The following reaction occurs in acidic medium
${K M n O}_{4} + 8 H^{+} + 5 e^{-} \to K^{+} + {M n}^{2 +} + 4 H_{2} O$
What is the equivalent weight of ${K M n O}_{4}$ ?

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$31.6$

$39.5$

$158.0$

$79.0$

answer is B.

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Detailed Solution

$E q u i v a l e n t w e i g h t = \frac{M o l e c u l a r w e i g h t}{n - f a c t o r}$

From the given reaction: ${K M n O}_{4} + 8 H^{+} + 5 e^{-} \to K^{+} + {M n}^{2 +} + 4 H_{2} O$

The oxidation state of Manganese changes from +7 to +2. So there is a change of 5. Thus the n-factor would be 5

Molecular weight of $K M n O_{4}$ = 158 (K=19, $M n$ = 55 O=16)

Putting the values in the formula:

Therefore equivalent weight = $\frac{158}{5} = 31.6$

Hence the correct option is (B).

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