AQA A-Level Chemistry Paper 3, 2024: Question 17

1 mark · Easy difficulty · Multiple Choice

Calculate the standard electrode potential of the PbO2 / PbSO4 electrode given the cell EMF and the standard potential of the PbSO4 / Pb electrode.

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Question

AQA A-Level Chemistry Paper 3, 2024: Question 17
Question text

17 A cell with EMF = +2.15 V is made from two electrodes.

The half-equations for the two electrodes are shown.

positive electrode:

PbO (s) + 3 H+(aq) + HSO –(aq) + 2 e– ⟶ PbSO (s) + 2 H O(l)

24 4 2

negative electrode:

PbSO (s) + H+(aq) + 2 e– ⟶ Pb(s) + HSO –(aq) Eo = –0.46 V

What is the standard electrode potential of the PbO2 / PbSO4 electrode?

[1 mark]

A –2.61 V

B –1.69 V

C +1.69 V

D +2.61 V

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 3, 2024: Question 17

17 C 1 (AO2) +1.69 V

How to answer it

Lead–acid cell: finding an unknown E°

What this question tests 1 mark • AO2
  • Using E°cell = E°(positive electrode) − E°(negative electrode) with standard electrode potentials written as reductions.
  • Correct sign handling when rearranging to find an unknown E° value.
  • Interpreting “positive electrode / negative electrode” in the context of a galvanic cell (positive = cathode).
Examiner focus: this is a one-step AO2 rearrangement question. Most lost marks from sign errors or flipping the subtraction the wrong way round.

Part (a): Standard electrode potential of the PbO₂ / PbSO₄ electrode

Given: Ecell = +2.15 V; E°(negative electrode) = −0.46 V

💡 Key knowledge (how AQA expects you to do it)

  • Standard electrode potentials (E°) are tabulated for reduction half-equations.
  • In a galvanic cell, the positive electrode is the cathode (reduction happens there).
  • Formula used at A-level: E°cell = E°cathode − E°anode .
  • “Negative electrode” in this question corresponds to the anode in a galvanic cell (even though its half-equation is written as a reduction for E°).

📐 Calculation (step-by-step, full-mark method)

  1. Write the relationship:
    E°cell = E°(positive electrode) − E°(negative electrode)
  2. Substitute the values:
    +2.15 = E°(PbO₂ / PbSO₄) − (−0.46)
  3. Simplify the double negative:
    +2.15 = E°(PbO₂ / PbSO₄) + 0.46
  4. Rearrange:
    E°(PbO₂ / PbSO₄) = 2.15 − 0.46 = 1.69 V
  5. State with sign and units:
    E° = +1.69 V
Significance: 2 d.p. matches data given (2.15 and 0.46). Units must be V.

✅ Correct answer (as per mark scheme)

Option C — +1.69 V

Mark scheme: C, 1 mark (AO2), +1.69 V.

🧠 Exam technique (how to secure the mark fast)

  • Use the phrase: “Ecell = Epositive − Enegative” (because the question labels them for you).
  • When you see subtraction of a negative, immediately rewrite: −(−0.46) = +0.46 .
  • Sanity check: a lead–acid cell has a large positive Ecell, so the positive electrode should have a fairly large positive E° (around +1 to +2 V). +1.69 V is sensible.

❌ Common errors (what loses the only mark)

  • Wrong subtraction order: doing Ecell = Enegative − Epositive gives a wrong sign/magnitude.
  • Forgetting the double negative: treating −(−0.46) as −0.46 .
  • Choosing +2.61 V by adding: 2.15 + 0.46 (this ignores rearrangement).
  • Dropping the + sign: writing 1.69 V is usually fine, but if options include negatives, be explicit: +1.69 V .
Examiner insight: because it’s multiple choice, a single sign slip immediately leads to a different option — no method marks are available.

Quick recap checklist (for similar 1-mark E° questions)

💡 Checklist

  • Identify cathode/positive electrode and anode/negative electrode.
  • Use E°cell = E°cathode − E°anode .
  • Be careful with negatives: subtracting a negative becomes addition.
  • Give the final answer with units (V) and an appropriate sign.

🧠 10-second mental check

  • If E°anode is negative, then E°cell should be larger than E°cathode (because you are subtracting a negative).
  • So E°cathode = E°cell − |E°anode| when the anode E° is negative.

Topics

Physical Chemistry · 3.1.11 Electrode Potentials · 3.1.7 Oxidation, Reduction and Redox Equations

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.