What is the quantity of electricity in coulombs needed to reduce 1 mol of
Question:

What is the quantity of electricity in coulombs needed to reduce 1 mol of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} ?$ Consider the reaction: $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+8 \mathrm{H}_{2} \mathrm{O}$

Solution:

The given reaction is as follows:

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}$

Therefore, to reduce 1 mole of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$, the required quantity of electricity will be:

=6 F

= 6 × 96487 C

= 578922 C