiply System 2 by 5 (10 electrons)
   Multiply System 4 by 2 (10 electrons)
4. Combine and cancel common species (H⁺ and H₂O):
   2MnO₄⁻ + 16H⁺ + 5SO₃²⁻ + 5H₂O → 2Mn²⁺ + 8H₂O + 5SO₄²⁻ + 10H⁺
   Cancelling 10H⁺ and 5H₂O from both sides gives:
   2MnO₄⁻ + 6H⁺ + 5SO₃²⁻ → 2Mn²⁺ + 3H₂O + 5SO₄²⁻

❌ Where Students Lost Marks

This is a 2-mark question. Top-level responses successfully cancelled out excess spectator ions like H⁺ and H₂O on both sides of the equation. Leaving unsimplified equations with H⁺ on both sides typically drops you to 1 mark.