CBSE 2022 · Region 4 · Set 3 · Q10 · 3 marks
Write the Nernst equation and calculate the emf of the following cell at $\displaystyle 298$ K : \[\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0 \cdot 1 \mathrm{M}) \| \mathrm{Cd}^{2+}(0 \cdot 01)\right| \mathrm{Cd} \] Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\ominus}=-0.76 \mathrm{~V}$ \[\mathrm{E}_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{\ominus}=-0 \cdot 40 \mathrm{~V} \] $\displaystyle (\log 10=1)$
Marking-scheme solution
\[\begin{aligned}
E_{\text {cell }} &=E_{\text {cell }}^{0}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cd}^{2+}\right]} \\
E_{\text {cell }}^{0} &=-0.40-(-0.76)=0.36 \mathrm{~V} \\
E_{\text {cell }} &=0.36-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cd}^{2+}\right]} \\
&=0.36-\frac{0.059}{2} \log \frac{(0.1)}{(0.01)} \\
&=0.36-0.0295 \log 10 \\
E_{\text {cell }} &=0.3305 \mathrm{~V}
\end{aligned}
\]
ElectrochemistryNernst EquationApplyshort_answermedium
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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.