Consider the reaction: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$. What is the quantity of electricity in coulombs needed to reduce $1 \ mol$ of $Cr_{2}O_{7}^{2-}$?

(D) The given balanced half-reaction is: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$.
From the stoichiometry of the reaction,$1 \ mol$ of $Cr_{2}O_{7}^{2-}$ requires $6 \ mol$ of electrons for reduction.
The quantity of electricity $(Q)$ is given by $Q = n \times F$,where $n$ is the number of moles of electrons and $F$ is Faraday's constant $(96487 \ C \ mol^{-1})$.
$Q = 6 \ mol \times 96487 \ C \ mol^{-1} = 578922 \ C$.