Question

The quantity of electricity of Faraday needed to reduce $1$ mol of $Cr_2O_7^{2-}$ to $Cr^{3+}$ is

Answer 1

$\overset{+6}{Cr}_2O^{2-}_7→ 2 Cr^{3+}$
∵ Each $Cr$is converting from $+6$ to $+3$
∴ $6$ faradays of charge is required.

So, the answer is $6$.