KMnO_4 oxidises I^{-} in acidic and neutral/f | vedclass

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Question

(A) In acidic medium,$KMnO_4$ acts as a strong oxidizing agent and oxidizes iodide ions $(I^{-})$ to iodine $(I_2)$: $2 MnO_4^{-} + 10 I^{-} + 16 H^{+

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$I_2$ and $IO_3^-$

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(A) In acidic medium,$KMnO_4$ acts as a strong oxidizing agent and oxidizes iodide ions $(I^{-})$ to iodine $(I_2)$: $2 MnO_4^{-} + 10 I^{-} + 16 H^{+} \rightarrow 2 Mn^{2+} + 5 I_2 + 8 H_2 O$ In neutral or faintly alkaline medium,$KMnO_4$ oxidizes iodide ions $(I^{-})$ to iodate ions $(IO_3^{-})$: $2 MnO_4^{-} + I^{-} + H_2 O \rightarrow 2 MnO_2 + IO_3^{-} + 2 OH^{-}$ Therefore,the products are $I_2$ and $IO_3^{-}$ respectively.

$KMnO_4$ oxidises $I^{-}$ in acidic and neutral/faintly alkaline solution,respectively to

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