Let x be the number of moles of KMnO ₄ required to completely oxidize one mole of iodide ion ( I ^- ) in faintly alkaline medium. Let y be the number of moles of K ₂ Cr ₂ O ₇ required to completely oxidize one mole of iodide ion in acidic medium. The value of the ratio x y is:

Step-by-step solution

In faintly alkaline medium, KMnO ₄ oxidizes I ^- to IO ₃^- and itself gets reduced to MnO ₂ . The oxidation half-reaction is: I ^- IO ₃^- + 6e^- (n-factor of I ^- = 6 ) The reduction half-reaction is: MnO ₄^- + 3e^- MnO ₂ (n-factor of KMnO ₄ = 3 ) Equating the equivalents: Moles of I ^- 6 = x 3 1 6 = 3x x = 2 In acidic medium, K ₂ Cr ₂ O ₇ oxidizes I ^- to I ₂ and itself gets reduced to Cr ³⁺ . The oxidation half-reaction is: 2 I ^- I ₂ + 2e^- (n-factor of I ^- = 1 ) The reduction half-reaction is: Cr ₂ O ₇²⁻ + 6e^