I^{-}(aq.) + MnO_4^{-}(aq.) xrightarrow{H^{+} | vedclass

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(A) $1$. In acidic medium,$MnO_4^{-}$ oxidizes $I^{-}$ to $I_2$: $2MnO_4^{-} + 10I^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$. Thus,$X = I_2$.

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$I_2, IO_3^{-}, MnO_2$

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(A) $1$. In acidic medium,$MnO_4^{-}$ oxidizes $I^{-}$ to $I_2$: $2MnO_4^{-} + 10I^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$. Thus,$X = I_2$.$2$. In neutral or weakly basic medium,$MnO_4^{-}$ oxidizes $I^{-}$ to $IO_3^{-}$: $2MnO_4^{-} + I^{-} + H_2O \rightarrow 2MnO_2 + IO_3^{-} + 2OH^{-}$. Thus,$Y = IO_3^{-}$.$3$. In the presence of $ZnSO_4$ (which acts as a buffer),$MnO_4^{-}$ reacts with $Mn^{2+}$ to form $MnO_2$: $2MnO_4^{-} + 3Mn^{2+} + 2H_2O \rightarrow 5MnO_2 + 4H^{+}$. Thus,$Z = MnO_2$.

$I^{-}(aq.) + MnO_4^{-}(aq.) \xrightarrow{H^{+}} X + Mn^{2+}(aq.)$
$I^{-}(aq.) + MnO_4^{-}(aq.) \xrightarrow{\text{Neutral or weakly } OH^{-}} Y + MnO_2$
$MnO_4^{-}(aq.) + Mn^{2+}(aq.) \xrightarrow{ZnSO_4} Z + 4H^{+}$
Products $X, Y$ and $Z$ are respectively :

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$I^{-}(aq.) + MnO_4^{-}(aq.) \xrightarrow{H^{+}} X + Mn^{2+}(aq.)$$I^{-}(aq.) + MnO_4^{-}(aq.) \xrightarrow{\text{Neutral or weakly } OH^{-}} Y + MnO_2$$MnO_4^{-}(aq.) + Mn^{2+}(aq.) \xrightarrow{ZnSO_4} Z + 4H^{+}$Products $X, Y$ and $Z$ are respectively :