Which from following expressions is used to calculate $\mathrm{E}_{\text {cell }}$ for the following cell at…
- $\cdot \mathrm{E}_{\text {cell }}=\left(\mathrm{E}_{\text {cell }}^{\circ}+0.0592\right) \mathrm{V}$
- $\mathrm{E}_{\text {cell }}=\left(\mathrm{E}_{\text {cell }}^{\circ}-0.0592\right) \mathrm{V}$
- $E_{\text {cell }}=\left(E_{\text {cell }}^{\circ}-0.0296\right) V$
- $\mathrm{E}_{\mathrm{cell}}=\left(\mathrm{E}_{\mathrm{cell}}^{\circ}+0.0296\right) \mathrm{V}$
Solution
Reaction: $\mathrm{Pb}_{(\mathrm{s})}+2 \mathrm{Ag}_{(10 \mathrm{M})}^{+} \rightarrow \mathrm{Pb}_{(1 \mathrm{M})}^{++}+2 \mathrm{Ag}_{(\mathrm{s})}$ Nernst equation at $25^{\circ} \mathrm{C}$ : $\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.0592}{2} \log _{10} \frac{\left[\mathrm{~Pb}^{++}\right]}{\left[\mathrm{Ag}^{+}\right]^2} \\ & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.0592}{2} \log _{10} \frac{1}{100} \\ & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.0592}{2} \log _{10} 10^{-2} \\ & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.0592}{2} \times(-2) \\ & \mathrm{E}_{\text {cell }}=\left(\mathrm{E}_{\text {cell }}^{\mathrm{o}}+0.0592\right) \mathrm{V}\end{aligned}$
Asked in: MHT CET 2024 (02 May Shift 1)