Which from following expressions is used to find the cell potential of $\mathrm{Cd}_{\text {(s)…

Which from following expressions is used to find the cell potential of $\mathrm{Cd}_{\text {(s) }}\left|\mathrm{Cd}_{\text {(aq) }}^{++}\right|\left|\mathrm{Cu}_{\text {(aq) }}^{++}\right| \mathrm{Cu}_{\text {(s) }}$ cell at $25^{\circ} \mathrm{C}$ ?
  1. $\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-0.0296 \log \frac{\left[\mathrm{Cd}^{++}\right]}{\left[\mathrm{Cu}^{++}\right]}$
  2. $\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}+0.0296 \log \frac{\left[\mathrm{Cd}^{++}\right]}{\left[\mathrm{Cu}^{++}\right]}$
  3. $\mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-0.0592 \log \frac{\left[\mathrm{Cu}^{++}\right]}{\left[\mathrm{Cd}^{++}\right]}$
  4. $\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}+0.0592 \log \frac{\left[\mathrm{Cu}^{++}\right]}{\left[\mathrm{Cd}^{++}\right]}$

Solution

Cell reaction: $\begin{aligned} & \mathrm{Cd}_{(\mathrm{s})} \longrightarrow \mathrm{Cd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \quad \text { (oxidation at anode) } \\ & \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})} \quad \text { (reduction at cathode) } \\ & \hline \mathrm{Cd}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \longrightarrow \mathrm{Cd}_{(\mathrm{aq})}^{2+}+2 \mathrm{Cu}_{(\mathrm{s})} \\ & \text { (overall reaction) } \end{aligned}$ Now, $\begin{aligned} \mathrm{E}_{\text {cell }} & =\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.0592 \mathrm{~V}}{\mathrm{n}} \log _{10} \frac{[\text { Product }]}{[\text { Reactant }]} \\ & =0.02-\frac{0.0592 \mathrm{~V}}{2} \log _{10} \frac{\left[\mathrm{Cd}^{++}\right]}{\left[\mathrm{Cu}^{++}\right]} \\ & =\mathrm{E}_{\text {cell }}^{\prime}-0.0296 \log \frac{\left[\mathrm{Cd}^{++}\right]}{\left[\mathrm{Cu}^{++}\right]} \end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 1)

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