Edexcel International A Level Chemistry: Redox chemistry (YCH11) – 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.

This practice set works through Topic 8A: Redox chemistry of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 8.1 to 8.10. Topic 8A belongs to Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols, an IAS unit. Learn the content first from the redox study guide and the redox revision notes, and track progress on the printable checklist. More resources are on the course hub.

Data for this set: H = 1.0, O = 16.0, Na = 23.0, S = 32.1, Br = 79.9, Sr = 87.6. Molar volume of a gas = 24.0 dm³ mol⁻¹ at room temperature and pressure.

Questions

1. Give the oxidation number of the element named in each part.

(a) N in NO₂⁺ [1] (b) S in SF₄ [1] (c) H in NaAlH₄ [1] (d) O in Rb₂O₂ [1]

2. This question is about names and formulae.

(a) Name N₂O and K₂SO₄, using Roman numerals to show oxidation numbers. [2] (b) Write the formulae of calcium chlorate(I), lithium nitrate(III) and aluminium sulfate(VI). [3]

3. Define these terms.

(a) Oxidation, in terms of electrons and in terms of oxidation number. [2] (b) Oxidising agent. [1] (c) Disproportionation. [1]

4. Four reactions are shown.

• Reaction W: BaCO₃ → BaO + CO₂
• Reaction X: 2KNO₃ → 2KNO₂ + O₂
• Reaction Y: 3HNO₂ → HNO₃ + 2NO + H₂O
• Reaction Z: Na₂O + H₂O → 2NaOH

(a) Classify each reaction as not redox, redox, or disproportionation. Give oxidation number evidence for each. [4] (b) Explain why reaction X is not a disproportionation, even though only one reactant is involved. [1]

5. In acidic solution, zinc reduces nitrate(V) ions to ammonium ions.

(a) State the change in oxidation number of nitrogen. [1] (b) Write the half-equation for the reduction of NO₃⁻ to NH₄⁺. [2] (c) Write the half-equation for the oxidation of zinc. [1] (d) Combine your half-equations into a full ionic equation. [2]

6. In acid, 0.00480 mol of sulfate(IV) ions, SO₃²⁻, exactly reduces 0.00192 mol of bromate(V) ions, BrO₃⁻. Each SO₃²⁻ is oxidised to SO₄²⁻.

(a) Write the half-equation for the oxidation of SO₃²⁻. [1] (b) Calculate the amount, in mol, of electrons released by the sulfate(IV) ions. [1] (c) Deduce the number of electrons gained by each BrO₃⁻ ion. [1] (d) Deduce the oxidation number of bromine in the product, and identify the product. [2] (e) Write the full ionic equation for the reaction. [2]

7. White phosphorus reacts with hot aqueous sodium hydroxide:

P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻

(a) Give the oxidation number of phosphorus in PH₃ and in H₂PO₂⁻. [2] (b) Explain why this is a disproportionation reaction. [2]

8. Potassium reacts with sulfur to form potassium sulfide, K₂S.

(a) Explain, in terms of electrons and oxidation numbers, which element is oxidised and which is reduced. [3] (b) Explain why metals generally form positive ions and non-metals generally form negative ions. [2]

9. Sodium hydride reacts with water: NaH + H₂O → NaOH + H₂

(a) Give the oxidation number of hydrogen in NaH and in H₂O. [2] (b) Identify the oxidising agent and the reducing agent. [2] (c) Explain why this is not a disproportionation reaction. [1] (d) Write two half-equations for the reaction, one for each hydrogen-containing reactant. [2] (e) Calculate the mass of sodium hydroxide formed from 0.300 g of sodium hydride. [2]

10. Strontium reacts with cold water.

(a) Write an ionic equation for the reaction, with state symbols. [2] (b) Write the oxidation and reduction half-equations. [2] (c) Identify the oxidising agent, using oxidation numbers. [2] (d) Calculate the volume of hydrogen, in cm³, formed at room temperature and pressure from 0.657 g of strontium. [3]

11. In one step of extracting bromine from seawater, bromine vapour is absorbed into water with sulfur dioxide.

(a) Write the two half-equations, for bromine and for sulfur dioxide in water. [2] (b) Combine them into a full ionic equation. [2] (c) State the oxidation number changes and identify the oxidising agent. [2] (d) Calculate the mass of sulfur dioxide that reacts with 4.00 g of bromine. Answer to three significant figures. [3]

Answers

1. (a) x + 2(−2) = +1, so +5 [1] (b) F is −1, so +4 [1] (c) Metal hydride, so −1 [1] (d) Peroxide, so −1 [1] Examiner insight: An oxidation number without its sign, or written as “5+”, is unlikely to be accepted; write the sign first.

2. (a) nitrogen(I) oxide [1]; potassium sulfate(VI) [1] (b) Ca(ClO)₂ [1]; LiNO₂ [1]; Al₂(SO₄)₃ [1] Examiner insight: Check each formula by adding up the charges; brackets are needed whenever more than one polyatomic ion is present.

3. (a) Loss of electrons [1]; increase in oxidation number [1] (b) A species that gains electrons (and is itself reduced) [1] (c) The same element in a single species is simultaneously oxidised and reduced [1] Examiner insight: A disproportionation definition that leaves out “single species” or “same element” describes ordinary redox and is unlikely to score.

4. (a) W: no changes (Ba +2, C +4, O −2 throughout), not redox [1]. X: N +5 → +3 and O −2 → 0, redox [1]. Y: N +3 → +5 in HNO₃ and +3 → +2 in NO, disproportionation [1]. Z: Na +1, O −2, H +1 throughout, not redox [1]. (b) Two different elements change (nitrogen is reduced, oxygen is oxidised); disproportionation needs one element to go both up and down [1]. Examiner insight: Each classification needs supporting oxidation numbers; a bare label with no numbers gives the examiner nothing to credit.

5. (a) +5 to −3 (a decrease of 8) [1] (b) NO₃⁻ to NH₄⁺, with 10H⁺ and 3H₂O [1]; 8e⁻ giving charge balance: NO₃⁻ + 10H⁺ + 8e⁻ → NH₄⁺ + 3H₂O [1] (c) Zn → Zn²⁺ + 2e⁻ [1] (d) Zinc half-equation × 4 [1]; 4Zn + NO₃⁻ + 10H⁺ → 4Zn²⁺ + NH₄⁺ + 3H₂O [1] Examiner insight: The electron count in (b) must equal the fall in oxidation number; checking that 8 = 5 − (−3) catches most errors.

6. (a) SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻ [1] (b) 2 × 0.00480 = 0.00960 mol [1] (c) 0.00960 ÷ 0.00192 = 5 [1] (d) +5 − 5 = 0 [1]; the product is bromine, Br₂ [1] (e) Correct species and ratio 2BrO₃⁻ : 5SO₃²⁻ [1]; 2BrO₃⁻ + 5SO₃²⁻ + 2H⁺ → Br₂ + 5SO₄²⁻ + H₂O [1] Examiner insight: Set out each step of a ratio deduction; a correct product stated without the electron arithmetic leaves the “deduce” command unanswered.

7. (a) PH₃: −3 [1]; H₂PO₂⁻: +1 [1] (b) Phosphorus in one species, P₄ (0), is reduced to −3 [1] and oxidised to +1 at the same time [1]. Examiner insight: Quote the starting oxidation number of 0 for P₄; the explanation needs both directions from that value.

8. (a) Each K atom loses one electron, 0 → +1, so potassium is oxidised [1]. Each S atom gains two electrons, 0 → −2, so sulfur is reduced [1]. Two K atoms supply the two electrons one S atom gains [1]. (b) Metal atoms have few outer-shell electrons, which they lose to form positive ions [1]; non-metal atoms have nearly full outer shells and gain electrons to form negative ions [1]. Examiner insight: Link the electron change to the direction of the oxidation number change for each element; one element alone gives an incomplete answer.

9. (a) NaH: −1 [1]; H₂O: +1 [1] (b) Oxidising agent: H₂O (H +1 → 0) [1]; reducing agent: NaH (H −1 → 0) [1] (c) The hydrogen that changes comes from two different species, so it is not one species being both oxidised and reduced [1]. (d) 2H⁻ → H₂ + 2e⁻ [1]; 2H₂O + 2e⁻ → H₂ + 2OH⁻ [1] (e) n(NaH) = 0.300 ÷ 24.0 = 0.0125 mol [1]; mass NaOH = 0.0125 × 40.0 = 0.500 g [1] Examiner insight: Questions like (c) test the “single species” part of the definition; quoting the formulae of both species makes the point clear.

10. (a) Sr(s) + 2H₂O(l) → Sr²⁺(aq) + 2OH⁻(aq) + H₂(g): correct balanced species [1]; state symbols [1] (b) Sr → Sr²⁺ + 2e⁻ [1]; 2H₂O + 2e⁻ → H₂ + 2OH⁻ [1] (c) Water is the oxidising agent [1], because H goes from +1 to 0 (reduced), while Sr goes from 0 to +2 [1]. (d) n(Sr) = 0.657 ÷ 87.6 = 0.00750 mol [1]; n(H₂) = 0.00750 mol (1 : 1) [1]; V = 0.00750 × 24.0 = 0.180 dm³ = 180 cm³ [1] Examiner insight: The question asks for cm³; an answer left as 0.180 with no unit conversion does not answer what was asked.

11. (a) Br₂ + 2e⁻ → 2Br⁻ [1]; SO₂ + 2H₂O → SO₄²⁻ + 4H⁺ + 2e⁻ [1] (b) Electrons already equal, add and cancel [1]: Br₂ + SO₂ + 2H₂O → 2Br⁻ + SO₄²⁻ + 4H⁺ [1] (c) Br 0 → −1 and S +4 → +6 [1]; Br₂ is the oxidising agent [1] (d) n(Br₂) = 4.00 ÷ 159.8 = 0.02503 mol [1]; n(SO₂) = 0.02503 mol (1 : 1) [1]; mass = 0.02503 × 64.1 = 1.60 g [1] Examiner insight: Use the molar mass of Br₂ (159.8), not Br (79.9); halving or doubling the amount here is a common slip.

Where marks are usually lost

• Writing oxidation numbers without a sign, or as ionic charges.
• Missing the hydride exception in NaH or NaAlH₄, or the peroxide exception in Rb₂O₂.
• Leaving brackets out of formulae such as Ca(ClO)₂ and Al₂(SO₄)₃.
• Calling any reaction with only one reactant a disproportionation without checking which elements change.
• Writing “the oxidising agent is oxidised”.
• Balancing atoms but not charge in a half-equation.
• Combining half-equations before equalising electrons, or leaving electrons in the final equation.
• Using the molar mass of a halogen atom instead of the diatomic molecule.
• Forgetting to convert dm³ to cm³ when the question asks for cm³.

Next steps

• Redox revision notes
• Redox study guide
• Edexcel International A Level Chemistry hub
• Printable Edexcel Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 2, Topic 8A: Redox chemistry, outcomes 8.1 to 8.10.