2024 O Level Chem P1 Q22 — Redox Reactions / KMnO₄ Tests Solution

Answer: A. Acidified potassium manganate(VII), \(\text{KMnO}_4\) (purple), is a strong oxidising agent. Potassium iodide (KI) is a classic reducing agent — when potassium iodide is oxidised, the colourless iodide ion \(\text{I}^{-}\) is converted to brown iodine \(\text{I}_2\). At the same time, \(\text{MnO}_4^{-}\) is reduced to \(\text{Mn}^{2+}\) (very pale pink, effectively colourless at low concentration).

So the colour change when iodide is oxidised is colourless → brown, while the colour change when manganate(VII) is reduced is purple → colourless.

With excess KI, all of the \(\text{KMnO}_4\) is consumed, so the purple colour disappears completely. The brown iodine produced is now the dominant species in solution, so the final colour is brown.

B (colourless) ignores the iodine produced. C (pale pink) would only appear if no iodine were present. D (purple) describes unreacted \(\text{KMnO}_4\), which is impossible with excess reducing agent.