These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. All reactions take place in aqueous solution. Unless a question says otherwise, the solution is acidic, so you may use H₂O and H⁺ to balance. The titration data in Question 5 are invented for practice.
Question 1 (multiple choice · foundation)
In acidic solution, chlorate ions, ClO₃⁻, are reduced to chloride ions, Cl⁻. Which half-reaction is correctly balanced?
- (A) ClO₃⁻ + 6H⁺ + 5e⁻ → Cl⁻ + 3H₂O
- (B) ClO₃⁻ + 3H⁺ + 6e⁻ → Cl⁻ + 3H₂O
- (C) ClO₃⁻ + 6H⁺ + 6e⁻ → Cl⁻ + 3H₂O
- (D) ClO₃⁻ + 6H⁺ → Cl⁻ + 3H₂O + 6e⁻
Answer and explanation
Answer: (C). Three O atoms need 3H₂O on the right. That puts 6 H on the right, so add 6H⁺ on the left. Charge: left −1 + 6 = +5, right −1, so add 6e⁻ to the left. Final charge: −1 on both sides. Check: Cl goes from +5 to −1, a gain of 6 electrons.
- (A) balances the atoms, but the charge is 0 on the left and −1 on the right. It uses 5 electrons, perhaps from the +5 oxidation number, but the chlorine ends at −1, not 0.
- (B) has only 3 H on the left and 6 H on the right, so hydrogen is not balanced (charge is also wrong: −4 against −1).
- (D) puts the electrons on the product side. That describes an oxidation, but Cl is reduced here. The charges (+5 and −7) do not match either.
Question 2 (multiple choice · core)
Solutions of iron(III) chloride are used to dissolve unwanted copper from circuit boards. The net ionic equation is:
Cu(s) + 2Fe³⁺(aq) → Cu²⁺(aq) + 2Fe²⁺(aq)
Which pair of half-reactions combines to give this equation?
- (A) Cu → Cu²⁺ + 2e⁻ and Fe³⁺ + e⁻ → Fe²⁺
- (B) Cu²⁺ + 2e⁻ → Cu and Fe²⁺ → Fe³⁺ + e⁻
- (C) Cu → Cu²⁺ + e⁻ and Fe³⁺ + e⁻ → Fe²⁺
- (D) Cu → Cu²⁺ + 2e⁻ and Fe³⁺ + 2e⁻ → Fe²⁺
Answer and explanation
Answer: (A). Copper’s oxidation number rises from 0 to +2 (oxidized, 2e⁻ lost). Iron’s falls from +3 to +2 (reduced, 1e⁻ gained per ion). Doubling the iron half-reaction gives 2e⁻ gained, which matches the 2e⁻ lost, and adding gives the equation shown.
- (B) is both half-reactions written backwards. Adding them gives the reverse reaction.
- (C) is not charge balanced: Cu → Cu²⁺ + e⁻ has charge 0 on the left and +1 on the right.
- (D) is not charge balanced: Fe³⁺ + 2e⁻ has charge +1, but Fe²⁺ has +2.
Question 3 (multiple choice · core)
Aluminium metal reacts with dilute hydrochloric acid to form Al³⁺ ions and hydrogen gas. Which net ionic equation is fully balanced?
- (A) Al(s) + 2H⁺(aq) → Al³⁺(aq) + H₂(g)
- (B) Al(s) + 3H⁺(aq) → Al³⁺(aq) + H₂(g)
- (C) 2Al(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂(g)
- (D) 2Al(s) + 3H⁺(aq) → 2Al³⁺(aq) + 3H₂(g)
Answer and explanation
Answer: (C). Half-reactions: Al → Al³⁺ + 3e⁻ and 2H⁺ + 2e⁻ → H₂. The lowest common multiple of 3 and 2 is 6, so multiply by 2 and 3. Then 2Al + 6H⁺ → 2Al³⁺ + 3H₂. Atoms: 2 Al and 6 H on each side. Charge: +6 on each side. Six electrons are transferred.
- (A) balances the atoms but not the charge (+2 against +3). This is the “looks balanced” trap.
- (B) balances the charge (+3 each side) but not hydrogen (3 H against 2 H).
- (D) balances aluminium but neither hydrogen (3 against 6) nor charge (+3 against +6).
Question 4 (multiple choice · core)
In acidic solution, hydrogen peroxide, H₂O₂, can be oxidized to oxygen gas. Which half-reaction represents this change?
- (A) H₂O₂ + 2e⁻ → O₂ + 2H⁺
- (B) H₂O₂ → O₂ + 2H⁺ + e⁻
- (C) H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O
- (D) H₂O₂ → O₂ + 2H⁺ + 2e⁻
Answer and explanation
Answer: (D). Oxygen goes from −1 in H₂O₂ to 0 in O₂. Two O atoms each rise by 1, so 2 electrons are lost, and they appear as products. Charge: 0 on the left; +2 − 2 = 0 on the right.
- (A) puts the electrons on the left, which would mean electrons are gained. The charge (−2 against +2) does not balance.
- (B) has the right idea but only one electron, so the charge is 0 against +1.
- (C) is a correctly balanced half-reaction, but it is the reduction of H₂O₂ to water, not its oxidation to O₂.
Question 5 (constructed response · core)
A student analyses a solution containing tin(II) ions, Sn²⁺, by titration with acidified potassium permanganate. Permanganate ions, MnO₄⁻, are reduced to Mn²⁺, and Sn²⁺ ions are oxidized to Sn⁴⁺. A 25.00 mL sample of the tin(II) solution needs 18.40 mL of 0.0200 M KMnO₄ to reach the end point.
(a) Write the balanced half-reaction for each process. (b) Write the balanced net ionic equation for the overall reaction. (c) Calculate the concentration of Sn²⁺ in the sample.
Worked solution and suggested Marlbridge rubric
(a) Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O Oxidation: Sn²⁺ → Sn⁴⁺ + 2e⁻
(b) The lowest common multiple of 5 and 2 is 10. Multiply the reduction by 2 and the oxidation by 5, then add:
2MnO₄⁻(aq) + 16H⁺(aq) + 5Sn²⁺(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 5Sn⁴⁺(aq)
Check: charge left −2 + 16 + 10 = +24; right +4 + 20 = +24. K⁺ is a spectator and is left out.
(c)
- n(MnO₄⁻) = 0.0200 mol L⁻¹ × 0.01840 L = 3.68 × 10⁻⁴ mol
- Mole ratio Sn²⁺ : MnO₄⁻ = 5 : 2, so n(Sn²⁺) = 3.68 × 10⁻⁴ × 5/2 = 9.20 × 10⁻⁴ mol
- [Sn²⁺] = 9.20 × 10⁻⁴ mol ÷ 0.02500 L = 0.0368 M
| Point | What earns it |
|---|---|
| 1 | Correctly balanced permanganate half-reaction, with 5e⁻ on the reactant side |
| 1 | Correctly balanced tin half-reaction, with 2e⁻ on the product side |
| 1 | Balanced overall equation with no electrons, using the 2 : 5 multipliers |
| 1 | Moles of MnO₄⁻ from concentration × volume in litres |
| 1 | 0.0368 M, using the 5 : 2 ratio from the student’s own equation in (b) |
Accept an equation from (b) carried forward into (c). Common errors: a 1 : 1 ratio gives 0.0147 M; using the ratio upside down (2/5) gives 0.00589 M.
Question 6 (constructed response · core)
When zinc metal is added to an acidic yellow solution containing VO₂⁺ ions, the solution turns blue as VO²⁺ ions form. Zinc dissolves as Zn²⁺.
(a) Give the oxidation number of vanadium in VO₂⁺ and in VO²⁺. (b) Write the two balanced half-reactions. (c) Write the balanced net ionic equation. (d) A student writes: VO₂⁺ + Zn + 2H⁺ → VO²⁺ + Zn²⁺ + H₂O. The student says it is balanced “because every atom is the same on both sides”. Evaluate the claim.
Worked solution and suggested Marlbridge rubric
(a) In VO₂⁺: V + 2(−2) = +1, so V = +5. In VO²⁺: V + (−2) = +2, so V = +4.
(b) Reduction: VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O (charge +2 on both sides) Oxidation: Zn → Zn²⁺ + 2e⁻
(c) Multiply the reduction by 2 so that 2e⁻ are gained and 2e⁻ are lost:
2VO₂⁺(aq) + 4H⁺(aq) + Zn(s) → 2VO²⁺(aq) + 2H₂O(l) + Zn²⁺(aq)
Check: charge +6 on both sides; V 2, O 4, H 4, Zn 1 on each side.
(d) The claim is incorrect. The atoms do balance, but the charge does not: the left side is +1 + 0 + 2 = +3 and the right side is +2 + 2 = +4. In electron terms, each Zn atom loses 2 electrons, while each VO₂⁺ ion gains only 1. Electrons lost must equal electrons gained, so two VO₂⁺ ions are needed for every Zn atom.
| Point | What earns it |
|---|---|
| 1 | Both oxidation numbers, +5 and +4 |
| 1 | Balanced vanadium half-reaction with H⁺, H₂O and 1e⁻ |
| 1 | Balanced zinc half-reaction |
| 1 | Balanced overall equation with 2 : 1 ratio and no electrons |
| 1 | States the claim is wrong and gives the charge mismatch or the unequal electron transfer as the reason |
Do not award the point in (d) for “the claim is wrong” without a reason based on charge or electrons.
Question 7 (constructed response · stretch)
Extension question: it uses the basic-solution method from the extension section of the study guide.
In basic solution, permanganate ions, MnO₄⁻, oxidize sulfite ions, SO₃²⁻, to sulfate ions, SO₄²⁻. The manganese forms solid manganese(IV) oxide, MnO₂. Write the balanced net ionic equation. Show your half-reactions.
Worked solution and suggested Marlbridge rubric
Reduction. Mn goes from +7 to +4: 3 electrons gained. In acid: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O Add 4OH⁻ to both sides and combine 4H⁺ + 4OH⁻ into 4H₂O, then cancel 2H₂O: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (charge −4 on both sides)
Oxidation. S goes from +4 to +6: 2 electrons lost. In acid: SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻ Add 2OH⁻ to both sides, combine, then cancel H₂O: SO₃²⁻ + 2OH⁻ → SO₄²⁻ + H₂O + 2e⁻ (charge −4 on both sides)
Combine. The lowest common multiple of 3 and 2 is 6. Multiply the reduction by 2 and the oxidation by 3, add, and cancel 6e⁻, 6OH⁻ and 3H₂O:
2MnO₄⁻(aq) + H₂O(l) + 3SO₃²⁻(aq) → 2MnO₂(s) + 2OH⁻(aq) + 3SO₄²⁻(aq)
Check: Mn 2, S 3, O 18, H 2 on each side; charge −8 on each side.
| Point | What earns it |
|---|---|
| 1 | Correct electron counts: 3 per Mn and 2 per S, shown in half-reactions |
| 1 | Both half-reactions balanced for atoms and charge (in acid or in base) |
| 1 | Correct 2 : 3 multipliers so that 6 electrons cancel |
| 1 | Final equation in basic form (OH⁻, no H⁺), fully simplified |
Alternative method: balance the whole equation in acid first (2MnO₄⁻ + 2H⁺ + 3SO₃²⁻ → 2MnO₂ + H₂O + 3SO₄²⁻), then add 2OH⁻ to both sides and simplify. This earns full credit. An answer left with H⁺ in it loses only the last point.
How did you do?
- Q1 or Q4 wrong: re-read “Step by step: permanganate to manganese(II)” in the study guide and practise checking charge.
- Q2, Q3 or Q6 wrong: go back to Worked example 1 and the rule that electrons lost must equal electrons gained.
- Q5 wrong: check your balanced equation first, then the mole ratio, as in “Using the balanced equation as a mole ratio”.
- Q7 incomplete: try the basic-solution extension in the study guide, one half-reaction at a time.
Then tick off the topic checklist.