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Redox Reactions and Couple

KMnO_4 acts a...

Question

$K M n O_{4}$ acts as an oxidising agent in the neutral medium and gets reduced to $M n O_{2}$ .The equivalent weight of $K M n O_{4}$ in neutral medium

mol. wt/2

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mol.wt/3

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mol. wt/4

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mol .wt/7

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Solution

The correct option is B mol.wt/3
In neutral medium $M n^{+ 7}$ oxidation state changes into +4 oxidation state, hence equivalent weight of $K M n O_{4} = \frac{M}{3}$

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K = 39.09

M n = 54.94

O = 16.0

2 K M n O_{4} + 2 K O H \to 2 K_{2} {M n O}_{4} + H_{2} O + [O]

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