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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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Similar questions

K M n O_{4}

M n O_{2}

K M n O_{4}

K M n O_{4}

= 158

C O_{2}

5 C_{2} O_{4}^{2 -} + 2 M n O^{-} + 16 H^{+} \to 10 C O_{2} + 2 M n^{2 +} + 8 H_{2} O

K M n O_{4}

M n O_{4}^{-}

M n^{2 +}

K M n O_{4}

= 158

250 m L

0.04 N

K M n O_{4}

K I O_{3}

2 C r (O H)_{3} + 4 O H + K I O_{3} \to 2 C r O_{4}^{2 -} + 5 H_{2} O + K I

{M n O}_{4}^{-}

{M n}^{+ 2}

{K M n O}_{4}

{K M n O}_{4}

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