How many mole of electron are required fro the reduction of 1 mole of $\mathrm{Cr}^3+$ to…
How many mole of electron are required fro the reduction of 1 mole of $\mathrm{Cr}^3+$ to $\mathrm{Cr}_{(\mathrm{s})}$ ?
- $3$
- $6$
- $1$
- $\frac{6.022 \times 10^{23}}{3}$
Solution
$\underset{1 \mathrm{mo1}}{\mathrm{Cr}^{+3}}+\underset{3 \mathrm{mo1}}{3 \mathrm{e}^{-}} \rightarrow \mathrm{Cr}(\mathrm{s})$
Asked in: MHT CET 2022 (05 Aug Shift 1)
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