Exam-style Question 6

Question (4 marks): Sulfur dioxide decolourises bromine water:SO₂ (aq) + Br₂ (aq) + 2 H₂O (l) → SO₄²⁻ (aq) + 4 H⁺ (aq) + 2 Br⁻ (aq)(a) Use oxidation numbers to show which element is oxidised and which is reduced. (b) Write the half-equation for the sulfur species. (c) Name the ion SO₃²⁻, which SO₂ forms in alkali.Answer (a): In SO₂, x + 2(−2) = 0, so sulfur is +4. In SO₄²⁻, x + 4(−2) = −2, so sulfur is +6. Sulfur is oxidised. Bromine goes from 0 in Br₂ to −1 in Br⁻, so bromine is reduced.Answer (b): Balance O with water, H with H⁺, then charge with electrons:SO₂ + 2 H₂O → SO₄²⁻ + 4 H⁺ + 2 e⁻Charge check: left 0, right (−2) + (+4) + (−2) = 0. The 2 electrons match the rise of 2 in oxidation number. The bromine half-equation, Br₂ + 2 e⁻ → 2 Br⁻, uses the same 2 electrons.Answer (c): Sulfur is +4 in SO₃²⁻, so it is the sulfate(IV) ion.Examiner tip: the observation (orange bromine water turning colourless) is evidence that Br₂ has been reduced to colourless Br⁻ ions. SO₂ is the reducing agent.