KMnO _ 4 oxidises I ^ - in acidic and neutral/faintly alkaline...

In the acidic medium, ${MnO}_{4}^{-}$ convert to ${Mn}^{2 +}$ and ${I}^{-}$ convert to ${I}_{2}$.

$2{MnO}_{4}^{-} + 10{I}^{-} + 16{H}^{+} \to 2{Mn}^{2 +} + 5{I}_{2} + 8{H}_{2}O$

In neutral/faintly alkaline solution, ${MnO}_{4}^{-}$ convert to ${MnO}_{2}$ and ${I}^{-}$ convert to ${IO}_{3}^{-}$.

$2{MnO}_{4}^{-} + {I}^{-} + {H}_{2}O \to 2{MnO}_{2} + 2{OH}^{-} + {IO}_{3}^{-}$