Question

A battery is constructed of $Cr$ and $N{a}_{2}C{r}_2{O}_7$. The unbalanced chemical equation when such a battery discharges is following:
$\mathrm{Na}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7+\mathrm{Cr}{}^{+}\mathrm{H}{}^{+}\to \mathrm{Cr}{}_3{}^{+}+\mathrm{H}{}_2\mathrm{O}+\mathrm{Na}{}^{+}$
If one Faraday of electricity is passed through the battery during the charging, the number of moles of $C{r}^{3+}$ removed from the solution is

• A. $\frac{4}{3}$
• B. $\frac{1}{3}$
• C. $\frac{3}{3}$
• D. $\frac{2}{3}$

Answer 1

Answer: (C)

Reduction half reaction :
$\mathrm{Cr}{}_2\mathrm{O}{}_7^{2-}+6\mathrm{e}{}^{-}+14\mathrm{H}{}^{+}\to 2\mathrm{Cr}{}^{3+}+7\mathrm{H}{}_2\mathrm{O}$
Oxidation half reaction:
$\mathrm{Cr}\to \mathrm{Cr}{}^{3+}+2\mathrm{e}{}^{-}$
Overall reaction:
$\mathrm{Cr}{}_2\mathrm{O}{}_7^{2-}+\mathrm{Cr}+14\mathrm{H}{}^{+}+3\mathrm{e}{}^{-}\to 3\mathrm{Cr}{}^{3+}+7\mathrm{H}{}_2\mathrm{O}$
$3F$ of electricity = 3 moles of $C{r}^{3+}$
$1F$ of electricity $=\frac{3}{3}$ moles of $C{r}^{3+}$