OCR A Level Chemistry A: Redox (H432) – 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.

Here you practise section 2.1.5, Redox, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They test every learning outcome, 2.1.5(a) to (f), from Module 2, Foundations in chemistry. H432 is the full A level. Questions 10 and 11 are longer problems that combine several outcomes.

Revise first with the Redox study guide and the Redox revision notes. Other Module 2 topics are on the OCR chemistry hub; mark off each outcome you are sure of on the printable checklist. For a wider check, try the free diagnostics.

Where a calculation needs data, it is given in the question.

Questions

1. For each species below, state the oxidation number of the element given.

(a) nitrogen in N₂O [1] (b) chlorine in HClO₄ [1] (c) hydrogen in BaH₂ [1] (d) oxygen in SrO₂, strontium peroxide [1]

2. Write the formula of each compound.

(a) tin(II) fluoride [1] (b) potassium chlorate(VII) [1] (c) iron(III) nitrate [1]

3. Name each compound, using a Roman numeral where one is needed.

(a) CuCl [1] (b) NaClO [1] (c) Na₂SO₃ [1]

4. Zinc granules react with dilute sulfuric acid.

(a) Write the full equation for the reaction, including state symbols. [2] (b) Using oxidation numbers, state which element is oxidised and which is reduced. [2] (c) Explain why the sulfate ion is neither oxidised nor reduced. [1]

5. Iron filings react with excess dilute hydrochloric acid.

(a) Give the name and formula of the salt formed. [2] (b) Write the full equation for the reaction. [1] (c) 0.837 g of iron reacts completely. Calculate the volume of hydrogen formed, in cm³, at room temperature and pressure. Use Ar(Fe) = 55.8; one mole of gas occupies 24.0 dm³ under these conditions. [2]

6. Chromium metal can be made by heating chromium(III) oxide with aluminium powder.

Cr₂O₃ + 2Al → Al₂O₃ + 2Cr

(a) Describe what happens to chromium in this reaction in terms of oxidation number and electrons. [2] (b) Show that the total number of electrons lost equals the total gained. [2]

7. Hot concentrated sulfuric acid reacts with zinc. One reaction that takes place is:

Zn + 2H₂SO₄ → ZnSO₄ + SO₂ + 2H₂O

(a) Give the oxidation number of sulfur in H₂SO₄ and in SO₂. [1] (b) Explain, in terms of electron loss and gain, what is oxidised and what is reduced. Refer to the two sulfur atoms separately. [3] (c) Suggest why this reaction does not produce hydrogen, unlike the reaction in question 4. [1]

8. Sodium peroxide reacts with water:

2Na₂O₂ + 2H₂O → 4NaOH + O₂

(a) State the oxidation number of oxygen in Na₂O₂. [1] (b) Explain why oxygen in Na₂O₂ is both oxidised and reduced in this reaction. [3]

9. A d-block element X forms an ion in which X has an oxidation number of +6. In acidic solution, 0.00400 mol of this ion reacts exactly with 0.00600 mol of zinc. All the zinc forms Zn²⁺ ions and all of X ends up in a single new oxidation number.

Deduce the new oxidation number of X. Show your working. [4]

10. A student adds a 0.250 g sample of impure aluminium powder to excess dilute hydrochloric acid. Only the aluminium in the sample produces a gas. The student collects 0.252 dm³ of hydrogen at room temperature and pressure.

(a) Write the full equation for the reaction of aluminium with hydrochloric acid. [1] (b) State the change in oxidation number of aluminium and of hydrogen. [2] (c) Calculate the percentage by mass of aluminium in the sample. Use Ar(Al) = 27.0; take 24.0 dm³ mol⁻¹ as the gas molar volume. [4]

11. Chlorine dioxide, ClO₂, reacts with aqueous sodium hydroxide:

2ClO₂ + 2NaOH → NaClO₂ + NaClO₃ + H₂O

(a) Name NaClO₂ and NaClO₃, using Roman numerals. [2] (b) Give the oxidation number of chlorine in ClO₂, NaClO₂ and NaClO₃. [2] (c) Explain, in terms of electrons, why this is a redox reaction even though only one element changes oxidation number. [2] (d) Write the formula of sodium chlorate(VII) and state the oxidation number of chlorine in it. [2]

Answers

1. (a) 2x + (−2) = 0, so +1 [1] (b) +1 + x + 4(−2) = 0, so +7 [1] (c) BaH₂ is a metal hydride; Ba is +2, so H is −1 [1] (d) SrO₂ is a peroxide; Sr is +2, so O is −1 [1] Examiner insight: Each oxidation number needs its sign; “1” or “7” alone is ambiguous, and a hydride or peroxide given the usual value of +1 or −2 is a common way to lose credit.

2. (a) Sn²⁺ with two F⁻: SnF₂ [1] (b) Cl is +7, so the anion is ClO₄⁻: KClO₄ [1] (c) Fe³⁺ with three NO₃⁻: Fe(NO₃)₃ [1] Examiner insight: In (c) the brackets matter; FeNO₃₃ or FeNO₃ is not a correct formula for iron(III) nitrate, so write Fe(NO₃)₃ in full.

3. (a) Cl is −1, so Cu is +1: copper(I) chloride [1] (b) +1 + x + (−2) = 0 gives Cl +1: sodium chlorate(I) [1] (c) 2(+1) + x + 3(−2) = 0 gives S +4: sodium sulfate(IV) [1] Examiner insight: “Sodium sulfate” alone for (c) describes Na₂SO₄, so it does not answer the question; the numeral is what identifies the ion.

4. (a) Zn + H₂SO₄ → ZnSO₄ + H₂ [1]; state symbols (s), (aq), (aq), (g) [1] (b) Zinc is oxidised: 0 → +2 [1]. Hydrogen is reduced: +1 → 0 [1] (c) S stays at +6 and O at −2 in sulfate on both sides, so no oxidation number changes [1] Examiner insight: For (b), give the element and both oxidation numbers; “zinc loses electrons” without the numbers does not use oxidation numbers as the question asks.

5. (a) Iron(II) chloride [1], FeCl₂ [1] (b) Fe + 2HCl → FeCl₂ + H₂ [1] (c) n(Fe) = 0.837 ÷ 55.8 = 0.0150 mol, so n(H₂) = 0.0150 mol [1]; volume = 0.0150 × 24.0 = 0.360 dm³ = 360 cm³ [1] Examiner insight: Iron(III) chloride in (a) is a common slip; dilute acids take iron only to +2, and a wrong salt in (a) usually brings a wrong equation in (b) with it.

6. (a) Chromium’s oxidation number falls from +3 in Cr₂O₃ to 0 in Cr [1]; each chromium gains 3 electrons, so chromium is reduced [1] (b) Electrons gained: 2 Cr × 3 = 6 [1]; electrons lost: Al goes 0 → +3, so 2 Al × 3 = 6, and the totals are equal [1] Examiner insight: A “show that” needs both totals stated with the working that produces them; writing “6 = 6” without the per-atom changes leaves the reasoning unseen.

7. (a) +6 in H₂SO₄ and +4 in SO₂ (both needed) [1] (b) Zinc is oxidised, 0 → +2, losing 2 electrons [1]. One sulfur atom is reduced, +6 → +4, gaining 2 electrons as it forms SO₂ [1]. The other sulfur atom stays +6 in ZnSO₄, so it is neither oxidised nor reduced [1] (c) Sulfur in the acid is reduced instead of hydrogen, so the reduced product is SO₂ rather than H₂ [1] Examiner insight: Part (b) asks about the two sulfur atoms separately; a single statement that “sulfur is reduced” misses the point about the unchanged sulfate.

8. (a) 2(+1) + 2x = 0, so −1 [1] (b) Some peroxide oxygen atoms go from −1 to 0 in O₂, losing electrons, so they are oxidised [1]. The others go from −1 to −2 in NaOH, gaining electrons, so they are reduced [1]. Two O atoms lose 1 electron each and two gain 1 each, so 2 electrons are lost and 2 gained [1] Examiner insight: Name both products with their oxidation numbers; an answer that only says oxygen “changes” in two directions, without the numbers −1, 0 and −2, is too vague for credit.

9. Each Zn loses 2 electrons, so electrons released = 2 × 0.00600 = 0.0120 mol [1]. Electrons per X = 0.0120 ÷ 0.00400 = 3 [1]. X gains electrons, so its oxidation number decreases [1]. New oxidation number = +6 − 3 = +3 [1] Examiner insight: Set out the electron ratio line by line; a bare “+3” leaves nothing to reward if one step slips.

10. (a) 2Al + 6HCl → 2AlCl₃ + 3H₂ [1] (b) Aluminium: 0 → +3 [1]. Hydrogen: +1 → 0 [1] (c) n(H₂) = 0.252 ÷ 24.0 = 0.0105 mol [1]; n(Al) = 0.0105 × 2/3 = 0.00700 mol [1]; mass of Al = 0.00700 × 27.0 = 0.189 g [1]; percentage = 0.189 ÷ 0.250 × 100 = 75.6% [1] Examiner insight: Aluminium and hydrogen react in a 2 : 3 mole ratio, which is easy to turn upside down; multiplying by 3/2 instead gives a percentage above 100%, a sign the ratio is upside down.

11. (a) NaClO₂: sodium chlorate(III) [1]; NaClO₃: sodium chlorate(V) [1] (b) ClO₂ +4 [1]; NaClO₂ +3 and NaClO₃ +5 (both needed) [1] (c) One chlorine atom gains one electron, +4 → +3, so it is reduced [1]; the other loses one electron, +4 → +5, so it is oxidised; one electron is lost and one gained [1] (d) NaClO₄ [1]; chlorine +7 [1] Examiner insight: In (c) state both directions with numbers; a redox reaction needs an oxidation and a reduction, and here both happen to the same element starting in the same compound.

Where marks are usually lost

• Giving peroxide oxygen −2 or hydride hydrogen +1, as in questions 1 and 8.
• Omitting the sign on an oxidation number, or writing it as a charge such as 3+.
• Leaving out the Roman numeral where it is needed (sulfate(IV), chlorate(I)) or adding one where it is not.
• Writing iron(III) salts as the product of iron with a dilute acid.
• Dropping brackets in formulae such as Fe(NO₃)₃.
• Calling the sulfate or chloride ion oxidised or reduced in a metal-acid reaction.
• Treating all atoms of an element as changing when only some do (questions 7 and 8).
• Inverting a mole ratio from the equation, as in question 10.
• Not checking that electrons lost equal electrons gained before using a ratio.

Next steps

• Recap the rules and tables: Redox revision notes
• Full explanations and worked examples: Redox study guide
• Check the whole course: all free 10-minute diagnostics
• Mole and gas-volume methods: Amount of substance practice
• More OCR chemistry topics: course hub
• Printable checklist: OCR A Level Chemistry checklist
• Related OCR topic: Acids
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 2.1.5 Redox, learning outcomes (a) to (f). Published by OCR.