OxfordAQA A-Level Chemistry: Oxidation, reduction and redox equations (9620) – 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.

These questions cover section 3.1.5, Oxidation, reduction and redox equations (3.1.5.1), of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. All of 3.1.5 is International AS content and is also part of the full International A-level. It belongs to Unit 1: Inorganic 1 and Physical 1, and Unit 5: Practical and synoptic can assess all content. Questions 9 and 10 also use mole calculations from section 3.1.2.

If you need to review first, read the study guide or the revision notes. The OxfordAQA A-Level Chemistry hub and printable checklist cover the rest of the course, and the free diagnostics show which topics need more work.

Questions

1. Explain, in terms of electrons, why the reducing agent in a redox reaction is itself oxidised. [2]

2. Deduce the oxidation state of the stated element in each species.

(a) sulfur in SOCl₂ [1] (b) nitrogen in Li₃N [1] (c) xenon in XeOF₄ [1] (d) bromine in BrF₅ [1]

3. Lithium aluminium hydride, LiAlH₄, contains lithium and aluminium in their usual oxidation states. Deduce the oxidation state of hydrogen in LiAlH₄, showing how you reached it. [2]

4. Magnesium continues to burn in carbon dioxide:

2Mg + CO₂ → 2MgO + C

(a) Give the change in oxidation state of magnesium and of carbon. [2] (b) Identify the oxidising agent. [1]

5. Four reactions are shown.

A: CaCO₃ → CaO + CO₂ B: H₂ + Cl₂ → 2HCl C: AgNO₃ + NaCl → AgCl + NaNO₃ D: 2Fe + 3Cl₂ → 2FeCl₃

Identify the two redox reactions. For each one, state which element is oxidised and give its change in oxidation state. [3]

6. A student writes this half-equation for the reduction of manganese(IV) oxide in acidic solution:

MnO₂ + 4H⁺ → Mn²⁺ + 2H₂O + 2e⁻

Explain what is wrong with it and write the correct half-equation. [2]

7. Phosphonic acid, H₃PO₃, is oxidised to phosphoric(V) acid, H₃PO₄, in aqueous solution.

(a) Deduce the oxidation state of phosphorus in H₃PO₃ and in H₃PO₄. [1] (b) Write a half-equation for the oxidation of H₃PO₃ to H₃PO₄ in acidic solution. [2]

8. One reaction of fluorine with water produces hydrogen fluoride and oxygen:

2F₂ + 2H₂O → 4HF + O₂

(a) Give the oxidation states of fluorine and of oxygen before and after the reaction. [2] (b) Write a half-equation for the reduction of F₂ to HF and a half-equation for the oxidation of H₂O to O₂, both in acidic solution. [2] (c) Show how your half-equations combine to give the overall equation. [1]

9. In acidic solution, cerium(IV) ions, Ce⁴⁺, can oxidise methanoic acid, HCOOH, to carbon dioxide. The cerium(IV) ions are reduced to cerium(III) ions, Ce³⁺.

(a) Deduce the oxidation state of carbon in HCOOH and in CO₂. [1] (b) Write a half-equation for the oxidation of HCOOH to CO₂. [1] (c) Write the overall ionic equation for the reaction. [2] (d) A 20.0 cm³ sample of methanoic acid solution reacts exactly with 24.00 cm³ of 0.0500 mol dm⁻³ Ce⁴⁺ solution. Calculate the concentration of methanoic acid in mol dm⁻³. [3]

10. A student proposes that magnesium reduces selenous acid, H₂SeO₃, to selenium, Se, in acidic solution, and that the magnesium is oxidised to Mg²⁺ ions. Use this proposal to answer the questions.

(a) Give the oxidation state of selenium in H₂SeO₃ and in Se. [1] (b) Write a half-equation for the reduction of H₂SeO₃ to Se in acidic solution. [2] (c) Write the overall ionic equation for the reaction. [2] (d) Calculate the mass of magnesium needed to reduce 0.0150 mol of H₂SeO₃. Use Ar of Mg = 24.3. [3]

11. Chlorine dioxide, ClO₂, can be made by reacting chlorine with chlorite ions, ClO₂⁻:

2ClO₂⁻ + Cl₂ → 2ClO₂ + 2Cl⁻

(a) Deduce the oxidation state of chlorine in each of the four species. [2] (b) Write the two half-equations for this reaction and label each as oxidation or reduction. [2] (c) Use your half-equations to show that 1 mol of Cl₂ reacts with 2 mol of ClO₂⁻. [1] (d) Explain how chlorine can be both oxidised and reduced in this reaction. [1]

Answers

1. A reducing agent is an electron donor: it gives electrons to the species it reduces [1]. Oxidation is the process of electron loss, so the reducing agent is oxidised [1]. [2] Examiner insight: Make the link explicit in two steps (donates electrons, then electron loss is oxidation); “because it reduces something” just repeats the name.

2. (a) x + (−2) + 2(−1) = 0, so +4 [1] (b) 3(+1) + x = 0, so −3 [1] (c) x + (−2) + 4(−1) = 0, so +6 [1] (d) x + 5(−1) = 0, so +5 [1] Examiner insight: Start from the atoms with fixed values (F at −1, then O at −2) and write each result with the sign in front, such as +6, so it cannot be mistaken for an ionic charge.

3. Li is +1 and Al is +3, so the four hydrogen atoms must total −4 for a neutral compound: (+1) + (+3) + 4x = 0 [1]. Hydrogen is −1 [1]. [2] Examiner insight: Assuming hydrogen is +1 here gives a total of +8 for a neutral compound, which is impossible; when the sum will not reach zero, check whether hydrogen is bonded to a metal.

4. (a) Magnesium: 0 → +2 [1]. Carbon: +4 → 0 [1]. (b) CO₂ (carbon dioxide) [1] Examiner insight: Magnesium gains oxygen here, but base (b) on electrons: the carbon atom in CO₂ gains four electrons, so CO₂ is the oxidising agent to name.

5. B and D are redox reactions (and no others) [1]. In B, hydrogen is oxidised from 0 to +1 [1]. In D, iron is oxidised from 0 to +3 [1]. [3] Examiner insight: Work out the oxidation state of every element in A and C as well; each element keeps the same value on both sides, which is why neither is redox.

6. Mn falls from +4 to +2, so this is a reduction and the electrons belong on the left; as written the charge does not balance (+4 on the left, 0 on the right) [1]. Correct: MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O [1] [2] Examiner insight: Use the oxidation state change to place the electrons: Mn falls by 2, so 2e⁻ go with the reactants, and both sides then total +2.

7. (a) +3 in H₃PO₃ and +5 in H₃PO₄ (both needed) [1] (b) Species with H₂O and H⁺ balanced: H₃PO₃ + H₂O → H₃PO₄ + 2H⁺ [1]; electrons added: H₃PO₃ + H₂O → H₃PO₄ + 2H⁺ + 2e⁻ [1] Examiner insight: Check the electron count against the oxidation state change: P rises by 2, so exactly 2e⁻ must appear, on the right.

8. (a) Fluorine: 0 in F₂ → −1 in HF [1]. Oxygen: −2 in H₂O → 0 in O₂ [1]. (b) Reduction: F₂ + 2H⁺ + 2e⁻ → 2HF [1]. Oxidation: 2H₂O → O₂ + 4H⁺ + 4e⁻ [1] (c) Double the fluorine half-equation to 2F₂ + 4H⁺ + 4e⁻ → 4HF, add it to the oxidation, and cancel 4e⁻ and 4H⁺ to give 2F₂ + 2H₂O → 4HF + O₂ [1] Examiner insight: In (c) the multiplying and cancelling must be shown; simply copying the given equation does not show how the half-equations combine.

9. (a) +2 in HCOOH and +4 in CO₂ (both needed) [1] (b) HCOOH → CO₂ + 2H⁺ + 2e⁻ [1] (c) Ce⁴⁺ + e⁻ → Ce³⁺ multiplied by 2 so electrons match [1]; 2Ce⁴⁺ + HCOOH → 2Ce³⁺ + CO₂ + 2H⁺ [1] (d) Amount of Ce⁴⁺ = 0.02400 dm³ × 0.0500 mol dm⁻³ = 1.20 × 10⁻³ mol [1] Amount of HCOOH = 1.20 × 10⁻³ ÷ 2 = 6.00 × 10⁻⁴ mol [1] Concentration = 6.00 × 10⁻⁴ ÷ 0.0200 dm³ = 0.0300 mol dm⁻³ [1] Examiner insight: Convert cm³ to dm³ before multiplying, and state the 2 : 1 ratio you are using so your method is clear even if a number slips.

10. (a) +4 in H₂SeO₃ and 0 in Se (both needed) [1] (b) Atoms balanced: H₂SeO₃ + 4H⁺ → Se + 3H₂O [1]; with electrons: H₂SeO₃ + 4H⁺ + 4e⁻ → Se + 3H₂O [1] (c) Mg → Mg²⁺ + 2e⁻ multiplied by 2 so 4 electrons transfer [1]; 2Mg + H₂SeO₃ + 4H⁺ → 2Mg²⁺ + Se + 3H₂O [1] (d) Mole ratio Mg : H₂SeO₃ = 2 : 1, so amount of Mg = 2 × 0.0150 = 0.0300 mol [1] Mass = 0.0300 × 24.3 [1] = 0.729 g [1] Examiner insight: Selenium falls by 4 (from +4 to 0), so (b) needs 4e⁻; that count, set against 2e⁻ per Mg, gives the 2 : 1 ratio you use in (d).

11. (a) ClO₂⁻ +3, ClO₂ +4 [1]; Cl₂ 0, Cl⁻ −1 [1] (b) Oxidation: ClO₂⁻ → ClO₂ + e⁻ [1]. Reduction: Cl₂ + 2e⁻ → 2Cl⁻ [1] (c) Each ClO₂⁻ loses 1 electron and each Cl₂ gains 2, so the oxidation is doubled: 2 mol ClO₂⁻ supply the 2 mol of electrons that 1 mol Cl₂ accepts [1] (d) The chlorine atoms are in different species: Cl in ClO₂⁻ is oxidised (+3 → +4) while Cl in Cl₂ is reduced (0 → −1) [1] Examiner insight: For (d), point to two different starting species, ClO₂⁻ and Cl₂, and quote the direction of each change; the element alone cannot show it.

Where marks are usually lost

• Explaining redox with gain or loss of oxygen instead of electrons.
• Calling the reducing agent “the species reduced”.
• Writing oxidation states without a sign, or with the sign after the number.
• Forgetting that fluorine is always −1, so getting the wrong value for xenon in XeOF₄ or bromine in BrF₅.
• Taking hydrogen as +1 in a metal hydride.
• Putting electrons on the wrong side of a half-equation, or balancing charge with H⁺.
• Not cancelling H⁺ that appear on both sides after combining, as in question 8.
• Using a 1 : 1 ratio in a calculation instead of the ratio from the overall equation.
• Giving “carbon” or “chlorine” when the question asks for a species such as CO₂ or Cl₂.

Next steps

• Revision notes for this section
• Study guide for this section
• Amount of substance study guide
• OxfordAQA A-Level Chemistry hub
• Printable OxfordAQA A-Level Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, section 3.1.5 Oxidation, reduction and redox equations (3.1.5.1). Published by OxfordAQA.