AQA A-Level Chemistry: Oxidation, reduction and redox equations (7405) – 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 3.1.7, Oxidation, reduction and redox equations, of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. The whole section is AS content and is also part of the full A-level; none of it is marked “A-level only”. The specification assesses it in Paper 1 at AS and A-level, and A-level Paper 3 can include any content. Question 10 also uses mole calculations from section 3.1.2.

Before you start, you may want the study guide or the revision notes. The AQA A-Level Chemistry hub and printable checklist cover the rest of the course.

Questions

1. Define oxidation in terms of electrons, and state what an oxidising agent does in a redox reaction. [2]

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

(a) phosphorus in H₃PO₄ [1] (b) nitrogen in N₂H₄ [1] (c) chromium in CrO₄²⁻ [1] (d) oxygen in OF₂ [1]

3. State the oxidation state of hydrogen in calcium hydride, CaH₂, and explain why it differs from the usual value for hydrogen. [2]

4. Magnesium reacts with dilute hydrochloric acid. The ionic equation is:

Mg + 2H⁺ → Mg²⁺ + H₂

(a) Identify the species that is oxidised. Give the change in its oxidation state. [2] (b) Identify the oxidising agent. [1]

5. In an iodine titration, thiosulfate ions react with iodine:

2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻

(a) Deduce the oxidation state of sulfur in S₂O₃²⁻ and in S₄O₆²⁻. [2] (b) Write a half-equation for the oxidation process and a half-equation for the reduction process. [2] (c) Identify the oxidising agent. [1]

6. Copper reacts with dilute nitric acid. Nitrate ions, NO₃⁻, are reduced to nitrogen monoxide, NO, and copper is oxidised to Cu²⁺ ions.

(a) Give the oxidation state of nitrogen in NO₃⁻ and in NO. [1] (b) Write a half-equation for the reduction of NO₃⁻ to NO in acidic solution. [2] (c) Use your answer to (b) to write an overall ionic equation for the reaction. [2]

7. Concentrated sulfuric acid oxidises bromide ions to bromine. The sulfuric acid is reduced to sulfur dioxide.

(a) Write a half-equation for the reduction of H₂SO₄ to SO₂. [2] (b) Write an overall equation for the reaction of H₂SO₄ with H⁺ and Br⁻ ions. [2] (c) Identify the reducing agent. [1]

8. Iodate(V) ions, IO₃⁻, react with iodide ions in acidic solution to form iodine.

(a) Show that the oxidation state of iodine in IO₃⁻ is +5. [1] (b) Write a half-equation for the reduction of IO₃⁻ to I₂ in acidic solution. [2] (c) Write the overall ionic equation for the reaction. [2] (d) Explain why iodine in this reaction is both a product of oxidation and a product of reduction. [1]

9. A student writes this equation for the reaction of hydrogen peroxide with iodide ions in acid:

H₂O₂ + I⁻ + 2H⁺ → I₂ + 2H₂O

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

10. An acidified solution of potassium dichromate(VI) is used to find the concentration of iron(II) ions. Dichromate(VI) ions are reduced to Cr³⁺ ions.

(a) Give the oxidation state of chromium in Cr₂O₇²⁻ and in Cr³⁺. [1] (b) Write a half-equation for the reduction of Cr₂O₇²⁻ to Cr³⁺ in acidic solution. [2] (c) Write the overall ionic equation for the reaction of Cr₂O₇²⁻ with Fe²⁺ ions in acid. [2] (d) A 25.0 cm³ sample of an iron(II) solution reacts exactly with 18.75 cm³ of 0.0200 mol dm⁻³ potassium dichromate(VI). Calculate the concentration of Fe²⁺ ions in mol dm⁻³. [4]

11. An ion XO₃⁻ contains an unknown element X. In acidic solution, 0.00200 mol of XO₃⁻ reacts exactly with 0.00300 mol of tin(II) ions. The tin(II) ions are oxidised to tin(IV) ions.

(a) Deduce the oxidation state of X in XO₃⁻. [1] (b) Write a half-equation for the oxidation of tin(II) ions. [1] (c) Calculate the amount, in moles, of electrons transferred. [2] (d) Deduce the oxidation state of X after the reaction. Show your reasoning. [2] (e) Explain why tin(II) ions act as the reducing agent. [1]

Answers

1. Oxidation is the loss of electrons [1]. An oxidising agent accepts electrons (and is itself reduced) [1]. [2] Examiner insight: “Gain of oxygen” alone does not earn the first mark, because the specification defines oxidation in terms of electrons.

2. (a) 3(+1) + x + 4(−2) = 0, so +5 [1] (b) 2x + 4(+1) = 0, so −2 [1] (c) x + 4(−2) = −2, so +6 [1] (d) F is always −1, so x + 2(−1) = 0, giving +2 [1] Examiner insight: Each answer needs a sign; “5” or “5+” for phosphorus is usually not credited as an oxidation state.

3. Hydrogen is −1 [1]. Calcium is a Group 2 metal and is +2, so in this metal hydride hydrogen takes the negative value (it is bonded to a less electronegative metal, not to a non-metal) [1]. [2] Examiner insight: The explanation mark needs a reason tied to the compound (metal hydride, Ca is +2), not just “it is an exception”.

4. (a) Magnesium (Mg) is oxidised [1]; its oxidation state goes from 0 to +2 [1]. (b) H⁺ (hydrogen ions) [1] Examiner insight: For (b), “hydrogen” alone is ambiguous between H⁺ and H₂; name the ion that accepts the electrons.

5. (a) S₂O₃²⁻: 2x + 3(−2) = −2, so +2 [1]; S₄O₆²⁻: 4x + 6(−2) = −2, so +2.5 [1] (b) Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻ [1]; reduction: I₂ + 2e⁻ → 2I⁻ [1] (c) I₂ (iodine) [1] Examiner insight: +2.5 is accepted as an average oxidation state; rounding it to +2 or +3 loses the mark.

6. (a) +5 in NO₃⁻ and +2 in NO (both needed) [1] (b) Correct species with H⁺ and H₂O balanced: NO₃⁻ + 4H⁺ → NO + 2H₂O [1]; electrons correct: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O [1] (c) Cu → Cu²⁺ + 2e⁻ is multiplied by 3 and the nitrate half-equation by 2, so that 6 electrons are transferred [1]; 3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O [1] Examiner insight: In (c) the second mark is for the fully balanced equation; any electrons left in the final equation lose it, though multiples are accepted.

7. (a) Species balanced with H₂O and H⁺: H₂SO₄ + 2H⁺ → SO₂ + 2H₂O [1]; with electrons: H₂SO₄ + 2H⁺ + 2e⁻ → SO₂ + 2H₂O [1] (b) Oxidation half-equation 2Br⁻ → Br₂ + 2e⁻ combined with (a), electrons equal [1]; H₂SO₄ + 2H⁺ + 2Br⁻ → SO₂ + Br₂ + 2H₂O [1] (c) Br⁻ (bromide ions) [1] Examiner insight: An error carried from (a) into (b) can still earn credit in (b) if the combining step is correct (error carried forward).

8. (a) x + 3(−2) = −1, so x = +5 [1] (b) Balanced atoms with H₂O and H⁺: 2IO₃⁻ + 12H⁺ → I₂ + 6H₂O [1]; 2IO₃⁻ + 12H⁺ + 10e⁻ → I₂ + 6H₂O [1] (c) Iodide half-equation 2I⁻ → I₂ + 2e⁻ multiplied by 5, then added and simplified [1]; IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O [1] (d) Iodine forms from I⁻ (−1 → 0, oxidation) and from IO₃⁻ (+5 → 0, reduction) [1] Examiner insight: For a “show that” in (a), the working must appear; writing “+5” alone repeats the question and earns nothing.

9. Iodine atoms do not balance: one I on the left and two on the right (the charge does not balance either: +1 on the left, 0 on the right) [1]. Correct equation: H₂O₂ + 2I⁻ + 2H⁺ → I₂ + 2H₂O [1] [2] Examiner insight: “It is not balanced” is too vague for the first mark; say which atom or charge is unbalanced.

10. (a) +6 in Cr₂O₇²⁻ and +3 in Cr³⁺ (both needed) [1] (b) Atoms balanced: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O [1]; Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O [1] (c) Fe²⁺ → Fe³⁺ + e⁻ multiplied by 6 [1]; Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O [1] (d) Amount of Cr₂O₇²⁻ = 0.01875 dm³ × 0.0200 mol dm⁻³ = 3.75 × 10⁻⁴ mol [1] Amount of Fe²⁺ = 6 × 3.75 × 10⁻⁴ = 2.25 × 10⁻³ mol [1] Concentration = 2.25 × 10⁻³ ÷ 0.0250 dm³ [1] = 0.0900 mol dm⁻³ [1] Examiner insight: In (d), using a 1 : 1 ratio instead of the 1 : 6 ratio from (c) loses the ratio mark, but the volume step can still be credited by error carried forward.

11. (a) x + 3(−2) = −1, so +5 [1] (b) Sn²⁺ → Sn⁴⁺ + 2e⁻ [1] (c) Each Sn²⁺ loses 2 electrons [1]; 2 × 0.00300 = 0.00600 mol of electrons [1] (d) Electrons gained per X = 0.00600 ÷ 0.00200 = 3 [1]; new oxidation state = +5 − 3 = +2 [1] (e) Sn²⁺ donates electrons (its oxidation state rises from +2 to +4), so it reduces X [1] Examiner insight: In (d) the reasoning mark needs electrons per X atom; the answer +2 with no working may not get both marks.

Where marks are usually lost

• Calling the oxidising agent “the species oxidised”; it is reduced.
• Leaving the sign off an oxidation state, or writing it as a charge (2+ instead of +2).
• Forgetting fluorine is always −1, so missing that oxygen is +2 in OF₂.
• Treating hydrogen as +1 in a metal hydride such as CaH₂.
• Rounding an average oxidation state such as +2.5.
• Balancing charge with H⁺ rather than electrons in a half-equation.
• Leaving e⁻, or H⁺ and H₂O on both sides, in the overall equation.
• Using the wrong mole ratio from the overall equation in a titration calculation.
• Writing “hydrogen” or “chlorine” when the question needs a named ion such as H⁺ or Br⁻.

Next steps

• Revision notes for this section
• Study guide for this section
• Mole practice questions
• AQA A-Level Chemistry hub
• Printable AQA A-Level Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards, section 3.1.7 Oxidation, reduction and redox equations. Published by AQA.