The questions are on the redox reactions test. Take it under real conditions before reading the mark scheme.
Quick answer: Redox reactions transfer electrons: oxidation is loss of electrons and an increase in oxidation number, reduction is gain of electrons and a decrease. This mark scheme gives the model answers and the marks for the 8 questions of the Chemistry 11-12 test on redox reactions, from oxidation numbers of sulfur to zinc and acid, with the equivalent answers to accept.
Use these model answers and the marking notes to score yourself out of 20. Award method marks where a numerical slip occurs, but equations must balance for both atoms and charge to earn the full mark.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R3.2
Question 1 Oxidation numbers of sulfur (3 marks)
- 2(+1) + x = 0, so S = −2.
- x + 2(−2) = 0, so S = +4.
- 2(+1) + x + 4(−2) = 0, so S = +6.
Marking: 1 mark for each correct value with sign.
Question 2 Ostwald process (2 marks)
- N: −3 → +2; O: 0 → −2.
- Nitrogen is oxidised (oxidation number rises) and oxygen is reduced; the reducing agent is NH₃.
Marking: 1 mark for each part; (a) needs both elements correct.
Question 3 Chlorate half-equation (3 marks)
Cl: +5 → −1, so 6 electrons are gained. Balance oxygen with H₂O and hydrogen with H⁺: ClO₃⁻ + 6H⁺ + 6e⁻ → Cl⁻ + 3H₂O. Charge: −1 + 6 − 6 = −1 on both sides.Marking: 1 mark for the correct species; 1 mark for H⁺ and H₂O balanced; 1 mark for electrons and charge balance.
Question 4 Chlorate and iron(II) (2 marks)
Multiply the iron half-equation by 6: ClO₃⁻ + 6H⁺ + 6Fe²⁺ → Cl⁻ + 3H₂O + 6Fe³⁺. Charge: −1 + 6 + 12 = +17 and −1 + 18 = +17.Marking: 1 mark for multiplying by 6; 1 mark for the correct overall equation.
Question 5 Dichromate titration (4 marks)
n(Cr₂O₇²⁻) = 0.01940 × 0.0100 = 1.94 × 10⁻⁴ mol. Ratio 1 : 6, so n(Fe²⁺) = 6 × 1.94 × 10⁻⁴ = 1.164 × 10⁻³ mol. c = 1.164 × 10⁻³ ÷ 0.0250 = 0.0466 mol/dm³.Marking: 1 mark for moles of dichromate; 1 mark for the ratio 1 : 6; 1 mark for moles of Fe²⁺; 1 mark for the concentration with unit.
Question 6 Hydrogen peroxide as two agents (2 marks)
Oxygen in H₂O₂ has oxidation number −1, between −2 (as in water) and 0 (as in O₂). It can be reduced to −2 (H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O), acting as an oxidising agent, or oxidised to 0 (H₂O₂ → O₂ + 2H⁺ + 2e⁻), acting as a reducing agent.Marking: 1 mark for −1 lying between −2 and 0; 1 mark for both reduction (oxidising agent) and oxidation (reducing agent).
Question 7 Copper(I) oxide in acid (2 marks)
Copper changes from +1 in Cu₂O to 0 in Cu (reduced) and to +2 in CuSO₄ (oxidised). The same element is both oxidised and reduced, so it is disproportionation.Marking: 1 mark for the oxidation numbers 0 and +2; 1 mark for the statement that one element is both oxidised and reduced.
Question 8 Zinc and acid (2 marks)
Zn: 0 → +2 (oxidised). H: +1 → 0 (reduced). Cl stays −1. The reducing agent is zinc.Marking: 1 mark for both oxidation number changes; 1 mark for zinc as reducing agent.
