Question

In the neutral or faintly alkaline medium, $KMn{O}_4$ oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from

Answer 1

Pemanganate ion as an oxidizing agent works most efficiently in acidic solution, because it is reduced to the greatest extent in this medium, from oxidation state $+7$ in $Mn{O}_4^{-}$ to $+2$ in $M{n}^{2+}$.

$8H^{+}$+ $Mn{O}_4^{-}$ + $5e^{-}$ $\to$ $M{n}^{2+}$ + $4H_{2}O$

The oxidation in neutral or slightly alkaline medium is less efficient, as $Mn$ $\left(VII\right)$ is only reduced to $Mn\left(IV\right)$, yielding insoluble $Mn{O}_2$ as a byproduct which needs to be removed from the reaction mixture during purification of the desired product.

$2H_{2}O$+$Mn{O}_4^{-}$+$3e^{-}$⟶ $Mn{O}_2$↓ +$4O{H}^{-}$


Hence, The reaction should be as follows: