The problems are on the redox reactions worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Redox reactions transfer electrons: oxidation is loss of electrons and an increase in oxidation number, reduction is gain of electrons and a decrease. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on redox reactions, from oxidation numbers of common species to oxidation numbers in electron count: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the oxidation numbers or half-equations first, then the combined equation or calculation, with the final result in bold. Check the charge balance of your own equations against these worked solutions.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R3.2
Problem 1 Oxidation numbers of common species
1. 2(+1) + x + 4(−2) = 0, so x = +6.2. (+1) + x + 4(−2) = 0, so x = +7.
3. x + 4(+1) = +1, so x = −3.
4. 2(+1) + 2x = 0, so x = −1 (peroxide).
5. x + 3(−2) = −1, so x = +5.
Problem 2 Sparkler chemistry
1. Mg: 0 → +2; O: 0 → −2.2. Magnesium is oxidised (0 → +2); oxygen is reduced (0 → −2).
3. Oxidising agent: O₂. Reducing agent: Mg.
Problem 3 Zinc coating on a nail
1. Zn → Zn²⁺ + 2e⁻.2. Oxidation: zinc loses electrons and its oxidation number rises from 0 to +2.
3. 3 mol × 2 = 6 mol of electrons.
Problem 4 Redox or not?
1. Not redox: no element changes its oxidation number (Ca +2, C +4, O −2, H +1, Cl −1 throughout).2. Redox: I goes −1 → 0 (oxidised) and Cl goes 0 → −1 (reduced).
3. Not redox: Ag stays +1, Cl stays −1, N stays +5. It is precipitation.
Problem 5 Roman numerals in names
1. Iron has +3 (2x = +6): iron(III) oxide.2. Copper has +2 (sulfate is 2−): copper(II) sulfate.
3. Cr³⁺ and O²⁻ combine as 2 : 3: Cr₂O₃.
4. Tin has +2: tin(II) chloride.
Problem 6 Harder oxidation numbers
1. 2x + 7(−2) = −2, so x = +6.2. 2x + 3(−2) = −2, so x = +2.
3. 2x + 6(+1) + (−2) = 0, so 2x = −4 and x = −2 (average).
4. 3x + 4(−2) = 0, so x = +8/3 (a mixture of Fe²⁺ and Fe³⁺).
Problem 7 Balancing two half-equations in acid
1. N: +5 → +2, so 3 electrons gained. NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O.2. H₂O₂ → O₂ + 2H⁺ + 2e⁻.
Check 1: charge left −1 + 4 − 3 = 0, right 0. Check 2: charge left 0, right +2 − 2 = 0. Final answers: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O and H₂O₂ → O₂ + 2H⁺ + 2e⁻.
Problem 8 Rhubarb leaves and permanganate
1. Lowest common multiple of 5 and 2 is 10: multiply the first by 2 and the second by 5.2. 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂.
3. Left: −2 + 16 − 10 = +4. Right: +4. The charges balance.
Problem 9 Thermite reaction
1. Al: 0 → +3. Fe: +3 → 0.2. Oxidising agent: Fe₂O₃ (Fe is reduced). Reducing agent: Al (oxidised).
3. Two Fe³⁺ each gain 3 electrons: 6 electrons.
Problem 10 Silver recovery from waste
1. Oxidation: Cu → Cu²⁺ + 2e⁻. Reduction: Ag⁺ + e⁻ → Ag.2. n(Cu) = 0.635 ÷ 63.5 = 0.0100 mol.
3. n(Ag) = 2 × 0.0100 = 0.0200 mol; m = 0.0200 × 107.9 = 2.16 g.
Problem 11 Making bleach
1. Cl₂: 0; NaCl: −1; NaClO: +1.2. The same element, chlorine, is both reduced (0 → −1) and oxidised (0 → +1).
3. One chlorine atom gains 1 electron and the other loses 1, so 1 electron is transferred per Cl₂.
Problem 12 Balancing by oxidation numbers
1. Cu: 0 → +2 (+2). N: +5 → +2 (−3).2. 3 Cu atoms (+6) balance 2 N atoms (−6).
3. 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O. Check N: 8 = 6 + 2; H: 8 = 8; O: 24 = 18 + 2 + 4.
Problem 13 Permanganate in alkali
1. MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O.2. Add 4OH⁻ to both sides; 4H⁺ + 4OH⁻ becomes 4H₂O: MnO₄⁻ + 4H₂O + 3e⁻ → MnO₂ + 2H₂O + 4OH⁻, which simplifies to MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻.
3. Charge: −1 − 3 = −4 on the left, −4 on the right. O: 4 + 2 = 6 and 2 + 4 = 6. H: 4 = 4.
Problem 14 Reading oxidation numbers from colours
1. MnO₄⁻ +7, MnO₄²⁻ +6, MnO₂ +4, Mn²⁺ +2.2. +7 → +4: 3 electrons. +7 → +2: 5 electrons.
3. Purple becomes (almost) colourless or very pale pink as Mn²⁺ forms.
Problem 15 Iron supplement tablet
1. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺.2. n(MnO₄⁻) = 0.02240 × 0.00200 = 4.48 × 10⁻⁵ mol.
3. n(Fe²⁺) in 25.0 cm³ = 5 × 4.48 × 10⁻⁵ = 2.24 × 10⁻⁴ mol; in 250 cm³ = 10 × that = 2.24 × 10⁻³ mol.
4. m = 2.24 × 10⁻³ × 55.8 = 0.1250 g = 125 mg.
Problem 16 Breathalyser chemistry
1. Cr: +6 → +3, a gain of 3 per Cr, 6 per Cr₂O₇²⁻. The carbon in CH₂OH goes from −1 to +3, a loss of 4.2. LCM of 6 and 4 is 12: dichromate half × 2, ethanol half × 3. 2Cr₂O₇²⁻ + 16H⁺ + 3C₂H₅OH → 4Cr³⁺ + 11H₂O + 3CH₃COOH (28H⁺ − 12H⁺ = 16H⁺ on the left; 14H₂O − 3H₂O = 11H₂O on the right).
3. H: 16 + 18 = 34 and 22 + 12 = 34. O: 14 + 3 = 17 and 11 + 6 = 17. Charge: −4 + 16 = +12 and +12.
Problem 17 Analysing bleach with thiosulfate
1. n(S₂O₃²⁻) = 0.02150 × 0.100 = 2.150 × 10⁻³ mol; n(I₂) = 1.075 × 10⁻³ mol.2. ClO⁻ : I₂ is 1 : 1, so 1.075 × 10⁻³ mol in 25.0 cm³; in the original 10.0 cm³ (×10) = 1.075 × 10⁻² mol.
3. c = 1.075 × 10⁻² ÷ 0.0100 = 1.08 mol/dm³; mass = 1.075 × 74.4 = 80.0 g/dm³.
Problem 18 Spot the error in a redox equation
1. Left: −1 + 8 + 2 = +9. Right: +2 + 3 = +5. The charges differ.2. Mn gains 5 electrons but each Fe²⁺ loses only 1, so five iron ions are needed: 5.
3. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Charge: −1 + 8 + 10 = +17 and +2 + 15 = +17.
Problem 19 Iodine in table salt
1. Atoms: I 1 + 5 = 6 and 3 × 2 = 6; O 3 and 3; H 6 and 6. Charge: −1 − 5 + 6 = 0 and 0.2. n(S₂O₃²⁻) = 0.01260 × 0.00500 = 6.30 × 10⁻⁵ mol; n(I₂) = 3.15 × 10⁻⁵ mol; n(IO₃⁻) = 1.05 × 10⁻⁵ mol.
3. m(I) = 1.05 × 10⁻⁵ × 126.9 = 1.33 × 10⁻³ g = 1.33 mg per 50.0 g, so × 20 = 26.6 mg/kg.
Problem 20 Oxidation numbers in electron count
1. Left: −2 + 14 − 6 = +6. Right: 2 × 3 = +6. Balanced.2. Cr: 2 × (+6 → +3) = −6. I: 6 × (−1 → 0) = +6. The totals are equal and opposite.
3. n(Cr₂O₇²⁻) = 0.0250 × 0.0200 = 5.00 × 10⁻⁴ mol; n(I₂) = 3 × 5.00 × 10⁻⁴ = 1.50 × 10⁻³ mol; m = 1.50 × 10⁻³ × 253.8 = 0.381 g.
