AQA A-Level Chemistry Paper 3, June 2022: Question 21

1 mark · Easy difficulty · Multiple Choice

Calculate the volume of 0.020 mol dm⁻³ KMnO₄ solution required to completely oxidise 0.10 mol of vanadium(IV) ions using the provided redox equation.

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Question

Question 21 asks: 'The reaction between vanadium(IV) ions and manganate(VII) ions in acidic solution can be represented by the equation: 5 V⁴⁺ + MnO₄⁻ + 8 H⁺ → 5 V⁵⁺ + Mn²⁺ + 4 H₂O. What volume, in dm³, of 0.020 mol dm⁻³ KMnO₄ is needed to oxidise 0.10 mol of vanadium(IV) ions completely? [1 mark]' Followed by four multiple choice options: A 0.10, B 0.50, C 1.0, D 5.0.
Question text

21 The reaction between vanadium(IV) ions and manganate(VII) ions in acidic solution

can be represented by the equation

5 V4+ + MnO − + 8 H+ → 5 V5+ + Mn2+ + 4 H O

What volume, in dm3, of 0.020 mol dm−3 KMnO is needed to oxidise

0.10 mol of vanadium(IV) ions completely?

[1 mark]

A 0.10

B 0.50

C 1.0

D 5.0

Mark scheme

Show the mark scheme Mark scheme table shows for question 21: correct answer C (AO2), 1 mark, with working indicating an answer of 1.0.

21 C (AO2) 1 1.0

How to answer it

Redox Titration: Vanadium(IV) and Manganate(VII) Stoichiometry

📋 WHAT THIS QUESTION TESTS

This question assesses your ability to apply quantitative chemical principles to redox systems:

  • Interpreting stoichiometric reacting ratios from a balanced ionic redox equation.
  • Calculating reacting amounts in moles.
  • Rearranging the concentration equation ( n = c × V ) to determine solution volume in dm³ .
  • Navigating multiple-choice distractors under timed exam conditions without unit-conversion errors.

Question 21 Walkthrough

Multiple Choice [1 Mark] • AO2 (Application of Knowledge)

5 V⁴⁺ + MnO₄⁻ + 8 H⁺ → 5 V⁵⁺ + Mn²⁺ + 4 H₂O

✅ Correct Answer

C — 1.0

Mark scheme: 1 mark awarded for selecting C (AO2).

💡 Key Knowledge

  • Stoichiometric Ratio: The equation shows that 5 moles of V⁴⁺ react with 1 mole of MnO₄⁻.
  • Relationship: Amount (mol) = Concentration (mol dm⁻³) × Volume (dm³)
  • Rearrangement: Volume (dm³) = Amount (mol) / Concentration (mol dm⁻³)
  • Because volume is asked in dm³ and concentration is given in mol dm⁻³, no division or multiplication by 1000 is needed!

📐 Step-by-Step Calculation

  1. 1 Identify moles of vanadium(IV) ions given:
    n(V⁴⁺) = 0.10 mol
  2. 2 Apply the molar ratio from the balanced equation:
    Ratio is 5 V⁴⁺ : 1 MnO₄⁻
    n(MnO₄⁻) = n(V⁴⁺) / 5 = 0.10 / 5 = 0.020 mol
  3. 3 Calculate the required volume of KMnO₄:
    c(KMnO₄) = 0.020 mol dm⁻³
    V = n / c = 0.020 mol / 0.020 mol dm⁻³ = 1.0 dm³

🧠 Exam Technique & Speed Tip

  • Spot the shortcut: Notice that n(MnO₄⁻) = 0.020 mol and the concentration is also 0.020 mol dm⁻³ . Since the numerical values are identical, V = 0.020 / 0.020 = 1.0 dm³ immediately!
  • Always check target units: The question asks for the volume in dm³, not cm³. Candidates who reflexively multiply by 1000 waste precious time or select the wrong magnitude.

❌ Distractor Analysis & Common Traps

  • Choosing D (5.0): Forgetting the 5 : 1 ratio. If you assume a 1 : 1 ratio, n = 0.10 mol , giving V = 0.10 / 0.020 = 5.0 dm³ .
  • Choosing B (0.50): Calculation slip caused by confusing 0.10 / 0.020 and dividing by 10, or inverted ratios.
  • Choosing A (0.10): Blindly guessing matching numbers or multiplying 0.10 × 1.0 incorrectly.

Topics

Physical Chemistry · Inorganic Chemistry · 3.1.2 Amount of Substance · 3.1.7 Oxidation, Reduction and Redox Equations · 3.2.5 Transition Metals

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.