OxfordAQA International GCSE Chemistry: Redox reactions (9202) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

Redox reactions, section 3.8.4 (statements a to c), is the focus of this set. The source is the OxfordAQA International GCSE Chemistry (9202) specification, in the version numbered 6.2, for exams May/June 2018 onwards. Nothing in this qualification is tiered, and section 3.8.4 has no required practical. As the specification asks, many questions apply the redox definitions to reactions from other sections: metals, electrolysis, Group 1 and the halogens. Questions 7, 10 and 11 are longer problems that combine several ideas.

Learn the content first with the Redox reactions study guide and the Redox reactions revision notes. To see how redox sits among the other topics, open the OxfordAQA chemistry hub; to track statements, print the course checklist. Unsure where to start? A free diagnostic points to your weakest topics.

Questions

1. State what is meant by:

(a) oxidation, in terms of electrons [1] (b) reduction, in terms of oxygen. [1]

2. Pure chromium metal can be made by igniting a mixture of aluminium powder and chromium(III) oxide. The reaction releases a large amount of heat.

2Al + Cr₂O₃ → Al₂O₃ + 2Cr

(a) Identify the substance that is oxidised. Explain your answer in terms of oxygen. [2] (b) Explain why this is described as a redox reaction. [1]

3. Complete each half equation.

(a) __Cl⁻ → Cl₂ + __ [1] (b) Sn²⁺ + __ → Sn [1] (c) State which of your two half equations shows oxidation. Give a reason. [1]

4. A coil of aluminium foil, cleaned with sandpaper, is left in colourless silver nitrate solution. Shiny crystals of silver grow on the foil.

Al + 3AgNO₃ → Al(NO₃)₃ + 3Ag

(a) Write the ionic equation for this reaction. [2] (b) Describe what happens to the aluminium and to the silver ions in terms of oxidation and reduction. [2]

5. Molten lead(II) iodide, PbI₂, is electrolysed using inert electrodes.

(a) Give the cathode half equation. [1] (b) Give the anode half equation. [1] (c) State the electrode at which oxidation takes place. Explain your answer. [2]

6. For each reaction, state whether it is a redox reaction. Give a reason for each answer.

	Reaction
P	AgNO₃ + NaCl → AgCl + NaNO₃
R	Zn + Cl₂ → ZnCl₂
S	2Li + 2H₂O → 2LiOH + H₂
T	2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O

[4]

7. Nickel can be obtained by heating nickel(II) oxide with carbon.

2NiO + C → 2Ni + CO₂

Relative atomic masses (Ar): Ni = 59, O = 16

(a) Which substance is reduced? Give a reason. [2] (b) Explain why oxidation and reduction must happen together in this reaction. [1] (c) Suggest why magnesium cannot be extracted by heating magnesium oxide with carbon. [1] (d) Calculate the maximum mass of nickel that can be made from 15.0 g of nickel(II) oxide. [3]

8. Chlorine water is added to pale green iron(II) chloride solution. The mixture turns yellow-brown as iron(III) ions form.

Cl₂ + 2Fe²⁺ → 2Cl⁻ + 2Fe³⁺

(a) Give the half equation for the change to the iron(II) ions. State whether they are oxidised or reduced. [2] (b) Give the half equation for the change to the chlorine. [1] (c) Explain why the oxygen definition cannot be used to classify this reaction. [1] (d) A student says iron(II) ions cannot be oxidised because they are already positive. Explain why the student is wrong. [2]

9. A small piece of lithium is added to water. It fizzes and slowly disappears.

(a) Write a half equation for the change to the lithium atoms and state whether they are oxidised or reduced. [2] (b) Name the element that is reduced in this reaction. [1] (c) Potassium reacts with water more vigorously than lithium. Explain this in terms of the outer electrons. [2]

10. Magnesium powder is stirred into tin(II) chloride solution. Grey tin metal forms.

Mg + SnCl₂ → MgCl₂ + Sn

Relative atomic masses (Ar): Mg = 24, Sn = 119

(a) Give the ionic equation, leaving out spectator ions. [1] (b) Split it into two half equations. [2] (c) Describe the reaction in terms of oxidation and reduction, and state what it shows about the reactivity of magnesium and tin. [2] (d) Calculate the maximum mass of tin formed when 1.20 g of magnesium reacts completely. [3]

11. Copper can be made from copper(I) oxide or from copper(II) chloride solution.

Method A: Cu₂O + C → 2Cu + CO Method B: electrolysis of copper(II) chloride solution with inert electrodes. Copper forms at one electrode and chlorine at the other.

Relative atomic masses (Ar): Cu = 63.5, O = 16

(a) In method A, identify what is oxidised and what is reduced. [2] (b) For method B, give the half equation at each electrode. [2] (c) Explain why method B is a redox process overall, even though oxidation and reduction happen at different electrodes. [2] (d) Calculate the maximum mass of copper that method A can produce from 14.3 g of copper(I) oxide. [3]

Answers

1. (a) Loss of electrons [1] (b) Loss of oxygen from a substance [1] Examiner insight: Each part asks for one specific definition, so an answer giving both definitions mixed together, or the reverse of OIL RIG, usually earns nothing for that part.

2. (a) Aluminium is oxidised [1] because it gains oxygen, becoming aluminium oxide [1]. (b) Chromium(III) oxide loses oxygen (is reduced) at the same time as aluminium is oxidised [1]. Examiner insight: “Aluminium is oxidised” on its own is only half of (a); the reason must mention gain of oxygen, because the question says “in terms of oxygen”.

3. (a) 2Cl⁻ → Cl₂ + 2e⁻ [1] (b) Sn²⁺ + 2e⁻ → Sn [1] (c) (a) is oxidation, because chloride ions lose electrons (electrons appear on the right) [1]. Examiner insight: Check charge as well as atoms; 2Cl⁻ → Cl₂ + e⁻ balances the atoms but not the charge and is not worth the mark.

4. (a) Correct species Al, Ag⁺, Al³⁺, Ag [1]; balanced: Al + 3Ag⁺ → Al³⁺ + 3Ag [1] (b) Aluminium atoms lose (three) electrons, so aluminium is oxidised [1]. Silver ions gain electrons to form silver atoms, so they are reduced [1]. Examiner insight: Leaving NO₃⁻ in the equation means it is not an ionic equation; nitrate ions are spectators and should be cancelled.

5. (a) Pb²⁺ + 2e⁻ → Pb [1] (b) 2I⁻ → I₂ + 2e⁻ (or 2I⁻ − 2e⁻ → I₂) [1] (c) The anode (positive electrode) [1]. Iodide ions lose electrons there, and loss of electrons is oxidation [1]. Examiner insight: Writing I⁻ → I + e⁻ is a common slip; iodine forms as I₂ molecules, so two iodide ions are needed.

6. P: not redox; the ions only swap partners and no particle gains or loses electrons [1]. R: redox; zinc atoms lose electrons to form Zn²⁺ and chlorine gains electrons to form Cl⁻ [1]. S: redox; lithium atoms lose electrons to form Li⁺ (and hydrogen forms from water) [1]. T: not redox; it is a neutralisation in which no electrons are transferred [1]. Examiner insight: A yes/no with no reason is unlikely to score; each line needs the evidence, such as which particle loses electrons.

7. (a) Nickel(II) oxide [1], because it loses oxygen to form nickel [1]. (b) The oxygen lost by nickel(II) oxide is gained by carbon, so one cannot happen without the other [1]. (c) Carbon is less reactive than magnesium, so it cannot take oxygen away from magnesium oxide [1]. (d) Mr of NiO = 59 + 16 = 75; moles of NiO = 15.0 ÷ 75 = 0.200 mol [1] Ratio NiO : Ni is 2 : 2 = 1 : 1, so moles of Ni = 0.200 mol [1] Mass of Ni = 0.200 × 59 = 11.8 g [1] Examiner insight: Show the moles and the ratio as separate lines; if the final answer is wrong, clearly set-out working is the only way an examiner can see which steps were right.

8. (a) Fe²⁺ → Fe³⁺ + e⁻ [1]; oxidised, because each ion loses an electron [1]. (b) Cl₂ + 2e⁻ → 2Cl⁻ [1] (c) No oxygen is gained or lost by any substance in the reaction [1]. (d) Oxidation is loss of electrons, not a matter of starting charge [1]. An Fe²⁺ ion can still give up an electron, and doing so makes it Fe³⁺, which is more positive [1]. Examiner insight: In (a) the electron goes on the right; writing Fe²⁺ + e⁻ → Fe³⁺ gives a charge of +1 on the left and +3 on the right, so the equation cannot be credited.

9. (a) Li → Li⁺ + e⁻ [1]; lithium is oxidised [1]. (b) Hydrogen (in the water) [1] (c) In potassium the outer electron occupies a higher energy level, so it is held less strongly by the nucleus [1]. Potassium therefore loses that electron, and is oxidised, more easily than lithium [1]. Examiner insight: Refer to the outer electron being lost; saying potassium “gains electrons more easily” reverses the chemistry and gains no credit.

10. (a) Mg + Sn²⁺ → Mg²⁺ + Sn [1] (b) Mg → Mg²⁺ + 2e⁻ [1]; Sn²⁺ + 2e⁻ → Sn [1] (c) Magnesium loses electrons and is oxidised; tin(II) ions gain electrons and are reduced [1]. Magnesium is more reactive than tin [1]. (d) Moles of Mg = 1.20 ÷ 24 = 0.0500 mol [1] Ratio Mg : Sn is 1 : 1, so moles of Sn = 0.0500 mol [1] Mass of Sn = 0.0500 × 119 = 5.95 g [1] Examiner insight: In (c) say “tin(II) ions are reduced”, not “tin chloride is reduced”; chloride ions do not change.

11. (a) Carbon is oxidised (gains oxygen to form CO) [1]; copper(I) oxide is reduced (loses oxygen) [1]. (b) Cathode: Cu²⁺ + 2e⁻ → Cu [1]; anode: 2Cl⁻ → Cl₂ + 2e⁻ [1] (c) Copper ions gain electrons (reduction) at the cathode while chloride ions lose electrons (oxidation) at the anode [1]. Both happen at the same time and the electrons lost at the anode equal those gained at the cathode [1]. (d) Mr of Cu₂O = (2 × 63.5) + 16 = 143; moles = 14.3 ÷ 143 = 0.100 mol [1] Ratio Cu₂O : Cu is 1 : 2, so moles of Cu = 0.200 mol [1] Mass of Cu = 0.200 × 63.5 = 12.7 g [1] Examiner insight: The 1 : 2 ratio in (d) is the step most often missed; using 1 : 1 gives 6.35 g, which goes wrong at the ratio step and at every step after it.

Where marks are usually lost

• Swapping the definitions: oxidation is loss of electrons and gain of oxygen.
• Naming the compound (silver nitrate, tin chloride) instead of the ion that is reduced.
• Writing halogens or halide products as single atoms instead of Cl₂, Br₂ or I₂.
• Unbalanced charge in half equations, especially when two ions form one molecule.
• Leaving spectator ions in an ionic equation.
• Saying oxidation happens at the cathode.
• Giving “more reactive” with no mention of electrons when the question asks about electrons.
• Classing neutralisation or precipitation as redox.
• Ignoring the mole ratio in mass calculations from a redox equation.

Next steps

• Revise the definitions again: redox revision notes
• Electrolysis practice questions for more electrode half equations
• Mole calculations: quantitative chemistry practice
• All free 10-minute diagnostics
• Every OxfordAQA International GCSE Chemistry topic
• Book a free trial class
• Statement-by-statement checklist to print
• Back to the full redox study guide

Official syllabus

International GCSE Chemistry (9202) specification published by OxfordAQA; Version 6.2, for exams May/June 2018 onwards. Unit covered: 3.8.4 Redox reactions, statements a to c, applied to reactions from 3.3.1.1, 3.3.2, 3.7.1 and 3.10.