Edexcel International A Level Chemistry: Redox equilibria (YCH11) – 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.

This set covers Topic 16: Redox Equilibria of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 16.1 to 16.19 with Core Practicals 12, 13a and 13b. It is in Unit 5: Transition Metals and Organic Nitrogen Chemistry, at International A2 (IA2) level. Learn the content first from the redox equilibria study guide and the redox equilibria revision notes. Tick outcomes off on the Edexcel Chemistry checklist; the course hub lists the rest of the course.

Use the E⦵ values given. Relative atomic mass: Fe 55.8.

Questions

1. State what is meant by the standard electrode potential of a half-cell. Include the standard conditions. [3]

2. The standard hydrogen electrode is the reference for all E⦵ values.

(a) Describe how a standard hydrogen electrode is set up. [3] (b) Explain why a reference electrode is needed. [1]

3. A student joins a cobalt half-cell to a silver half-cell. E⦵(Co²⁺/Co) = −0.28 V and E⦵(Ag⁺/Ag) = +0.80 V.

(a) Calculate E⦵cell. [1] (b) Write the cell diagram. [1] (c) Give the equation for the reaction when the cell supplies current. [1] (d) The student made the salt bridge with potassium chloride solution. Explain why this is unsuitable and suggest a better salt. [2] (e) State why a high-resistance voltmeter is used. [1]

4. Describe how you would set up a half-cell to measure E⦵ for MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇌ Mn²⁺(aq) + 4H₂O(l), and how you would connect it to obtain the value. [3]

5. Use these values: Sn⁴⁺ + 2e⁻ ⇌ Sn²⁺, +0.15 V; I₂ + 2e⁻ ⇌ 2I⁻, +0.54 V; Fe³⁺ + e⁻ ⇌ Fe²⁺, +0.77 V.

(a) Show by calculation that iodine can oxidise Sn²⁺ ions, and write the equation. [2] (b) Predict whether Sn⁴⁺ ions will oxidise iodide ions. Give a reason. [1] (c) Identify the strongest reducing agent among Sn²⁺, I⁻ and Fe²⁺. [1]

6. In the cell from question 3, the cobalt(II) ion concentration is raised to 1.50 mol dm⁻³. Predict and explain the effect on the cobalt electrode potential and on the cell emf. [3]

7. Use these values: MnO₂ + 4H⁺ + 2e⁻ ⇌ Mn²⁺ + 2H₂O, +1.23 V; Cl₂ + 2e⁻ ⇌ 2Cl⁻, +1.36 V.

(a) Calculate E⦵cell for manganese(IV) oxide oxidising chloride ions, and state what it predicts. [2] (b) Chlorine is in fact made when manganese(IV) oxide is warmed with concentrated hydrochloric acid. Explain why the prediction fails. [2]

8. Use these values: Fe³⁺ + e⁻ ⇌ Fe²⁺, +0.77 V; Fe²⁺ + 2e⁻ ⇌ Fe, −0.44 V.

(a) Write an equation for the disproportionation of Fe²⁺ ions and use E⦵ values to show it is not feasible. [2] (b) A technician keeps a clean iron nail in a stock solution of iron(II) sulfate. Explain, using E⦵cell, how this keeps Fe³⁺ out of the solution. [2]

9. Reaction X has E⦵cell = +0.62 V and reaction Y has E⦵cell = +0.31 V. Both transfer the same number of electrons.

(a) State which reaction has the larger total entropy change. [1] (b) Deduce how the equilibrium constant of X compares with that of Y. [2]

10. A supplier states that its iron wire is 99.5% iron by mass. 1.250 g of the wire was dissolved in excess dilute sulfuric acid, forming Fe²⁺ ions, and the solution was diluted to 250.0 cm³. Portions of 25.0 cm³ were titrated with 0.0150 mol dm⁻³ potassium manganate(VII). The mean titre was 29.20 cm³.

(a) Write the ionic equation for the titration reaction. [1] (b) State the end-point observation. [1] (c) Calculate the percentage by mass of iron in the wire. [4] (d) Each burette reading has an uncertainty of ±0.05 cm³ and the pipette ±0.06 cm³. Calculate the total percentage uncertainty from these two items. [2] (e) Use your answers to comment on the supplier’s claim. [1]

11. A student prepared crystals of potassium iodate(VII), KIO₄ (Mr = 230.0). 0.250 g of the crystals was dissolved to make 100.0 cm³ of solution. Portions of 10.0 cm³ were acidified and added to excess potassium iodide:

IO₄⁻ + 7I⁻ + 8H⁺ → 4I₂ + 4H₂O

The iodine needed a mean titre of 16.85 cm³ of 0.0500 mol dm⁻³ sodium thiosulfate. Name a suitable indicator and its colour change, and calculate the percentage purity of the crystals. [7]

12. Fuel cells can run on hydrogen or on ethanol.

(a) Explain the principle on which a fuel cell works. [1] (b) Write the equations for the reactions at both electrodes of a hydrogen-oxygen fuel cell with an acidic electrolyte. [2] (c) Write the equation for the reaction at the negative electrode when the electrolyte is alkaline. [1] (d) In an ethanol fuel cell with an acidic electrolyte, ethanol is oxidised to carbon dioxide at the negative electrode. Write the half-equation. [2] (e) Write the overall equation for the ethanol fuel cell. [1]

Answers

1. The voltage (emf) of a half-cell measured against a standard hydrogen electrode [1], at 298 K and 100 kPa pressure of any gases [1], with all ions at 1.00 mol dm⁻³ [1]. Examiner insight: “room temperature” in place of 298 K is unlikely to be accepted for that condition.

2. (a) A platinum electrode coated with platinum black [1], with H₂ gas at 100 kPa passed over it [1], dipping into H⁺(aq) of concentration 1.00 mol dm⁻³, all at 298 K [1]. (b) The potential of a single electrode cannot be measured; only a difference between two electrodes can [1]. Examiner insight: name platinum; “an inert electrode” alone is too vague for a description.

3. (a) E⦵cell = +0.80 − (−0.28) = +1.08 V [1] (b) Co(s) | Co²⁺(aq) || Ag⁺(aq) | Ag(s) [1] (c) Co + 2Ag⁺ → Co²⁺ + 2Ag [1] (d) Chloride ions react with silver ions to form a silver chloride precipitate, lowering [Ag⁺] [1]; use potassium nitrate [1]. (e) It draws almost no current, so it measures the maximum emf [1]. Examiner insight: in (c), balance the electrons; the E⦵ value stays +0.80 V even though the silver half-equation is doubled.

4. Use a platinum electrode [1] in a solution containing MnO₄⁻, Mn²⁺ and H⁺ ions, each at 1.00 mol dm⁻³ [1]. Connect it to a standard hydrogen electrode with a salt bridge and a high-resistance voltmeter, at 298 K [1]. Examiner insight: H⁺ is in the half-equation, so it must also be at 1.00 mol dm⁻³.

5. (a) E⦵cell = +0.54 − 0.15 = +0.39 V, positive so feasible [1]; I₂ + Sn²⁺ → 2I⁻ + Sn⁴⁺ [1]. (b) No: for Sn⁴⁺ with I⁻, E⦵cell is 0.15 − 0.54 = −0.39 V, a negative value [1]. (c) Sn²⁺, the right-hand species of the least positive half-equation [1]. Examiner insight: on a “show” question, write the subtraction with signs; a bare “+0.39 V” shows nothing.

6. The equilibrium Co²⁺ + 2e⁻ ⇌ Co shifts to the right [1], so the cobalt electrode potential becomes less negative [1]. The gap between the two electrode potentials narrows, so the cell emf falls below 1.08 V [1]. Examiner insight: link the change to the equilibrium shift; “the emf goes down” alone explains nothing.

7. (a) E⦵cell = +1.23 − 1.36 = −0.13 V [1], so the reaction is predicted not to be feasible under standard conditions [1]. (b) Concentrated acid gives [H⁺] and [Cl⁻] far above 1.00 mol dm⁻³, making E for MnO₂ more positive and E for Cl₂/Cl⁻ less positive [1]. E⦵cell is only slightly negative, so these non-standard conditions make E cell positive [1]. Examiner insight: “warming speeds it up” misses the point; rate does not change feasibility.

8. (a) 3Fe²⁺ → Fe + 2Fe³⁺ [1]; E⦵cell = −0.44 − 0.77 = −1.21 V, negative, so not feasible [1]. (b) Iron reduces any Fe³⁺ formed: Fe + 2Fe³⁺ → 3Fe²⁺ [1]; E⦵cell = +0.77 − (−0.44) = +1.21 V, so this is feasible [1]. Examiner insight: for disproportionation, use the two half-equations that both contain the species in question.

9. (a) X, because ΔStotal is proportional to E⦵cell [1]. (b) ln K is proportional to E⦵cell, so ln K for X is 0.62/0.31 = 2 times ln K for Y [1]; K for X equals (K for Y)² [1]. Examiner insight: saying “K for X is twice K for Y” confuses K with ln K and is wrong.

10. (a) 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O [1] (b) The first permanent pale pink colour [1]. (c) Moles MnO₄⁻ = 0.0150 × 29.20/1000 = 4.38 × 10⁻⁴ mol [1] Moles Fe²⁺ in 25.0 cm³ = 5 × 4.38 × 10⁻⁴ = 2.19 × 10⁻³ mol [1] Moles in 250.0 cm³ = 2.19 × 10⁻² mol [1] Mass of iron = 2.19 × 10⁻² × 55.8 = 1.222 g; percentage = 1.222/1.250 × 100 = 97.8% [1] (d) Titre: ±0.10/29.20 × 100 = 0.34% [1]; total = 0.34 + 0.24 = 0.58% [1] (e) 0.58% of 97.8% is about 0.6 percentage points, much smaller than the 1.7 point gap from 99.5%, so measurement uncertainty cannot explain it: the claim is too high or a systematic error (such as air oxidising Fe²⁺) lowered the titre [1]. Examiner insight: in (c), show the 1 : 5 ratio and the ×10 scale-up; a bare final answer leaves no route to partial credit.

11. Starch, added when the solution is pale yellow [1]; blue-black to colourless [1]. Moles S₂O₃²⁻ = 0.0500 × 16.85/1000 = 8.425 × 10⁻⁴ mol [1] 1 IO₄⁻ gives 4I₂, which needs 8S₂O₃²⁻, so IO₄⁻ : S₂O₃²⁻ = 1 : 8 [1] Moles IO₄⁻ in 10.0 cm³ = 1.053 × 10⁻⁴ mol [1] Moles in 100.0 cm³ = 1.053 × 10⁻³ mol; mass = 1.053 × 10⁻³ × 230.0 = 0.242 g [1] Purity = 0.242/0.250 × 100 = 96.9% [1] Examiner insight: in an unstructured calculation, write the overall ratio as a separate line; stopping at 1 : 4 forgets the thiosulfate step.

12. (a) It uses the energy released when a fuel reacts with oxygen to generate a voltage [1]. (b) Negative: H₂ → 2H⁺ + 2e⁻ [1]; positive: O₂ + 4H⁺ + 4e⁻ → 2H₂O [1] (c) H₂ + 2OH⁻ → 2H₂O + 2e⁻ [1] (d) C₂H₅OH + H₂O → CO₂ + H⁺ + e⁻ with correct species [1]; balanced: C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺ + 12e⁻ [1] (e) C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O [1] Examiner insight: check every half-equation for charge as well as atoms; an equation with 12H⁺ but 6e⁻ is not balanced.

Where marks are usually lost

• Subtracting E⦵ values in the wrong order and reporting a negative emf for a working cell.
• Scaling E⦵ up along with its half-equation.
• Leaving out platinum where no solid is in the half-equation.
• Calling the salt bridge a path for electrons.
• Missing the 1 : 5 or 1 : 2 ratios, or the scale-up to the volumetric flask.
• Using a single burette reading for the titre uncertainty.
• Putting H⁺ in alkaline fuel cell equations, or OH⁻ in acid ones.

Next steps

• Recap with the redox equilibria revision notes.
• Reread the worked examples in the redox equilibria study guide.
• See every Edexcel International A Level Chemistry resource on the course hub.
• Mark your progress on the Edexcel Chemistry checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Unit 5: Transition Metals and Organic Nitrogen Chemistry, Topic 16: Redox Equilibria, outcomes 16.1 to 16.19.