$S O_{2}$ reduces $M n O_{4}^{-}$ to $M n^{2 +}$ . Number of electrons involved in the reduction of $M n O_{4}^{-}$ to $M n^{2 +}$ is :

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Solution

Let x be the oxidation no. of $M n$ in $M n O_{4}^{-}$

$x + 4 \times (- 2) = - 1$

$x = + 7$
The oxidation no. of $M n$ in $M n O_{4}^{-}$ is +7.
The oxidation number of Mn in $M n^{2 +}$ is +2.

Thus oxidation no. of manganese changes from +7 to +2.Number of electrons involved in the reduction of $M n O_{4}^{-}$ to $M n^{2 +}$ is 5.

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