AQA A-Level Chemistry Paper 1, 2022: Question 8

✅ Correct Answer

MnO₂

💡 Key Knowledge

• An oxidising agent oxidises another species and is itself reduced (gains electrons).
• The half-cell with the more positive E° value has the greater tendency to proceed in the forward direction (reduction).
• E°(MnO₂/Mn₂O₃) = +0.52 V vs E°(Zn²⁺/Zn) = −0.76 V. Therefore, the MnO₂ half-cell proceeds in the forward direction, meaning MnO₂ acts as the oxidising agent.

❌ Common Errors

• Writing Mn⁴⁺ or Mn : Specify the actual chemical species present in the half-equation ( MnO₂ ).
• Naming the entire left-hand side (e.g. MnO₂ + NH₄⁺ ): Only the species being reduced is the oxidising agent.
• Confusing the oxidising agent with the reducing agent (Zn).

✅ Correct Answer

Any one of the following:

• Allows ions to move / flow / transfer
• To complete the circuit
• Acts as a salt bridge

🧠 Exam Technique

The separator in a commercial battery serves the exact same role as a salt bridge in a laboratory beaker cell. It prevents physical mixing of the electrode pastes (which would short the cell chemically) while maintaining electrical neutrality through ion movement.

❌ Common Errors

• Fatal Error: Stating that it "allows electrons to flow/pass". Electrons flow strictly through the external circuit! An internal separator allowing electrons would cause an internal short circuit.
• Vague answers like "allows current / charge to flow" are ignored by examiners. Be specific: mention ions.

✅ Correct Answer

2Ag + ZnO → Zn + Ag₂O

(State symbols are not required)

💡 Discharge vs. Recharge

• Discharging (Spontaneous Cell Reaction):
  More positive E° (+0.34 V) goes forwards:
  Ag₂O + H₂O + 2e⁻ → 2Ag + 2OH⁻
  More negative E° (−1.26 V) goes backwards:
  Zn + 2OH⁻ → ZnO + H₂O + 2e⁻
  Overall discharge: Ag₂O + Zn → 2Ag + ZnO
• Recharging: An external power supply drives the exact reverse reaction:
  2Ag + ZnO → Zn + Ag₂O

❌ Common Errors

• Writing the discharge equation ( Ag₂O + Zn → 2Ag + ZnO ) instead of the recharge equation. Always re-read whether the question specifies discharging or recharging!
• Leaving spectator species like H₂O and OH⁻ uncancelled in the final equation.

✅ Correct Answer

Half-equation:

O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)

Standard electrode potential:

E° = +0.40 V (or +0.4 V )

📐 Calculations & Deduction

Step 1: Identify the missing half-reaction

In any hydrogen-oxygen fuel cell, hydrogen is oxidised at one electrode, and oxygen is reduced at the other. Under alkaline conditions, O₂ reacts with H₂O to form OH⁻:

O₂ + 2H₂O + 4e⁻ → 4OH⁻

Step 2: Calculate E° using Cell EMF

The hydrogen electrode is the negative electrode (oxidation):

E°(negative) = −0.83 V

EMF = E°(positive) − E°(negative)

+1.23 V = E°(O₂/OH⁻) − (−0.83 V)

E°(O₂/OH⁻) = +1.23 V − 0.83 V = +0.40 V

🧠 Exam Technique

• Remember both sets of fuel cell half-equations: acidic (involves H⁺ and H₂O) and alkaline (involves OH⁻ and H₂O).
• Always include the sign ( + ) for standard electrode potentials unless explicitly instructed otherwise.
• Multiples of the half-equation (e.g., ½O₂ + H₂O + 2e⁻ → 2OH⁻ ) are fully credited.

✅ Correct Answer

Same overall reaction

OR

2H₂ + O₂ → 2H₂O (or H₂ + ½O₂ → H₂O )

💡 Key Knowledge

Regardless of whether the electrolyte is acidic or alkaline:

• Acidic:
  Anode: H₂ → 2H⁺ + 2e⁻ (E° = 0.00 V)
  Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O (E° = +1.23 V)
  EMF = +1.23 − 0.00 = +1.23 V
• Alkaline:
  Anode: H₂ + 2OH⁻ → 2H₂O + 2e⁻ (E° = −0.83 V)
  Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻ (E° = +0.40 V)
  EMF = +0.40 − (−0.83) = +1.23 V

Since the initial reactants ( H₂ + O₂ ) and final products ( H₂O ) are identical, ΔG° is identical, giving the exact same EMF via ΔG° = −nFE° .

❌ Common Errors

• Saying "they use the same reactants" without specifying that the overall reaction or products are the same.
• Vague references to "both produce water" without stating that the overall chemical equation is identical.