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Topic 3.1.7: Redox - Exam Practice

AQA A-Level Chemistry • Physical Chemistry • Exam Practice

AQA A-LEVEL CHEMISTRY

Topic 3.1.7: Redox Equations Exam Practice

Test your understanding of oxidation states, half-equations, and redox titrations with exam-style questions.

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Structured Questions

Answer the questions below in the spaces provided. Check your answers with the mark schemes and examiner tips.

Question 1: Oxidation States & Redox Equations

6 marks

Nitrogen forms a range of oxides and ions with different oxidation states.

(a) Calculate the oxidation state of nitrogen in each of the following species: NO3^-, NH4^+, and N2O. [3]

(b) When copper metal reacts with dilute nitric acid, copper(II) nitrate, nitrogen monoxide gas (NO), and water are formed. Write a balanced overall ionic redox equation for this reaction. [3]

Show Mark Scheme

(a)

  • NO3^-: +5 [1]
  • NH4^+: -3 [1]
  • N2O: +1 [1]

(b)

  • Oxidation: Cu → Cu^2+ + 2e^- [1]
  • Reduction: NO3^- + 4H^+ + 3e^- → NO + 2H2O [1]
  • Overall equation: 3Cu + 2NO3^- + 8H^+ → 3Cu^2+ + 2NO + 4H2O [1]
Examiner Tip: Always show the separate half-equations first, balance the electrons lost and gained, then combine them. Make sure that both atoms and overall charges are balanced on both sides of the final equation.

Question 2: Manganate(VII) Titration Calculations

9 marks

In a laboratory experiment, a student dissolved an iron tablet containing iron(II) sulfate in dilute sulfuric acid. The solution was titrated against 0.0200 mol dm^-3 potassium manganate(VII) solution.

(a) Write the half-equation for the reduction of manganate(VII) ions (MnO4^-) to manganese(II) ions (Mn^2+) in acidic conditions. [2]

(b) Explain, in terms of electron transfer, why Fe^2+ acts as a reducing agent in this reaction. [2]

(c) The student found that 25.0 cm3 of the iron(II) solution required 22.50 cm3 of the MnO4^- solution for complete reaction. The overall redox equation is: 5Fe^2+ + MnO4^- + 8H^+ → 5Fe^3+ + Mn^2+ + 4H2O. Calculate the concentration, in mol dm^-3, of the iron(II) ions in the solution. [5]

Show Mark Scheme

(a)

  • MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O [2]
    • (1 mark for correct species and balancing atoms; 1 mark for correct number of electrons)

(b)

  • Fe^2+ loses an electron to form Fe^3+ [1]
  • A reducing agent is an electron donor / is oxidised during the reaction [1]

(c)

  • Moles of MnO4^- = 0.0200 × (22.50 / 1000) = 4.50 × 10^-4 mol [1]
  • Ratio of Fe^2+ to MnO4^- is 5:1 [1]
  • Moles of Fe^2+ = 5 × 4.50 × 10^-4 = 2.25 × 10^-3 mol [1]
  • Concentration of Fe^2+ = 2.25 × 10^-3 / (25.0 / 1000) [1]
  • Concentration = 0.0900 mol dm^-3 [1]
Examiner Tip: Keep all intermediate values in your calculator to avoid rounding errors. State the final answer to an appropriate number of significant figures (usually 3, matching the input data). Don't forget to include units.

Question 3: Disproportionation of Chlorine

5 marks

Chlorine gas reacts with cold, dilute aqueous sodium hydroxide in a redox reaction.

(a) Write a balanced chemical equation for this reaction. [2]

(b) State the oxidation states of chlorine in the reactants and products, and explain why this reaction is described as disproportionation. [3]

Show Mark Scheme

(a)

  • Cl2 + 2NaOH → NaCl + NaClO + H2O [2]
    • (1 mark for correct products, 1 mark for balanced equation)

(b)

  • Chlorine in Cl2 has an oxidation state of 0 [1]
  • Chlorine in NaCl is -1 (reduced) and in NaClO is +1 (oxidised) [1]
  • It is disproportionation because the same element (chlorine) is simultaneously oxidised and reduced in the reaction [1]
Examiner Tip: In definitions of disproportionation, make sure to state that the same element is both oxidised and reduced, and refer to specific oxidation states to back up your explanation.
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