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Q.

When 0.1 mole $M n O_{4}^{2 -}$ is oxidised the quantity of electricity required to completely oxidise $M n O_{4}^{2 -}$ to $M n O_{4}^{-}$ is

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a

96500 C

b

2 × 96500C

c

96.50 C

d

9650 C

answer is C.

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Detailed Solution

$\underset{0.1 m o l e}{\overset{(+ 6)}{M n}} O_{4}^{2 -} \to \overset{(+ 7)}{M n} O_{4}^{-} + e^{-}$

Quantity of electricity required = 0.1F = 0.1 × 96500 = 9650 C

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