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Q. How many faradays are required to reduce $1$ mol of $MnO_{4}^{-}$ to $Mn^{2+}$?

Electrochemistry

Solution:

$8H^{+} \underset{(1 mole)}{+ 5e^{-} +} MnO_{4}^{-} \to Mn^{+2} + 4H_{2}O$
$5$ moles of $e^{-} = 5$ Faradays