Exam-style Question 1

Question (3 marks): In acidic solution, manganate(VII) ions oxidise iron(II) ions to iron(III) ions. The half-equations are:MnO₄⁻ (aq) + 8 H⁺ (aq) + 5 e⁻ → Mn²⁺ (aq) + 4 H₂O (l)
Fe²⁺ (aq) → Fe³⁺ (aq) + e⁻(a) Write the overall ionic equation. (b) State the change in oxidation number of manganese.Answer (a): The manganese half-equation uses 5 electrons, but the iron half-equation releases only 1. Multiply the iron half-equation by 5:5 Fe²⁺ → 5 Fe³⁺ + 5 e⁻Now add the two half-equations. The 5 electrons appear on both sides and cancel:MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺Answer (b): In MnO₄⁻, x + 4(−2) = −1, so Mn is +7. In Mn²⁺ it is +2. The oxidation number decreases by 5, which matches the 5 electrons gained.Check the charges: left side (−1) + (+8) + (+10) = +17; right side (+2) + (+15) = +17. The equation balances.Examiner tips: never leave electrons in the final equation, and keep the H⁺ and H₂O from the half-equation. The mole ratio MnO₄⁻ : Fe²⁺ is 1 : 5, which you will use in titration calculations.