IB DP Chemistry Reactivity 3.2: Electron transfer: redox, cells and electrolysis – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This set practises Reactivity 3.2, Electron transfer reactions, with the February 2023 IB Diploma Programme Chemistry guide (first assessment 2025) as the checklist. Questions follow understandings Reactivity 3.2.1 to 3.2.11 in order (question 12 joins 3.2.10 and 3.2.11); all are SL and HL.

New to this content? Begin with the study guide; to refresh it quickly, read the revision notes. For other units, use the printable checklist or IB DP chemistry hub.

Where needed, use Ar(H) = 1.01, Ar(C) = 12.01, Ar(O) = 16.00, Ar(Mg) = 24.31, and a molar gas volume at STP of 22.7 dm³ mol⁻¹.

Questions

1. Deduce the oxidation state of:

(a) phosphorus in P₄O₁₀ [1] (b) oxygen in Na₂O₂ [1] (c) chlorine in HClO₄ [1] (d) Name Fe₂(SO₄)₃ and PbO₂, including oxidation numbers. [2]

2. Copper(II) oxide is reduced when heated in ammonia gas:

3CuO(s) + 2NH₃(g) → 3Cu(s) + N₂(g) + 3H₂O(g)

(a) Give the change in oxidation state of copper and of nitrogen. [2] (b) Identify the oxidizing agent and the reducing agent. [2]

3. In acidic solution, bromate(V) ions, BrO₃⁻, oxidize iron(II) ions to iron(III) ions and are reduced to bromine, Br₂.

(a) Deduce the reduction half-equation. [2] (b) Write the oxidation half-equation. [1] (c) Deduce the overall ionic equation. [2]

4. Aluminium foil is placed in aqueous nickel(II) sulfate and becomes coated with nickel. Write the two half-equations and the overall ionic equation. [3]

5. Four metals were each placed in solutions of the other three metals’ 2+ ions. The results are shown (✓ = metal deposit forms, ✗ = no change).

Metal	Mn²⁺(aq)	Fe²⁺(aq)	Sn²⁺(aq)	Cu²⁺(aq)
Mn	–	✓	✓	✓
Fe	✗	–	✓	✓
Sn	✗	✗	–	✓
Cu	✗	✗	✗	–

(a) List the metals in order of decreasing ease of oxidation, and explain how the data show this. [2] (b) Write the ionic equation for the reaction of iron with Sn²⁺(aq), and identify the reducing agent. [2]

6. Bromine water is added separately to aqueous potassium chloride and aqueous potassium iodide.

(a) Describe what is seen in each case. [2] (b) Give an ionic equation for the one reaction that takes place. [1] (c) Explain why bromine is reduced less easily than chlorine. [2]

7. A 0.365 g piece of magnesium ribbon reacts completely with excess dilute hydrochloric acid.

(a) Give a balanced equation including state symbols. [1] (b) Write the ionic equation. [1] (c) Calculate the volume of hydrogen formed at STP, in cm³. [3]

8. A voltaic cell is built from a magnesium half-cell and a tin half-cell, joined by a wire and a salt bridge of potassium nitrate. Magnesium is oxidized more easily than tin.

(a) Identify the anode and give its sign. [2] (b) Give the anode and cathode half-equations. [2] (c) State the direction of electron flow in the external circuit. [1] (d) Describe the movement of ions in the salt bridge and explain why it is needed. [2]

9. A nickel–metal hydride cell is rechargeable. During discharge:

• negative electrode: MH + OH⁻ → M + H₂O + e⁻ (M is a metal alloy that stores hydrogen)
• positive electrode: NiO(OH) + H₂O + e⁻ → Ni(OH)₂ + OH⁻

(a) Deduce the reaction at each electrode during charging. [2] (b) State which electrode acts as the anode during charging. [1] (c) Deduce the overall equation for discharge. [1] (d) Give one advantage of this cell over a primary cell, and one disadvantage of a hydrogen fuel cell compared with this cell. [2]

10. Molten strontium bromide is electrolysed using graphite electrodes.

(a) Explain why solid strontium bromide does not conduct but molten strontium bromide does. [2] (b) For each electrode, give its name and its half-equation. [2] (c) State how current is carried in the external circuit and in the electrolyte. [1] (d) State the energy change in this cell, and how it differs from a voltaic cell. [1]

11. 2,2-Dimethylpropan-1-ol, (CH₃)₃CCH₂OH, is a primary alcohol. It can be oxidized to an aldehyde or to a carboxylic acid.

(a) Describe the set-up used to obtain the aldehyde, and explain why it works. [2] (b) Using [O], give the equation in which the aldehyde forms. [1] (c) Using [O], give one equation for the conversion of the alcohol straight to the carboxylic acid. [1] (d) A student oxidizes 4.41 g of the alcohol under reflux and obtains 3.83 g of 2,2-dimethylpropanoic acid. Calculate the percentage yield. [3] (e) Name the product when heptan-4-ol is oxidized, and explain why 2-methylpropan-2-ol is not oxidized under the same conditions. [2]

12. This question is about reduction reactions.

(a) Write two equations, using [H], for the reduction of propanoic acid, CH₃CH₂COOH, to propan-1-ol. [2] (b) Name the product of reducing hexan-2-one, and write the equation using [H]. [2] (c) State the role of the hydride ion in these reductions. [1] (d) Hex-1-yne, CH≡CCH₂CH₂CH₂CH₃, is fully hydrogenated over a nickel catalyst. Calculate the volume of hydrogen at STP that reacts with 4.11 g of hex-1-yne. [3]

Answers

1. (a) 4x + 10(−2) = 0, so +5 [1] (b) 2(+1) + 2x = 0, so −1 (a peroxide) [1] (c) +1 + x + 4(−2) = 0, so +7 [1] (d) iron(III) sulfate [1]; lead(IV) oxide [1] Examiner insight: Write out the sum you set up, so that a slip in the arithmetic is easy to spot when you check.

2. (a) Copper: +2 → 0 [1]; nitrogen: −3 → 0 [1] (b) Oxidizing agent: CuO (or Cu²⁺) [1]; reducing agent: NH₃ [1] Examiner insight: Name the whole species, NH₃, as the agent; “nitrogen” alone is ambiguous because N₂ is a product.

3. (a) 2BrO₃⁻ → Br₂ with 6H₂O added to the right and 12H⁺ to the left [1]; 2BrO₃⁻ + 12H⁺ + 10e⁻ → Br₂ + 6H₂O [1] (b) Fe²⁺ → Fe³⁺ + e⁻ [1] (c) Multiply (b) by 10 to match electrons [1]; 2BrO₃⁻ + 12H⁺ + 10Fe²⁺ → Br₂ + 6H₂O + 10Fe³⁺ (charge +30 on each side) [1] Examiner insight: Check charge as well as atoms on both sides of the final equation; an equation that balances atoms only is not fully balanced.

4. Al → Al³⁺ + 3e⁻ [1]; Ni²⁺ + 2e⁻ → Ni [1]; 2Al + 3Ni²⁺ → 2Al³⁺ + 3Ni [1] Examiner insight: The six electrons must cancel exactly; writing Al + Ni²⁺ → Al³⁺ + Ni leaves the charges unbalanced.

5. (a) Mn > Fe > Sn > Cu [1]; Mn reduces all three other ions and Cu none: the more ions a metal reduces, the more easily it is oxidized [1] (b) Fe + Sn²⁺ → Fe²⁺ + Sn [1]; reducing agent: Fe [1] Examiner insight: An explanation that refers to the data (“reacts with three solutions”) earns more than a memorized reactivity series.

6. (a) With KCl(aq): no change, the solution stays yellow-orange [1]; with KI(aq): the solution turns brown [1] (b) Br₂ + 2I⁻ → 2Br⁻ + I₂ [1] (c) A bromine atom is larger, so the incoming electron is further from the nucleus [1]; with more shielding, it is attracted less strongly, so bromine gains an electron less readily [1] Examiner insight: Refer to the incoming electron being attracted to the nucleus; “bromine is less reactive” restates the question.

7. (a) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) [1] (b) Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g) [1] (c) n(Mg) = 0.365 ÷ 24.31 = 0.01501 mol [1]; n(H₂) = n(Mg) = 0.01501 mol [1]; V = 0.01501 × 22.7 = 0.341 dm³ = 341 cm³ [1] Examiner insight: Check the unit asked for: cm³ here, not dm³.

8. (a) Magnesium is the anode [1]; it is negative [1] (b) Anode: Mg → Mg²⁺ + 2e⁻ [1]; cathode: Sn²⁺ + 2e⁻ → Sn [1] (c) From magnesium to tin through the wire [1] (d) NO₃⁻ moves toward the magnesium half-cell and K⁺ toward the tin half-cell [1]; this keeps both solutions neutral as Mg²⁺ forms and Sn²⁺ is removed, and completes the circuit [1] Examiner insight: State that ions, not electrons, move in the salt bridge; answers suggesting electrons cross it lose credit.

9. (a) Negative electrode: M + H₂O + e⁻ → MH + OH⁻ [1]; positive electrode: Ni(OH)₂ + OH⁻ → NiO(OH) + H₂O + e⁻ [1] (b) The positive electrode (oxidation happens there on charging) [1] (c) MH + NiO(OH) → M + Ni(OH)₂ [1] (d) Advantage: it can be recharged many times, so less waste and lower long-term cost [1]; fuel cell disadvantage: hydrogen is hard to store or catalysts are expensive [1] Examiner insight: A charging reaction should be the exact reverse of the discharge reaction, with every species and the electron moved across the arrow.

10. (a) Solid: the ions sit in a rigid lattice and cannot move [1]; molten: ions are free to move and carry charge [1] (b) Cathode: Sr²⁺ + 2e⁻ → Sr [1]; anode: 2Br⁻ → Br₂ + 2e⁻ [1] (c) Electrons in the external circuit and ions in the electrolyte [1] (d) Electrical to chemical energy (non-spontaneous reaction); a voltaic cell does the reverse [1] Examiner insight: “Free electrons in the melt” is wrong for an ionic compound; the explanation needs mobile ions.

11. (a) Heat with the oxidizing agent and distil off the aldehyde as it forms [1]; with no hydrogen bonding between its molecules it boils below the alcohol, so it leaves before further oxidation [1] (b) (CH₃)₃CCH₂OH + [O] → (CH₃)₃CCHO + H₂O [1] (c) (CH₃)₃CCH₂OH + 2[O] → (CH₃)₃CCOOH + H₂O [1] (d) M(C₅H₁₂O) = 88.17 g mol⁻¹, so n = 4.41 ÷ 88.17 = 0.05002 mol [1]; M(acid) = 102.15 g mol⁻¹, so theoretical mass = 0.05002 × 102.15 = 5.109 g [1]; yield = 3.83 ÷ 5.109 × 100 = 75.0% [1] (e) Heptan-4-one [1]; in 2-methylpropan-2-ol the carbon bonded to OH carries no hydrogen atom, so it cannot be oxidized this way [1] Examiner insight: Keep unrounded intermediate values until the last line; rounding the moles early can move the final yield.

12. (a) CH₃CH₂COOH + 2[H] → CH₃CH₂CHO + H₂O [1]; CH₃CH₂CHO + 2[H] → CH₃CH₂CH₂OH [1] (b) Hexan-2-ol [1]; CH₃COC₄H₉ + 2[H] → CH₃CH(OH)C₄H₉ [1] (c) It acts as a source of hydrogen with two electrons, adding to the δ+ carbonyl carbon (it is the reducing species) [1] (d) M(C₆H₁₀) = 82.16 g mol⁻¹, so n = 4.11 ÷ 82.16 = 0.05002 mol [1]; two H₂ per molecule, so n(H₂) = 0.1000 mol [1]; V = 0.1000 × 22.7 = 2.27 dm³ [1] Examiner insight: A triple bond takes two H₂ to reach the alkane; using a 1 : 1 ratio halves the answer.

Where marks are usually lost

• Giving oxidation states without a sign, or as a total for several atoms.
• Calling the species that is oxidized the oxidizing agent.
• Balancing atoms but not charge in redox equations.
• Saying electrons move through the salt bridge or the molten electrolyte.
• Writing a charging reaction that is not the full reverse of discharge.
• Using reflux when the aldehyde is the target product.
• Using one mole of H₂ per mole of alkyne when the alkane is formed.

Next steps

• Reactivity 3.2 revision notes for a final read-through
• Reactivity 3.2 study guide for full explanations
• Course hub for IB DP chemistry
• Printable topic checklist
• Take all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Reactivity 3.2 (Electron transfer reactions), understandings 3.2.1 to 3.2.11 for SL and HL, practised from the Chemistry guide of the International Baccalaureate Diploma Programme, February 2023, first assessment 2025.