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
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