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Chemistry · 2024 · 3 marks
CBSE 2024 · Region 2 · Set 2 · Q23
Calculate emf of the following cell: Zn ( s )| Zn^2+(0 · 1 M )|| Sn^2+(0 · 001 M )| Sn ( s ) Given: E_Zn^2+ / Zn^0=-0 · 76 V, E_Sn^2+ / Sn^0=-0 · 14…
Calculate emf of the following cell : $\displaystyle \mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(0 \cdot 1 \mathrm{M})\right|\left|\mathrm{Sn}^{2+}(0 \cdot 001 \mathrm{M})\right| \mathrm{Sn}(\mathrm{s})$ Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0 \cdot 76 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{0}=-0 \cdot 14 \mathrm{~V}$ \[[\log 10=1] \]$\displaystyle 3$
Marking-scheme solution
\[\begin{aligned}
\text { Exell } & =\left(\mathrm{E}^{o}{ }_{c}-\mathrm{E}^{o}{ }_{a}\right)-\frac{0.059}{2} \log \left[\frac{Z n^{2+}}{S n^{2+}}\right] \\
& =[(-0 \cdot 14)-(-0 \cdot 76)]-\frac{0.059}{2} \log \frac{0.1}{0.001} \\
& =+0.62-\frac{0.059}{2} \times 2 \\
& =(0.62-0.059) \mathrm{V} \\
& =0.561 \mathrm{~V}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.