Edexcel A-Level Chemistry Paper 3, June 2023: Question 5

12 marks · Hard difficulty · Open Response

Construct electrochemical cells, use electrode potentials to determine reaction feasibility, and explain electron flow and redox processes in a hydrogen-oxygen fuel cell.

Practise this question

Question

A 5-part exam question about electrochemical cells. Part (a) features a diagram of an electrochemical cell with a zinc electrode and a space for labels Y and Z for a manganese(II)/manganate(VII) electrode system. Part (b) provides a table of standard electrode potentials and asks to explain the final oxidation state of chromium formed when excess zinc is added to acidified sodium dichromate(VI). Part (c) gives a cell diagram and asks for the reduction half-equation. Part (d) shows a diagram of a hydrogen-oxygen fuel cell and asks about electron flow and redox processes. Part (e) asks for an advantage of the hydrogen-oxygen fuel cell over petrol.
Question text

5 This question is about electrochemical cells.

(a) A diagram is shown of the apparatus that is used to measure the emf of a cell

with a zinc/zinc(II) electrode and an acidified manganese(II)/manganate(VII)

electrode system.

Complete the labels Y and Z by naming the substances needed.

Temperature and concentrations are not required.

(3)

voltmeter

V

Zn

salt bridge

Y …

Z …

solution of …

zinc nitrate (aq)

(b) Excess zinc is added to an acidified solution of sodium dichromate(VI).

Some electrode data are given in the table.

Electrode system E d / V

Cr2+(aq) + 2e– Cr(s) –0.91

Cr3+(aq) + e– Cr2+(aq) –0.41

½Cr O2–(aq) + 7H+(aq) + 3e– Cr3+(aq) + 3½H O(l) +1.33

27 2

Zn2+(aq) + 2e– Zn(s) –0.76

Explain, using only the data in the table, the final oxidation state of chromium

that is formed when zinc is added to acidified dichromate(VI) ions.

Include E d values where appropriate. Equations are not required.

cell

(5)

… *P71914A01132*

(c) A cell diagram is shown.*P71914A01232*

Ni(s) ½ Ni2+(aq) ½½ [NO–(aq) + 2H+(aq)], [NO (g) + H O(l)] ½ Pt(s) E d = +1.06 V

32 2 cell

Deduce the reduction half-equation.

State symbols are not required.

(1)

(d) State the direction of the electron flow in the hydrogen-oxygen fuel cell shown.

Justify your answer by reference to the redox processes in the cell.

hydrogen

acidic electrolyte

negative electrode

membrane cell load

positive electrode

water oxygen

(2)

(e) State one advantage of the hydrogen-oxygen fuel cell over the use of petrol as

fuel in a vehicle.

(1)

(Total for Question 5 = 12 marks)

Mark scheme

Show the mark scheme The mark scheme corresponding to the electrochemical cells question, giving detailed marking points for each part (a) through (e), including accepted chemical formulas, electrode potential calculations, Ecell values, and explanations for fuel cell operation.

Question

Answer Additional Guidance Mark

Number

5(a) An answer which makes reference to the following points: Names or formulae accepted but if both given then (3)

both must be correct

All three are standalone marks

• (Y) platinum / Pt (1) Ignore reference to (platinum) black

(Z)

• manganese(II) nitrate / Mn(NO3)2 Allow MnSO4

and

potassium manganate(VII) / KMnO4 (solution) (1) Allow sodium manganate(VII)/ NaMnO4

Allow potassium permanganate for KMnO4

Oxidation numbers essential if only the names are

given

• sulfuric acid (1) Allow nitric acid

Do not award use of hydrochloric acid

Ignore concentrations throughout

Penalise use of hydrochloric acid or manganese

halides once only

Question

Answer Additional Guidance Mark

Number

5(b) An explanation that makes reference to the following points: (5)

• (chromium(VI) reduced to) chromium +2 / (II) (1) Allow Cr2+

Allow TE on candidate EƟ values, e.g. all

cell

EƟ values positive then Cr(0) is the result

cell

because

Ɵ EƟ = (+1.33 − −0.76 =) (+) 2.09 (V)

• E cell value for the reduction of chromium(VI) to chromium(III) (1) cell

Ɵ EƟ = (−0.41 − −0.76 =) (+) 0.35 (V)

• E cell value for the reduction of chromium(III) to chromium(II) (1) cell

Ɵ EƟ = (−0.91 − −0.76 =) −0.15 (V)

• E cell value for the reduction of chromium(II) to chromium (1) cell

• first two reductions occur (because EƟcell is positive in both cases) Accept feasible for occur

and

final reaction does not occur (because EƟ is negative) (1) Ignore equations even if incorrect

cell

Penalise reference to Zn2+ reacting in the written

answer once only for M2 and M3

Question

Answer Additional Guidance Mark

Number

5(c) Example of equation (1)

• half-equation NO − + 2H+ + e(−) → NO + H O

32 2

Allow multiples / ⇌

Ignore state symbols even if incorrect

Question

Answer Additional Guidance Mark

Number

5(d) An answer that makes reference to the following points: (2)

• (electrons move) from the negative to the positive electrode (1) Allow annotation on diagram, see below

Allow move from the top electrode to the bottom

electrode

Ignore just electrons move down/clockwise

Do not allow movement through the middle of the fuel

cell

Allow anode for negative electrode and cathode for

positive electrode

• (because) the hydrogen is being oxidised / losing electrons Allow half-equations such as

and the oxygen is being reduced / gaining electrons (1) (Oxidation) H → 2H+ + 2e(−)

and

(Reduction) ½O +2H+ + 2e(−) → H O

Do not award formation of O2− ions

M2 is not dependent on M1

Question

Answer Additional Guidance Mark

Number

5(e) An answer which makes reference to any one of the following points (1)

• harmless product/water compared to pollutants Accept named pollutants e.g. CO/CO2/SO2/NOx

Allow hydrogen fuel cell only produces water

or

less reliant on fossil fuels/non-renewable fuels Allow hydrogen (fuel) is renewable/sustainable

Allow no use of fossil fuels

or Allow less/no green house gases produced

more efficient energy production Ignore just ‘more efficient’

or

(can be) smaller and lighter fuel

cell

(Total Question 5 = 12 marks)

How to answer it

Electrochemical Cells Study Guide

Edexcel A-Level Chemistry • Redox & Electrode Potentials

What this question tests

This comprehensive question assesses your understanding of electrochemical cells, standard cell diagrams, predicting redox feasibility using standard electrode potentials (E-theta values), identifying sequential reduction steps, interpreting fuel cell schematics, and evaluating the environmental benefits of hydrogen-oxygen fuel cells.

Part (a): Setting up an Electrochemical Cell

Labelling a manganese(II)/manganate(VII) half-cell

✅ Correct Answers

  • Y: Platinum / Pt (electrode)
  • Z: Manganese(II) nitrate / Mn(NO₃)₂ AND Potassium manganate(VII) / KMnO₄ (solution)
  • Acid: Sulfuric acid (H₂SO₄)

💡 Key Knowledge

  • A platinum electrode is required because both the oxidized and reduced species in the half-cell are aqueous ions (solution phase), meaning they cannot act as an electrical contact themselves.
  • The half-cell requires both Mn²⁺ ions and MnO₄⁻ ions alongside a source of H⁺ ions to function.

❌ Common Errors

  • Using hydrochloric acid instead of sulfuric acid (Cl⁻ ions would be undesirably oxidized by manganate(VII)).
  • Omitting either of the required ionic components for solution Z. Both manganese(II) and manganate(VII) must be present.

🧠 Exam Technique

  • Names or chemical formulae are both accepted by examiners, but ensure formulas have correct ionic charges if used.
  • Concentrations do not need to be specified unless standard conditions are explicitly demanded.
Allocated marks: 3 marks

Part (b): Feasibility and Sequential Reductions

Determining the final oxidation state of chromium with excess zinc

✅ Correct Answer & Mark Breakdown (5 Marks)

  • Final State: Chromium(II) / Cr²⁺ (1 mark)
  • Step 1 (Cr(VI) to Cr(III)): E-theta cell = (+1.33) - (-0.76) = +2.09 V (Positive, feasible) (1 mark)
  • Step 2 (Cr(III) to Cr(II)): E-theta cell = (-0.41) - (-0.76) = +0.35 V (Positive, feasible) (1 mark)
  • Step 3 (Cr(II) to Cr(0)): E-theta cell = (-0.91) - (-0.76) = -0.15 V (Negative, not feasible) (1 mark)
  • Conclusion link: First two reductions occur because E-theta cell is positive, but the final reduction does not occur because E-theta cell is negative. (1 mark)

📐 Step-by-Step E-theta Calculation Strategy

  1. Calculate E-theta cell = E-theta (reduction) - E-theta (oxidation) for each successive half-equation. Remember that zinc acts as the reducing agent and is itself oxidized (E-theta = -0.76 V).
  2. Check if the calculated E-theta cell value is positive (indicating a feasible reaction).
  3. Stop the sequence when the calculated E-theta cell value becomes negative.

❌ Common Errors

  • Forgetting to test the final step (Cr²⁺ to Cr) and incorrectly assuming all transition metal ions are reduced completely to metal atoms.
  • Incorrectly subtracting electrode potential values, leading to wrong signs for E-theta cell.

🧠 Exam Technique

  • You must quote the specific E-theta cell values for each stage to gain full credit. Simply stating "the numbers are positive" will not score top marks.
Allocated marks: 5 marks

Part (c): Cell Diagrams to Half-Equations

Deducing the reduction half-equation from cell notation

✅ Correct Answer

NO₃⁻ + 2H⁺ + e⁻ ⇌ NO₂ + H₂O (or using = )

💡 Key Knowledge

In standard cell representations, the right-hand side of the double vertical line ( || ) represents the reduction process (cathode). You must balance the half-equation using electrons and hydrogen ions as indicated by the species present in the diagram.

❌ Common Errors

  • Including state symbols when the question explicitly states: "State symbols are not required."
  • Incorrect stoichiometry for electrons or protons.

🧠 Exam Technique

Multiples of the balanced equation are accepted as long as all stoichiometric coefficients are correctly proportioned.

Allocated marks: 1 mark

Part (d): Fuel Cell Electron Flow & Redox

Explaining electron movement and redox processes in a hydrogen-oxygen fuel cell

✅ Correct Answers

  • Direction: Electrons move from the negative electrode to the positive electrode through the external circuit. (1 mark)
  • Justification: Hydrogen is oxidized (loses electrons) at the negative electrode, while oxygen is reduced (gains electrons) at the positive electrode. (1 mark)

💡 Key Knowledge

  • Negative electrode (Anode): H₂ → 2H⁺ + 2e⁻
  • Positive electrode (Cathode): ½O₂ + 2H⁺ + 2e⁻ → H₂O

❌ Common Errors

  • Stating that electrons flow through the internal membrane (electrons cannot cross the proton-exchange membrane; only ions do).
  • Mixing up oxidation and reduction locations.

🧠 Exam Technique

To secure both marks, link the physical direction of flow directly to the chemical terms oxidation and reduction.

Allocated marks: 2 marks

Part (e): Advantages of Hydrogen-Oxygen Fuel Cells

Evaluating green energy alternatives to petrol combustion

✅ Correct Answers (Any one of)

  • Produces harmless products / water only (compared to toxic exhaust pollutants like CO, CO₂, NOₓ, or unburned hydrocarbons).
  • Less reliant on finite fossil fuels / renewable energy source.
  • More efficient energy production.
  • Can be smaller and lighter as a power unit.

❌ Common Errors

  • Vague answers such as just saying "more efficient" without context (though "more efficient energy production" is credited, simple single-word descriptions lack rigor).
  • Ignoring that hydrogen production itself can require fossil fuels unless specified as green hydrogen. Stick closely to the mark scheme points regarding the vehicle operation.
Allocated marks: 1 mark

Topics

Physical Chemistry · Core Practicals · Core Practical 10: Construct electrochemical cells and measure electrode potentials · Topic 14: Redox II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.