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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
Options:
$$
\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{Cr}+\mathrm{H}^{+} \rightarrow \mathrm{Cr}^{3+}+\mathrm{H}_2 \mathrm{O}+\mathrm{Na}^{+}
$$
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
Solution:
1094 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{3}$
$\frac{3}{3}$
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 $\mathrm{Cr}^{3+}$
$$
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 e^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
$$
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 $\mathrm{Cr}^{3+}$
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