The coulombs of electricity required for reduction of 1 mole of $M n O_{4}^{-}$ to $M n^{2 +}$ are

96500 C

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1.93 \times 10^{5} C

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4.83 \times 10^{5} C

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9365 \times 10^{6} C

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Solution

The correct option is C $4.83 \times 10^{5} C$
Reduction of $M n O_{4}^{-}$ to $M n^{2} f$ is
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
$∴$ Reduction of d mole of $M n O_{4}^{-}$
Requires = 5 $\times$ 96500 C = 4.825 $\times 10^{5} C$

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