Exam-style Question 5

Question (4 marks): Hydrogen peroxide oxidises iodide ions in acidic solution:H₂O₂ (aq) + 2 H⁺ (aq) + 2 I⁻ (aq) → I₂ (aq) + 2 H₂O (l)(a) Give the oxidation number of oxygen in H₂O₂ and in H₂O. (b) Write the two half-equations. (c) Identify the oxidising agent and explain your answer.Answer (a): Hydrogen peroxide is a peroxide, so oxygen is −1 (2(+1) + 2x = 0). In water, oxygen is −2. A common error is to assume oxygen is always −2.Answer (b):H₂O₂ + 2 H⁺ + 2 e⁻ → 2 H₂O
2 I⁻ → I₂ + 2 e⁻Both half-equations involve 2 electrons, so they add directly to give the overall equation.Answer (c): H₂O₂ is the oxidising agent. It accepts electrons, and the oxidation number of its oxygen falls from −1 to −2, so it is reduced. Iodide ions are the reducing agent: iodine rises from −1 to 0.Observation: the colourless solution turns yellow-brown as iodine forms.Examiner tip: hydrogen stays at +1 throughout, so H⁺ is not the oxidising agent even though it appears on the left. Only species whose atoms change oxidation number are oxidised or reduced.