How many Faraday of electricity is required to reduce $1$ mole of $M n O_{4}^{-}$ ions to $M n^{2 +}$ ions?

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Solution

$5$ - Faraday of electricity is required to reduce $1$ mole of $M n O_{4}^{-}$ ions to $M n^{2 +}$ ions.
$M n O_{4}^{-} + 5 e^{-} \to M n^{2 +}$
The oxidation number of Mn in $M n O_{4}^{-}$ is $+ 7$ .
The oxidation number of Mn in $M n^{2 +}$ is $+ 2$ .
The change in the oxidation number is $7 - 2 = 5$ .
This suggests five moles of electrons for oxidation of one mole of $M n O_{4}^{-}$ to one mole of $M n^{2 +}$ .
The number of moles of electrons is equal to the number of Faraday of electricity required.

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$M n {O_{4}}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{o} = 1.51 V$
The quantity of electricity required in Faraday to reduce five moles of $M n O_{4}^{-}$ is ____.

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