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Q.
How many coulombs of electricity are required for the reduction of $1\, mole$ of $MnO_{4}$ to $Mn^{2+}$?
Electrochemistry
Solution:
$MnO^{-}_{4} + 5e^{-} \Rightarrow Mn^{2+}$
It is obvious from above equation that reduction of $1$ mole of $MnO^{-}_{4}$ to $Mn^{2+}$ requires $5$ Faraday i.e., $5 \times 96500 \approx 4.83 \times 10^{5} C$