CBSE 2022 · Region 4 · Set 1 · Q10 · 3 marks
Write the Nernst equation and calculate emf of the following cell at $\displaystyle 298$ K : \[\mathrm{Cr}\left|\mathrm{Cr}^{3+}(0 \cdot 1 \mathrm{M}) \| \mathrm{Fe}^{2+}(0 \cdot 01 \mathrm{M})\right| \mathrm{Fe} \] Given : $\displaystyle \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\ominus}=-0.75 \mathrm{~V}$ \[\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\ominus}=-0.45 \mathrm{~V} \] $\displaystyle (\log 10=1)$
Marking-scheme solution
\[\begin{aligned}
E_{\text {cell }} & =E_{\text {cell }}^{0}-\frac{0.059}{6} \log \frac{\left[\mathrm{Cr}^{3+}\right]^{2}}{\left[\mathrm{Fe}^{2+}\right]^{3}} \\
E_{\text {cell }}^{0} & =-0.45-(-0.75) \\
& =0.30 \mathrm{~V} \\
E_{\text {cell }} & =0.3-\frac{0.059}{6} \log \frac{(0.1)^{2}}{(0.01)^{3}}
\end{aligned}
\]
\(\displaystyle =0 \cdot 3-.00985 \log \frac{\left(10^{-1}\right)^{2}}{\left(10^{-2}\right)^{3}}\)
\(\displaystyle =0 \cdot 3-.00985 \times 4 \log 10\)
\(\displaystyle =0.3-0.0394\)
\(\displaystyle =0 \cdot 2606 \mathrm{~V}\)
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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.