# For an electrochemical cell

Question:

For an electrochemical cell$\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(a q, 1 \mathrm{M}) \| \mathrm{Pb}^{2+}(a q, 1 \mathrm{M})\right| \mathrm{Pb}(\mathrm{s})$ the ratio $\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}$ when this cell attains equilibrium is ________.

(Given: $\mathrm{E}_{\mathrm{Sn}^{2+} \mid \mathrm{Sn}}^{0}=-0.14 \mathrm{~V}$,$\left.\mathrm{E}_{\mathrm{Pb}} \quad \mid \mathrm{Pb}=-0.13 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{=0.06}\right)$

Solution:

(2.15)

At equilibrium state $\mathrm{E}_{\text {cell }}=0 ; \mathrm{E}_{\text {cell }}^{0}=0.01 \mathrm{~V}$

$\mathrm{Sn}+\mathrm{Pb}^{2+} \longrightarrow \mathrm{Sn}^{2+}+\mathrm{Pb}$

$\mathrm{E}=\mathrm{E}_{\text {cell }}^{0}-\frac{0.06}{n} \log \frac{[\mathrm{P}]}{[\mathrm{R}]}$

$0=0.01-\frac{0.06}{2} \log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}$

$0.01=\frac{0.06}{2} \log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}$

$\frac{1}{3}=\log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}$

$\frac{\left[\mathrm{Sb}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=10^{1 / 3}=2.15$