AS Chemistry: Redox Processes — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Redox Processes revision notes

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Section A

1. Define oxidation and reduction in terms of electron transfer. [2]

2. Deduce the oxidation number of the named element in each species, showing your working: (a) Cr in Cr₂O₇²⁻, (b) S in H₂SO₄, (c) N in NH₄⁺, (d) Cl in ClO₃⁻. [4]

3. State what is meant by a disproportionation reaction, and give an example species that can undergo one. [2]

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Section B

4. Acidified potassium manganate(VII) oxidises iron(II) ions to iron(III) ions.

(a) Write the half-equation for the reduction of MnO₄⁻ to Mn²⁺ in acidic solution. [2]

(b) Write the half-equation for the oxidation of Fe²⁺. [1]

(c) Combine them into the overall ionic equation. [2]

(d) State the colour change at the end point and explain why no indicator is needed. [2]

5. 25.0 cm³ of an iron(II) solution required 22.4 cm³ of 0.0200 mol dm⁻³ KMnO₄ solution for complete reaction, using the equation from Question 4(c).

(a) Calculate the moles of MnO₄⁻ used. [2]

(b) Deduce the moles of Fe²⁺ present. [2]

(c) Calculate the concentration of the iron(II) solution. [2]

6. Chlorine reacts with cold dilute sodium hydroxide: Cl₂ + 2NaOH → NaCl + NaClO + H₂O

(a) Deduce the oxidation number of chlorine in each chlorine-containing species. [3]

(b) Explain why this is a disproportionation reaction. [2]

7. Balance the equation MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O by tracking oxidation-number changes, showing each step, and check your answer balances for both atoms and charge. [5]

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Answers

1. Oxidation is loss of electrons [1]; reduction is gain of electrons [1] — remembered by the mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain).

2. (a) +6 [1] (b) +6 [1] (c) −3 [1] (d) +5 [1].

3. A reaction in which the same element is simultaneously oxidised and reduced [1] [1]. Example: chlorine, Cl₂, in cold dilute sodium hydroxide (Question 6).

4. (a) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O [1] for species, [1] for balancing.

(b) Fe²⁺ → Fe³⁺ + e⁻ [1].

(c) MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ [1] [1].

(d) Colourless to pale pink / purple persisting [1]. Manganate(VII) is self-indicating — it is intensely coloured and is decolourised as it reacts, so the first permanent pink marks the end point of the titration [1].

5. (a) n = cV = 0.0200 × (22.4 ÷ 1000) [1] = 4.48 × 10⁻⁴ mol [1].

(b) Ratio MnO₄⁻ : Fe²⁺ = 1 : 5 [1] n(Fe²⁺) = 5 × 4.48 × 10⁻⁴ = 2.24 × 10⁻³ mol [1].

(c) c = n ÷ V = 2.24 × 10⁻³ ÷ 0.0250 [1] = 0.0896 mol dm⁻³ [1].

6. (a) Cl₂ = 0 [1]; NaCl = −1 [1]; NaClO = +1 [1].

(b) Chlorine is reduced from 0 to −1 and oxidised from 0 to +1 in the same reaction [1], so the same element undergoes both processes [1].

7. Manganese goes from +7 to +2, a decrease of 5, so each Mn gains 5 electrons [1]; iron goes from +2 to +3, an increase of 1, so each Fe loses 1 electron [1]. Equalising electrons lost and gained requires 5 Fe²⁺ per Mn [1]: MnO₄⁻ + 5Fe²⁺ + H⁺ → Mn²⁺ + 5Fe³⁺ + H₂O. Balancing oxygen needs 4H₂O, which needs 8H⁺ [1]: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Check: charge on the left = (−1) + 5(+2) + 8(+1) = +17; charge on the right = (+2) + 5(+3) + 0 = +17 — balanced [1].

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Where marks are usually lost

• Forgetting H⁺ and H₂O when balancing half-equations in acidic solution, or forgetting them when balancing by the oxidation-number method instead.
• Using a 1:1 ratio for manganate(VII) and iron(II) — the balanced equation shows it is 1:5.
• Not converting cm³ to dm³ before substituting into c = n/V.
• Saying an indicator is needed for a manganate(VII) titration.
• Forgetting to check a balanced half-equation or overall equation against both atom count and total charge on each side — a rushed answer can look balanced for atoms while still being wrong on charge.
• Mixing up the two balancing approaches — balancing electrons lost/gained via oxidation-number changes, versus writing and combining separate half-equations — rather than picking one method and following it through consistently.

Two routes to the same balanced equation

The overall equation for manganate(VII) oxidising iron(II) can be reached either by writing and combining two half-equations (Question 4), or by tracking the change in oxidation number for each element directly and equalising electrons lost and gained (Question 7) — both are valid AS Chemistry methods, and examiners accept either provided the working is shown clearly. The oxidation-number method is often faster for a simple two-species reaction, while half-equations are more reliable once H⁺, OH⁻ or H₂O also need balancing on both sides.

For condensed recall notes on this topic, see the Redox Processes revision notes; for the full explanation with additional worked examples, see the Redox Processes study guide.