Chemistry - Electrochemistry Question with Solution | TestHub

$\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{Sn}}^{2+}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Sn}\left(\mathrm{s}\right)$

Given: $\frac{2.303\mathrm{RT}}{\mathrm{F}} = 0.059 \mathrm{V}$

${\mathrm{E}}_{\mathrm{cell}}^0 = \frac{2.303\mathrm{RT}}{2\mathrm{F}} \log {\mathrm{K}}_{\mathrm{c}}$

$\Rightarrow {\mathrm{E}}_{\mathrm{cell}}^0 = \frac{0.059}{2} \log \left({10}^{20}\right)$

$\Rightarrow {\mathrm{E}}_{\mathrm{cell}}^0 = \frac{0.059}{2} \times 20$

$\Rightarrow {\mathrm{E}}_{\mathrm{cell}}^0 = 0.59 \mathrm{V}$

${\mathrm{E}}_{\mathrm{cell}}^0 = {\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^0 - {\mathrm{E}}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}^0$

$\Rightarrow {\mathrm{E}}_{\mathrm{cell}}^0 = {\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^0 - (-0.76)$

$\Rightarrow {\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^0 = -0.76 + 059 = -0.17$

$\Rightarrow {\mathrm{E}}_{\mathrm{Sn}/{\mathrm{Sn}}^{2+}}^0 = 0.17 = 17 \times {10}^{-2}$

So, answer is 17.