OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2021: Question 19

✅ Correct Answer: Part (i) - Cell Diagram

Must show two separate beakers connected by a salt bridge and linked externally with wires leading to a voltmeter.

• Left half-cell: Chromium metal ( Cr(s) ) electrode dipping into chromium(III) ions ( Cr³⁺(aq) ).
• Right half-cell: Platinum ( Pt ) inert electrode dipping into a mixture of MnO₄⁻(aq) , Mn²⁺(aq) , and H⁺(aq) .

✅ Correct Answer: Part (ii) - Overall Equation

5Cr + 3MnO₄⁻ + 24H⁺ → 5Cr³⁺ + 3Mn²⁺ + 12H₂O

(Multiples are accepted, state symbols can be ignored based on mark scheme guidance).

💡 Key Knowledge

• Inert Electrodes: Platinum is mandatory when both the oxidized and reduced species in a half-cell are aqueous ions (e.g., MnO₄⁻ and Mn²⁺ ).
• Electron Balancing: To combine half-equations (System 1: 3 electrons; System 4: 5 electrons), cross-multiply to get a lowest common multiple of 15 electrons.

🧠 Exam Technique

• When drawing half-cells, explicitly label the chemical formulas of the ions and solids present in solution.
• Double-check that electrons cancel out completely when writing the final overall redox equation.

❌ Common Errors

• Using a reactive metal like copper or zinc instead of platinum for the ion-ion half-cell.
• Forgetting to include H⁺(aq) ions in the acidic permanganate half-cell setup or overall equation.

✅ Correct Answer: Part (i) - Oxidation Numbers

Mn is oxidised from +6 (in MnO₄²⁻ ) to +7 (in MnO₄⁻ ).

Mn is reduced from +6 (in MnO₄²⁻ ) to +4 (in MnO₂ ).

✅ Correct Answer: Part (ii) - Electrode Potentials & Equilibrium

System 3 ( MnO₄⁻/MnO₄²⁻ ) is less positive / more negative than System 5 ( MnO₄²⁻/MnO₂ ).

Therefore, System 3 shifts left and System 5 shifts right, causing disproportionation.

💡 Key Knowledge

• Disproportionation: A redox process in which the same element is simultaneously oxidized and reduced.
• Electrode Potential Rule: The system with the more positive E-value will proceed in the forward direction (reduction), forcing the system with the less positive E-value to go backwards (oxidation).

❌ Common Errors

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• Omitting signs (+ or -) alongside oxidation numbers (e.g. writing '6' instead of '+6').
• Using vague terminology like "higher" or "lower" instead of "more positive" or "less positive" when comparing standard electrode potentials.

✅ Correct Answer: Parts (i), (ii) & (iii)

(i) Negative half-equation:
H₂ + 2OH⁻ → 2H₂O + 2e⁻

(ii) Electrode potential calculation:
(0.40 - 1.23 =) -0.83 V

(iii) Difference from storage cell:
Fuel cells require a continuous supply of reactants (fuel and oxygen/oxidant) to generate electrical energy continuously.

📐 Step-by-Step Calculation (Part ii)

1. Identify the formula connecting cell potential: E(cell) = E(positive) - E(negative)
2. Substitute known values: 1.23 = 0.40 - E(negative)
3. Rearrange to solve: E(negative) = 0.40 - 1.23
4. Calculate final value with correct units and sign: -0.83 V

💡 Key Knowledge

• In alkaline fuel cells, hydroxide ions ( OH⁻ ) are both reactants in the anode half-equation and species present within the electrolyte.
• Storage cells store chemical energy internally and run out when reactants are depleted, whereas fuel cells need an uninterrupted external input of fuel.

❌ Common Errors

• Reversing the polarity subtraction in calculation part (ii), resulting in +0.83 V instead of -0.83 V .
• Writing acidic half-equations ( H⁺ ) instead of alkaline half-equations ( OH⁻ ) for the fuel cell.