CBSE 2024 · Region 5 · Set 1 · Q25 · 3 marks
Calculate the emf of the following cell : $\displaystyle 3$ $\displaystyle \mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.01 \mathrm{M}) \longrightarrow \mathrm{Ni}^{2+}(0.1 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})$ Given that $\displaystyle \mathrm{E}_{\text {cell }}^{\circ}=1.05 \mathrm{~V}, \log 10=1$
Marking-scheme solution
\(\displaystyle \mathrm{E}^{\circ}{ }_{\text {cell }}=1.05 \mathrm{~V}\)
\[\begin{aligned}
& \mathrm{E}_{\text {cell }}=1.05-\frac{0.059}{2} \log \frac{[\mathrm{Ni}]^{2+}}{[\mathrm{Ag}+]^{2}} \\
& =1.05-0.059 / 2 \log \frac{(0.1)}{(0.01)^{2}} \\
& =1.05-0.0825 \\
& =0.9615 \mathrm{~V}(\text { Deduct } 1 / 2 \text { mark for incorrect unit or no unit })
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.