A2 · Practice questions · Variable oxidation states and redox

Count the electrons.

Six original Cambridge-style questions on transition metal redox and titrations.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Ratios, indicators, acids.

MnO₄⁻ : Fe²⁺ 1 : 5; MnO₄⁻ : C₂O₄²⁻ 2 : 5.
Cr₂O₇²⁻ : Fe²⁺ 1 : 6.
Cu²⁺ : S₂O₃²⁻ 1 : 1 (via I₂, starch indicator).
Acidify with H₂SO₄; E^⊖_cell positive means feasible.

Recall

[2 marks]

Write the ionic equation for the reaction of acidified manganate(VII) ions with iron(II) ions, and give the colour change at the end point.

MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. ✓
Colourless (very pale yellow) to the first permanent pale pink. ✓

Calculation

[4 marks]

6.00 g of impure iron(II) sulfate crystals, FeSO₄.7H₂O (M_r 278.0), is dissolved in dilute H₂SO₄ and made up to 250 cm³. 25.0 cm³ portions need 20.00 cm³ of 0.0200 mol dm⁻³ KMnO₄. Calculate the percentage purity.

n(MnO₄⁻) = 0.0200 × 0.02000 = 4.00 × 10⁻⁴ mol. ✓
n(Fe²⁺) in 25.0 cm³ = 5 × 4.00 × 10⁻⁴ = 2.00 × 10⁻³ mol; in 250 cm³ = 0.0200 mol. ✓
Mass = 0.0200 × 278.0 = 5.56 g. ✓
Purity = 5.56 ÷ 6.00 × 100 = 92.7%. ✓

Calculation

[3 marks]

25.0 cm³ of 0.0500 mol dm⁻³ sodium ethanedioate is acidified and warmed. Calculate the volume of 0.0200 mol dm⁻³ KMnO₄ needed.

n(C₂O₄²⁻) = 0.0500 × 0.0250 = 1.25 × 10⁻³ mol. ✓
2MnO₄⁻ : 5C₂O₄²⁻, so n(MnO₄⁻) = 1.25 × 10⁻³ × 2/5 = 5.00 × 10⁻⁴ mol. ✓
V = 5.00 × 10⁻⁴ ÷ 0.0200 = 0.0250 dm³ = 25.0 cm³. ✓

Calculation

[3 marks]

Excess KI is added to 25.0 cm³ of copper(II) sulfate solution. The iodine released needs 21.20 cm³ of 0.100 mol dm⁻³ sodium thiosulfate. Calculate the concentration of Cu²⁺.

2Cu²⁺ + 4I⁻ → 2CuI + I₂; I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, so Cu²⁺ : S₂O₃²⁻ = 1 : 1. ✓
n(Cu²⁺) = 0.100 × 0.02120 = 2.12 × 10⁻³ mol. ✓
[Cu²⁺] = 2.12 × 10⁻³ ÷ 0.0250 = 0.0848 mol dm⁻³. ✓

Explain

[2 marks]

Use E^⊖ values to explain why potassium manganate(VII) titrations are acidified with sulfuric acid rather than hydrochloric acid.

E^⊖(MnO₄⁻/Mn²⁺) = +1.52 V is more positive than E^⊖(Cl₂/Cl⁻) = +1.36 V, so MnO₄⁻ oxidises Cl⁻ to Cl₂. ✓
Extra manganate(VII) would be used, giving too large a titre; sulfate ions are not oxidised. ✓

Analysis

[2 marks]

Write the half-equation for the reduction of dichromate(VI) ions in acid, and use it to deduce the ratio Cr₂O₇²⁻ : Fe²⁺.

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (each Cr goes +6 to +3). ✓
Six electrons are gained and each Fe²⁺ gives one, so the ratio is 1 : 6. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.