Chemistry - Electrochemistry Question with Solution | TestHub

For the given cell reaction, $E_{\text{cell }}^{\circ }=-0.14\mathrm{V}+0.13\mathrm{V}=-0.01\mathrm{V}$ According to Nerst equation, $\begin{array}{rl}{E}_{\text{cell }}& =E_{\text{cell }}^{\circ }-\frac{0.0591}{2}\log \frac{\left[{\mathrm{Pb}}^{2+}\right]}{\left[{\mathrm{Sn}}^{2+}\right]}\\ \Rightarrow 0& =-0.01=\frac{0.0591}{2}\log \frac{\left[{\mathrm{Pb}}^{2+}\right]}{\left[{\mathrm{Sn}}^{2+}\right]}\end{array}$ $\begin{array}{rl}& \Rightarrow \log \frac{\left[{\mathrm{Pb}}^{2+}\right]}{\left[{\mathrm{Sn}}^{2+}\right]}=\frac{0.01}{-0.0296}=-0.3\\ & \Rightarrow \frac{\left[{\mathrm{Pb}}^{2+}\right]}{\left[{\mathrm{Sn}}^{2+}\right]}=\mathrm{antilog}(-0.3)\\ & \Rightarrow \frac{\left[{\mathrm{Pb}}^{2+}\right]}{\left[{\mathrm{Sn}}^{2+}\right]}=0.5\\ & \end{array}$