What is the ratio of the number of moles of KMnO₄ required to completely oxidise one mole of KI in an acidic medium to that in a faintly alkaline medium?

Step-by-step solution

In an acidic medium, KMnO₄ oxidises iodide ( I^- ) to iodine ( I₂ ) and itself gets reduced to Mn²⁺ . The n-factor for KI is 1 (since 2I^- I₂ + 2e^- , so 1 mole of I^- loses 1 mole of e^- ). The n-factor for KMnO₄ is 5 ( Mn⁺⁷ Mn⁺² ). Equating the number of equivalents: Moles of KMnO₄ 5 = Moles of KI 1 n₁ 5 = 1 1 n₁ = 1 5 moles. In a faintly alkaline medium, KMnO₄ oxidises iodide ( I^- ) to iodate ( IO₃^- ) and itself gets reduced to MnO₂ . The n-factor for KI is 6 ( I^- IO₃^- , change in oxidation state from -1 to