OCR A Level Chemistry A: Redox and electrode potentials (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 5.2.3, Redox and electrode potentials, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017: outcomes 5.2.3(a) to (k). The specification assigns Module 5 to Components 01 and 03.

Learn the content in the study guide first. See also the OCR chemistry hub, the printable checklist and a free diagnostic. Use the data given in each question.

Questions

1. Hydrogen peroxide oxidises iron(II) ions in acid: H₂O₂ + 2Fe²⁺ + 2H⁺ → 2Fe³⁺ + 2H₂O. Identify the oxidising agent and the reducing agent, using oxidation numbers to explain. [2]

2. Acidified dichromate(VI) ions oxidise tin(II) ions to tin(IV) ions. The reduction half-equation is Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Write the oxidation half-equation and construct the overall ionic equation. [3]

3. In one reaction, hydrazine, N₂H₄, is oxidised to nitrogen gas by iodate(V) ions, IO₃⁻, which are reduced to iodide ions. Water is the only other product. Use oxidation numbers to construct the balanced equation. [3]

4. Acidified iron(II) solution is titrated with potassium manganate(VII). Explain why no indicator is added, state the end-point colour, and explain why dilute sulfuric acid, not hydrochloric acid, is used. [4]

5. 5.880 g of impure ammonium iron(II) sulfate, (NH₄)₂Fe(SO₄)₂·6H₂O (Mr = 392.0), is used to prepare 250.0 cm³ of solution in dilute sulfuric acid. Samples of 25.0 cm³ react with 0.0200 mol dm⁻³ KMnO₄. The titres are: rough 15.35 cm³; then 14.70, 14.95 and 14.80 cm³. Only the iron(II) reacts.

(a) Select the titres you would use and calculate the mean titre. [2] (b) Calculate the percentage purity of the sample. [5]

6. 10.0 cm³ of a bleach is diluted to 250.0 cm³. A 25.0 cm³ portion is added to excess acidified potassium iodide: ClO⁻ + 2I⁻ + 2H⁺ → Cl⁻ + I₂ + H₂O. The iodine reacts with 14.20 cm³ of 0.100 mol dm⁻³ sodium thiosulfate.

(a) State when starch is added and the colour change at the end point. [2] (b) Calculate the concentration of ClO⁻ in the undiluted bleach. [4]

7. A 1.012 g sample of hydrated vanadyl sulfate, VOSO₄·xH₂O, is used to prepare 100.0 cm³ of solution in dilute sulfuric acid. 20.0 cm³ portions need a mean titre of 16.00 cm³ of 0.0100 mol dm⁻³ KMnO₄. Use these half-equations and Ar values (V 50.9, S 32.1, O 16.0, H 1.0) to find x.

VO²⁺ + H₂O → VO₂⁺ + 2H⁺ + e⁻ MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O [5]

8. A student plans to measure the standard electrode potential of the Fe³⁺/Fe²⁺ half-cell.

(a) Describe the half-cell containing the iron ions. [2] (b) Describe the reference half-cell it is connected to. [3] (c) Name two other items needed and explain why one must have a high resistance. [2]

9. Use these values: Fe²⁺ + 2e⁻ ⇌ Fe, E⦵ = −0.44 V; Ag⁺ + e⁻ ⇌ Ag, E⦵ = +0.80 V. A standard cell is built from the two half-cells.

(a) Calculate the standard cell potential. [1] (b) Identify the positive electrode and the direction of electron flow in the external circuit. [2] (c) Give the equation for the reaction in the cell. [1] (d) The silver nitrate solution is replaced by 0.10 mol dm⁻³ silver nitrate. Explain the effect on the cell potential. [2]

10. Use these values:

Half-equation	E⦵ / V
Zn²⁺ + 2e⁻ ⇌ Zn	−0.76
Sn²⁺ + 2e⁻ ⇌ Sn	−0.14
2H⁺ + 2e⁻ ⇌ H₂	0.00
Fe³⁺ + e⁻ ⇌ Fe²⁺	+0.77
Br₂ + 2e⁻ ⇌ 2Br⁻	+1.09
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O	+1.33

(a) Identify the strongest oxidising agent and the strongest reducing agent in the table. [2] (b) Show that bromine can oxidise Fe²⁺ ions and write the equation. [2] (c) Predict whether Fe³⁺ ions oxidise Br⁻ ions. Explain. [1] (d) Tin is predicted to react with 1 mol dm⁻³ H⁺(aq), yet little reaction may be seen. Give the cell potential and suggest a reason. [2] (e) Dichromate(VI) is predicted to oxidise bromide ions. Explain why this may not happen if the solution is only weakly acidic. [2]

11. A silver oxide button cell is not rechargeable. Use these values: Ag₂O + H₂O + 2e⁻ ⇌ 2Ag + 2OH⁻, E⦵ = +0.34 V; Zn(OH)₂ + 2e⁻ ⇌ Zn + 2OH⁻, E⦵ = −1.25 V.

(a) Calculate the cell potential. [1] (b) Write the negative-electrode half-equation for the cell in use. [1] (c) Write the overall cell reaction. [1] (d) Give one benefit and one risk of lithium-based cells. [2]

12. A direct methanol fuel cell has an acidic electrolyte. Use these values: CO₂ + 6H⁺ + 6e⁻ ⇌ CH₃OH + H₂O, E⦵ = +0.02 V; O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O, E⦵ = +1.23 V.

(a) Identify the negative electrode and write the half-equation for the reaction there. [2] (b) Construct the overall equation for the cell. [2] (c) Calculate the cell potential. [1] (d) Explain how a fuel cell differs from a storage cell in the way it supplies energy. [2]

Answers

1. H₂O₂ is the oxidising agent: O goes from −1 to −2 [1]. Fe²⁺ is the reducing agent: Fe goes from +2 to +3 [1] [2] Examiner insight: Name the species with its oxidation number change; “oxygen” or “iron” alone is imprecise.

2. Sn²⁺ → Sn⁴⁺ + 2e⁻ [1]. Multiply by 3 to match six electrons [1]. Cr₂O₇²⁻ + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O [1] [3] Examiner insight: Check charges match (+18 each side); no electrons may remain in the final equation.

3. Each N rises from −2 to 0, so each N₂H₄ loses 4 electrons [1]. I falls from +5 to −1, gaining 6 [1]. Lowest common multiple 12: 3N₂H₄ + 2IO₃⁻ → 3N₂ + 2I⁻ + 6H₂O [1] [3] Examiner insight: N₂H₄ has two N atoms, so it loses 4 electrons, not 2.

4. Manganate(VII) is purple and is reduced to almost colourless Mn²⁺, so it acts as its own indicator [1]. End point: first permanent pale pink [1]. H⁺ ions are needed for the reduction of MnO₄⁻ [1]. HCl is not used because MnO₄⁻ would oxidise Cl⁻ to chlorine, giving a titre that is too large [1] [4] Examiner insight: “Purple to colourless” is wrong; the end point is the first permanent pink.

5. (a) Concordant titres: 14.70 and 14.80 cm³ (within 0.10 cm³) [1]; mean = 14.75 cm³ [1] (b) n(MnO₄⁻) = 0.0200 × 14.75 ÷ 1000 = 2.95 × 10⁻⁴ mol [1]. n(Fe²⁺) in 25.0 cm³ = 5 × 2.95 × 10⁻⁴ = 1.475 × 10⁻³ mol [1]. In 250.0 cm³: 1.475 × 10⁻² mol [1]. Mass of pure salt = 1.475 × 10⁻² × 392.0 = 5.782 g [1]. Purity = 5.782 ÷ 5.880 × 100 = 98.3% [1] [5] Examiner insight: State which titres you averaged; including 14.95 cm³ or the rough titre gives the wrong mean.

6. (a) Starch is added when the iodine colour has faded to pale yellow [1]. End point: blue-black to colourless [1] (b) n(S₂O₃²⁻) = 0.100 × 14.20 ÷ 1000 = 1.42 × 10⁻³ mol [1]. n(I₂) = n(ClO⁻) = 7.10 × 10⁻⁴ mol in 25.0 cm³ [1]. In 250.0 cm³, from 10.0 cm³ of bleach: 7.10 × 10⁻³ mol [1]. Concentration = 7.10 × 10⁻³ ÷ 0.0100 = 0.710 mol dm⁻³ [1] [4] Examiner insight: 0.0284 mol dm⁻³ is the diluted value; the question asks for the undiluted bleach.

7. Ratio MnO₄⁻ : VO²⁺ = 1 : 5 [1]. n(MnO₄⁻) = 0.0100 × 16.00 ÷ 1000 = 1.60 × 10⁻⁴ mol [1]. n(VO²⁺) in 20.0 cm³ = 8.00 × 10⁻⁴ mol; in 100.0 cm³ = 4.00 × 10⁻³ mol [1]. Mr = 1.012 ÷ 4.00 × 10⁻³ = 253.0; Mr(VOSO₄) = 163.0 [1]. 18.0x = 90.0, so x = 5 [1] [5] Examiner insight: Label each line (moles of what, in what volume); a bare x = 5 shows no method.

8. (a) A platinum electrode [1] dipping into a solution containing Fe³⁺ and Fe²⁺ ions at 1 mol dm⁻³ each (or equal concentrations) [1] (b) Hydrogen gas at 100 kPa [1] bubbled over a platinum electrode [1] in 1 mol dm⁻³ H⁺(aq), at 298 K [1] (c) A salt bridge and a voltmeter [1]. The voltmeter has a high resistance so that almost no current flows, giving the maximum potential difference [1] [7] Examiner insight: In (b), list each condition; “standard conditions” alone is too vague.

9. (a) 0.80 − (−0.44) = 1.24 V [1] (b) The silver electrode is positive [1]. Electrons flow from the iron electrode to the silver electrode through the wire [1] (c) Fe + 2Ag⁺ → Fe²⁺ + 2Ag [1] (d) Lower [Ag⁺] shifts Ag⁺ + e⁻ ⇌ Ag to the left, so the silver electrode potential becomes less positive [1]. The cell potential decreases below 1.24 V [1] [6] Examiner insight: In (d), link the equilibrium shift to the electrode potential, then to the cell.

10. (a) Strongest oxidising agent: Cr₂O₇²⁻ [1]. Strongest reducing agent: Zn [1] (b) E⦵cell = 1.09 − 0.77 = +0.32 V, positive so feasible [1]. Br₂ + 2Fe²⁺ → 2Br⁻ + 2Fe³⁺ [1] (c) No: Fe³⁺/Fe²⁺ (+0.77 V) is less positive than Br₂/Br⁻ (+1.09 V), so E⦵cell would be −0.32 V [1] (d) E⦵cell = 0.00 − (−0.14) = +0.14 V [1]. The reaction may have a high activation energy, so it is very slow [1] (e) H⁺ is on the left of the dichromate half-equation, so lower [H⁺] shifts it left and its electrode potential becomes less positive [1]. It may fall below +1.09 V, so the reaction is no longer feasible [1] [9] Examiner insight: Oxidising agents are on the left, reducing agents on the right; naming Cr³⁺ or Zn²⁺ is a common slip.

11. (a) 0.34 − (−1.25) = 1.59 V [1] (b) Zn + 2OH⁻ → Zn(OH)₂ + 2e⁻ [1] (c) Zn + Ag₂O + H₂O → Zn(OH)₂ + 2Ag [1] (d) Benefit: e.g. high energy for their mass, or rechargeable [1]. Risk: e.g. toxic materials, or fire if damaged [1] [5] Examiner insight: In (c), cancel the 2OH⁻ that appears on both sides.

12. (a) The methanol electrode is negative (less positive E⦵) [1]. CH₃OH + H₂O → CO₂ + 6H⁺ + 6e⁻ [1] (b) Methanol half-equation × 2, oxygen half-equation × 3 (12 electrons), then add [1]. Cancel H⁺ and H₂O: 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O [1] (c) 1.23 − 0.02 = 1.21 V [1] (d) A fuel cell uses energy from fuel and oxygen supplied continuously [1]; a storage cell has a fixed store of chemicals, so it runs down and must be recharged or replaced [1] [7] Examiner insight: Fuel cell equations are given, not recalled; reverse the half-equation with the less positive E⦵.

Where marks are usually lost

• Averaging a non-concordant titre or including the rough titre.
• Forgetting the dilution factor, or the volume of the original sample, in multi-step titrations.
• Writing “colourless to blue-black” for the thiosulfate end point.
• Using an iron electrode for an Fe³⁺/Fe²⁺ half-cell.
• Multiplying E⦵ when a half-equation is doubled.
• Treating a positive E⦵cell as proof that a reaction will be fast.

Next steps

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Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, published by OCR. Section 5.2.3, Redox and electrode potentials.