WJEC A-Level Chemistry Unit 3, June 2025: Question 4

1 mark · Medium difficulty · Short Answer

Write the overall ionic equation for the oxidation of ethanedioic acid by acidified manganate(VII) ions using the provided half-equations.

Practise this question

Question

Question 4 states that ethanedioic acid, (COOH)2, can be oxidised by acidified manganate(VII) ions. Two half-equations are given: MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and (COOH)₂ → 2CO₂ + 2H⁺ + 2e⁻. The question asks students to use the half-equations to write the overall ionic equation for the reaction, worth 1 mark.
Question text

4. Ethanedioic acid, (COOH)2, can be oxidised by acidified manganate(VII) ions.

Use the half-equations below to write the ionic equation for the reaction. [1]

_ _

MnO + 8H+ + 5e Mn2+ + 4H O

_

(COOH) 2CO + 2H+ + 2e

Mark scheme

Show the mark scheme Mark scheme for Question 4 providing the correct balanced overall ionic equation: 2MnO4⁻ + 5(COOH)₂ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O, awarded 1 mark.

4 2MnO – + 5(COOH) + 6H+ → 2Mn2+ + 10CO + 8H O

42 2 2 1 1

How to answer it

Oxidation of Ethanedioic Acid by Manganate(VII)

📌 What this question tests

This question assesses your ability to combine two redox half-equations into a balanced full ionic equation. Key skills include:

  • Finding the lowest common multiple (LCM) of electrons transferred to balance oxidation and reduction.
  • Multiplying half-equations by integers so that electrons fully cancel out.
  • Simplifying species that appear on both sides of the reaction (specifically H⁺ ions).

Question 4

Combining Half-Equations to Form a Full Ionic Equation [1 Mark]

✅ Correct Answer

The fully balanced ionic equation is:

2MnO₄⁻ + 5(COOH)₂ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Mark Breakdown:
• [1 mark] for the fully balanced equation with all species, coefficients, and charges correct.

📐 Step-by-Step Method

1 Identify electrons in each half-equation:

  • Reduction: MnO₄⁻ gains 5e⁻
  • Oxidation: (COOH)₂ loses 2e⁻

2 Equalise electrons (LCM = 10):

  • Multiply MnO₄⁻ equation by 2:
    2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O
  • Multiply (COOH)₂ equation by 5:
    5(COOH)₂ → 10CO₂ + 10H⁺ + 10e⁻

3 Combine and cancel:

  • Electrons cancel completely: 10e⁻ on each side.
  • H⁺ ions cancel: 16H⁺ on left minus 10H⁺ on right leaves 6H⁺ on the left.

💡 Key Knowledge

  • Conservation of Charge: The total charge must be identical on both sides:
    Left: 2(−1) + 6(+1) = +4
    Right: 2(+2) = +4
  • Acidic Medium: Manganate(VII) requires excess H⁺ (typically dilute H₂SO₄) to reduce Mn(VII) down to nearly colourless Mn²⁺.
  • No Electrons in Final Equation: An overall ionic redox equation should never contain unreacted electrons (e⁻).

🧠 Exam Technique

  • Double-check charges: Even if species balance elementally, a missing charge on MnO₄⁻ or Mn²⁺ will instantly cost the mark.
  • Look for cancellations: Always scan both sides for spectator species or common ions like H⁺ or H₂O to ensure the equation is in its simplest form.
  • Keep formulae intact: Ethanedioic acid is given as (COOH)₂, so write 5(COOH)₂ rather than altering it to H₂C₂O₄ unless prompted.

❌ Common Errors

  • Failing to cancel H⁺ ions: Leaving 16H⁺ on the left and 10H⁺ on the right (e.g. ... + 16H⁺ → ... + 10H⁺ ... ). WJEC does not award the mark if identical species appear on both sides.
  • Leaving electrons in the equation: Retaining 10e⁻ on both sides.
  • Incorrect scaling: Multiplying one half-equation correctly but forgetting to multiply all terms (e.g. writing 2CO₂ instead of 10CO₂).

Topics

Physical Chemistry · 1.1 Formulae and equations · 3.2 Redox reactions

Question and mark scheme from the WJEC A-Level Chemistry examination, Unit 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.