A battery is constructed of $\mathrm{Cr}$ and $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$. The unbalanced…

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
  1. $\frac{4}{3}$
  2. $\frac{1}{3}$
  3. $\frac{3}{3}$
  4. $\frac{2}{3}$

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 $\mathrm{Cr}^{3+}$

Asked in: JEE Main 2012 (07 May Online)

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