Which from following expressions is used to calculate \(\mathrm{E}_{\text {cell }}\) for the following cell at \(25^{\circ} \mathrm{C}\) ?
\(\mathrm{Pb}_{(\mathrm{s})}\left|\mathrm{Pb}_{(\mathrm{IM})}^{++}\right|\left|\mathrm{Ag}_{(10 \mathrm{M})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})}\)

\(\mathrm{Pb}_{(\mathrm{s})}\left|\mathrm{Pb}_{(1 \mathrm{M})}^{++}\right|\left|\mathrm{Ag}_{(10 \mathrm{M})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})}\)
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 \(25^{\circ} \mathrm{C}\) :
\(\begin{aligned} & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.0592}{2} \log _{10} \frac{\left[\mathrm{~Pb}^{++}\right]}{\left[\mathrm{Ag}^{+}\right]^2} \\ & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.0592}{2} \log _{10} \frac{1}{100} \\ & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.0592}{2} \log _{10} 10^{-2} \\ & \mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.0592}{2} \times(-2) \\ & \mathrm{E}_{\text {cell }}=\left(\mathrm{E}_{\text {cell }}^{\mathrm{o}}+0.0592\right) \mathrm{V}\end{aligned}\)