Edexcel A-Level Chemistry AS Paper 1, June 2024: Question 9

9 marks · Medium difficulty · Calculations

Deduce redox equations, identify oxidising/reducing agents and disproportionation criteria, and calculate the volume of oxygen gas produced using the ideal gas equation.

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Question

Question text

9 This question is about some redox reactions.

(a) Iodine is reduced by thiosulfate ions. The relevant half‑equations are shown.

½I + e– → I–

2S O2– → S O62– + 2e–

23 4

Deduce an overall equation for this reaction.

State symbols are not required.

(1)

(b) In a different redox reaction, the chlorate(I) ion, ClO– , can react with

the chloride ion as shown in this equation.

ClO– + Cl– + 2H+ → Cl + H O

(i) State a reason why this is not a disproportionation reaction.

(1)

(ii) Identify the reducing agent in this reaction.

(1)

(iii) Which is the half‑equation for the chlorate(I) ion, ClO– , in this reaction?

(1)

A ClO– + Cl– → Cl + ½O + 2e–

B ClO– + H+ + e– → ½Cl + OH–

C ClO– + 2H+ + e– → ½Cl + H O

D ClO– → ½Cl + ½O + e–

The equation for this reaction is shown.

2KClO3(s) → 2KCl(s) + 3O2(g)

22 3

Calculate the volume,*P76893A02224*in cm, of oxygen produced at a temperature of 30°C and

pressure of 110000 Pa.

[The ideal gas equation is pV = nRT

Molar mass of KClO = 122.6 g mol–1

Gas constant (R) = 8.31 J mol–1 K–1]

(5)

(Total for Question 9 = 9 marks)

Mark scheme

Show the mark scheme

Question

Acceptable Answer Additional Guidance Mark

Number

9(a) Example of equation (1)

• balanced equation 2S O 2− + I → S O 2− + 2I−

23 2 4 6

Ignore state symbols even if incorrect

Inclusion of electron scores 0

Question

Acceptable Answer Additional Guidance Mark

Number

9(b)(i) An answer that makes reference to the following point: (1)

• a single species is not oxidised and reduced Allow reaction identified as ‘reverse

or disproportionation’ / comproportionation

two different species are not oxidised and reduced Allow ions for species

(to form the same species) Ignore one species is oxidised and one species is

reduced

Question

Acceptable Answer Additional Guidance Mark

Number

9(b)(ii) An answer that makes reference to the following point: (1)

• (reducing agent is the) chloride ion / Cl−

Question

Answer Mark

Number

9(b)(iii) The only correct answer is C (ClO− + 2H+ + e− → ½Cl + H O) (1)

A is not correct because this equation shows both oxidant and reductant

B is not correct because this equation produces hydroxide ions which would not be possible in acid conditions

D is not correct because oxygen is not a product of the overall reaction

How to answer it

Redox Reactions & Ideal Gas Calculations

What this question tests

This question assesses your mastery of redox chemistry fundamentals: combining half-equations to form overall redox equations, identifying disproportionation and comproportionation, recognizing oxidizing and reducing agents via oxidation numbers, constructing ionic half-equations, and applying the ideal gas equation ( pV = nRT ) alongside stoichiometry.

Question 9 (a)

Combining Half-Equations

✅ Correct Answer

2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻

Mark: (1) mark for the fully balanced overall equation.

💡 Key Knowledge

To combine half-equations, balance the number of electrons lost in oxidation with those gained in reduction.
Multiply the first half-equation ( ½I₂ + e⁻ → I⁻ ) by 2 so the electrons cancel out.

❌ Common Errors

Including electrons in the final overall equation (this loses the mark instantly).
Failing to multiply all coefficients in the first half-equation by 2.

Question 9 (b)(i)

Disproportionation Analysis

✅ Correct Answer

A single species is not both oxidized and reduced (or two different species are not oxidized and reduced to form the same species).

Mark: (1) mark. Note: "comproportionation" or "reverse disproportionation" is also accepted.

🧠 Exam Technique

Always check oxidation numbers of all elements across the reaction arrow. Disproportionation requires one element in a single reactant to simultaneously increase and decrease in oxidation number.

Question 9 (b)(ii)

Identifying the Reducing Agent

✅ Correct Answer

Chloride ion / Cl⁻ (or chlorine)

Mark: (1) mark.

💡 Key Knowledge

A reducing agent donates electrons and is itself oxidized. Here, Cl⁻ (oxidation state -1) increases to Cl₂ (oxidation state 0).

Question 9 (b)(iii)

Selecting the Correct Half-Equation

✅ Correct Answer

C: ClO⁻ + 2H⁺ + e⁻ → ½Cl₂ + H₂O

Mark: (1) mark.

🧠 Exam Technique & Distractor Analysis

A: Incorrect because it shows both oxidant and reductant together.
B: Incorrect as it produces hydroxide ions ( OH⁻ ), which cannot coexist in acidic conditions containing H⁺ .
D: Incorrect because oxygen gas ( O₂ ) is not a product in the given overall equation.

Question 9 (c)

Ideal Gas Calculation & Stoichiometry

📐 Step-by-Step Calculation

Moles of KClO₃:
Molar mass = 122.6 g mol⁻¹
Moles = 5.00 g / 122.6 g mol⁻¹ = 0.04078 mol
Moles of O₂ produced:
Ratio KClO₃ : O₂ is 2 : 3
Moles O₂ = 0.04078 × (3 / 2) = 0.06118 mol
Convert temperature to Kelvin:
T = 30 + 273.15 = 303.15 K
Rearrange ideal gas equation ( V = nRT / p ):
V = (0.06118 × 8.31 × 303.15) / 110000
V = 153.97 × 10⁻⁶ m³
Convert m³ to cm³ (× 10⁶):
V = 1400 cm³ (to 2 or 3 significant figures)

Marks: (5) marks total (Moles of solid, stoichiometric ratio, unit conversions, ideal gas rearrangement, final answer with correct sig figs).

❌ Common Calculation Traps

Unit Conversion Failures: Forgetting to convert m³ to cm³ by multiplying by 10⁶.
Temperature Slip: Using 30°C directly instead of converting to Kelvin (303 K).
Ratio Errors: Inverting or forgetting the 3/2 stoichiometric multiplier from the balanced equation.

Topics

Physical Chemistry · Inorganic Chemistry · Topic 3: Redox I · Topic 5: Formulae, Equations and Amounts of Substance · Topic 4: Inorganic Chemistry and the Periodic Table

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.