CBSE 2022 · Region 3 · Set 1 · Q5 · 3 marks
Calculate $\displaystyle \Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ and $\displaystyle \log \mathrm{Kc}$ for the following cell : \[\begin{aligned} & \ \mathrm{Ni}(\mathrm{~s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{~s}) \\ & \text { Given that } \mathrm{E}^{\circ} \text { cell }=1.05 \mathrm{~V}, \mathrm{IF}=96,500 \mathrm{Cmol}^{-1} . \end{aligned} \]Calculate the e.m.f. of the following cell at 298K : \[\mathrm{Fe}(\mathrm{~s})\left|\mathrm{Fe}^{2+}(0.001 \mathrm{M})\right|\left|\mathrm{H}^{+}(0.01 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})(1 \mathrm{bar}) \mid \mathrm{Pt}(\mathrm{~s}) \] Given that $\displaystyle \mathrm{E}^{\circ}$ cell $\displaystyle =+0.44 \mathrm{~V}$ \[[\log 2=0.3010 \quad \log 3=0.4771 \quad \log 10=1] \]
Calculate $\displaystyle \Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ and $\displaystyle \log \mathrm{Kc}$ for the following cell : \[\begin{aligned} & \ \mathrm{Ni}(\mathrm{~s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{~s}) \\ & \text { Given that } \mathrm{E}^{\circ} \text { cell }=1.05 \mathrm{~V}, \mathrm{IF}=96,500 \mathrm{Cmol}^{-1} . \end{aligned} \]
Calculate the e.m.f. of the following cell at 298K : \[\mathrm{Fe}(\mathrm{~s})\left|\mathrm{Fe}^{2+}(0.001 \mathrm{M})\right|\left|\mathrm{H}^{+}(0.01 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})(1 \mathrm{bar}) \mid \mathrm{Pt}(\mathrm{~s}) \] Given that $\displaystyle \mathrm{E}^{\circ}$ cell $\displaystyle =+0.44 \mathrm{~V}$ \[[\log 2=0.3010 \quad \log 3=0.4771 \quad \log 10=1] \]
Marking-scheme solution
\[\begin{aligned}
\Delta_{\mathrm{r}} G^{\circ} & =-n F E_{\text {cell }}^{\circ} \\
& =-2 \times 96500 \times 1.05 \mathrm{Jmol}^{-1} \\
& =-202,650 \mathrm{~J} \mathrm{~mol}^{-1} \text { Or }-202.65 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\]
\[\begin{aligned}
\log K_{C} & =\frac{n E_{\text {cell }}^{\circ}}{0.059} \\
& =\frac{2 \times 1.05}{0.059} \\
& =35.6
\end{aligned}
\]
\[\begin{aligned}
& E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^{2}} \\
& E_{\text {cell }}=0.44 \mathrm{~V}-\frac{0.059}{2} \log \frac{[0.001]}{[0.01]^{2}} \\
& E_{\text {cell }}=0.44 \mathrm{~V}-\frac{0.059}{2} \log 10 \\
& E_{\text {cell }}=0.44 \mathrm{~V}-0.0295 \mathrm{~V}=0.4105 \mathrm{~V}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.