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Chemistry · 2022 · 3 marks
CBSE 2022 · Region 4 · Set 2 · Q10
Write the Nernst equation and calculate the emf of the following cell at 298 K: 3 Zn | Zn^2+(0 · 001 M ) ∥ H^+(0 · 01 M )| H_2( g )(1 bar ) ∣ Pt ( s…
Write the Nernst equation and calculate the emf of the following cell at $\displaystyle 298$ K : $\displaystyle 3$ \[\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0 \cdot 001 \mathrm{M}) \| \mathrm{H}^{+}(0 \cdot 01 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})(1 \text { bar }) \mid \mathrm{Pt}(\mathrm{~s}) \] Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}=-0.76 \mathrm{~V}$ \[\mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}}^{\ominus}=0.00 \mathrm{~V} \] $\displaystyle [\log 10=1]$
Marking-scheme solution
\[\begin{aligned}
& E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^{2}} \\
& E_{\text {cell }}^{0}=0.0-(-0.76)=0.76 \mathrm{~V}
\end{aligned}
\]
\[\begin{aligned}
& =0.76-\frac{0.059}{2} \log \frac{[0.001]}{[0.01]^{2}} \\
& =0.76-0.0295 \times 1 \quad \text { (Deduct } \frac{1}{2} \text { mark for no or incorrect unit) } \\
& =0.7305 \mathrm{~V}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.