Q3 / 90 · JEE Main 2012 (07 May Online) · Chemistry
A battery is constructed of $\mathrm{Cr}$ and $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$. The unbalanced chemical equation when such a battery discharges is following:
$$
\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{Cr}+\mathrm{H}^{+} \rightarrow \mathrm{Cr}^{3+}+\mathrm{H}_2 \mathrm{O}+\mathrm{Na}^{+}
$$
If one Faraday of electricity is passed through the battery during the charging, the number of moles of $\mathrm{Cr}^{3+}$ removed from the solution is
- A$\frac{4}{3}$
- B$\frac{1}{3}$
- Correct answerC$\frac{3}{3}$
- D$\frac{2}{3}$
Solution
Reduction half reaction :
$$
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 e^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
$$
Oxidation half reaction :
$$
\mathrm{Cr} \longrightarrow \mathrm{Cr}^{3+}+3 e^{-}
$$
Overall reaction :
$$
\begin{aligned}
& \mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{Cr}+14 \mathrm{H}^{+}+3 e^{-} \\
& 3 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
\end{aligned}
$$
$3 \mathrm{~F}$ of electricity $=3$ moles of $\mathrm{Cr}^{3+}$
$1 \mathrm{~F}$ of electricity $=\frac{3}{3}$ moles of $\mathrm{Cr}^{3+}$
Hence, the correct answer is option: 'C'.