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Chemistry · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 1 · Q33
(a) Calculate emf and Δ G for the following cell at 298 K: Mg ( s ) / Mg^2+(0.01 M ) / / Ag^+(0.001 M ) / Ag ( s ); Given: E_Mg^2+ / Mg^°=-2.37 V…
Calculate emf and $\displaystyle \Delta \mathrm{G}$ for the following cell at $\displaystyle 298$ K : \[\begin{aligned} & \mathrm{Mg}(\mathrm{~s}) / \mathrm{Mg}^{2+}(0.01 \mathrm{M}) / / \mathrm{Ag}^{+}(0.001 \mathrm{M}) / \mathrm{Ag}(\mathrm{~s}) \\ & \text { Given }: \mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.37 \mathrm{~V} \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=+0.80 \mathrm{~V} \\ & {\left[1 \mathrm{~F}=96500 \mathrm{C} \mathrm{~mol}^{-1}, \log 10=1\right]} \end{aligned} \]
For the reaction : \[2 \mathrm{AgCl}(\mathrm{~s})+\mathrm{H}_{2}(\mathrm{~g})(0.4 \mathrm{~atm}) \longrightarrow 2 \mathrm{Ag}(\mathrm{~s})+2 \mathrm{H}^{+}(0.1 \mathrm{M})+2 \mathrm{Cl}^{-}(0.2 \mathrm{M}) \] Calculate emf of the cell at $\displaystyle 25$ °C. \[\begin{aligned} & \text { Given }: \Delta \mathrm{G}^{\circ}=-43500 \mathrm{~J} \mathrm{~mol}^{-1} \\ & {\left[\log 10=1,1 \mathrm{~F}=96500 \mathrm{C} \mathrm{~mol}^{-1}\right]} \end{aligned} \]
Marking-scheme solution
\[\begin{array}{l}
\Delta \mathrm{G}^{\circ}=-\mathrm{nF} \mathrm{E}_{\text {cell }}^{\circ} \\
-43500=-2 \times 96500 \times \mathrm{E}_{\text {cell }}^{\circ} \\
\mathrm{E}_{\text {cell }}^{\circ}= 0.225 \mathrm{~V} \\
\mathrm{E}_{\text {cell }}= \\
\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{H}^{+}\right]^{2}\left[\mathrm{Cl}^{-}\right]^{2}}{\mathrm{P}_{\mathrm{H}_{2}}} \\
=0.225-\frac{0.059}{2} \log \frac{[0.1]^{2}[0.2]^{2}}{0.4} \\
=0.225-\frac{0.059}{2} \log 10^{-3} \\
=0.225+\frac{0.059}{2} \times 3 \\
=0.225+0.0885 \\
=0.3135 \mathrm{~V}
\end{array}
\]
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CBSE Class 12 Chemistry past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.