Chemistry - Electrochemistry Question with Solution | TestHub

At equilibrium
${\mathrm{E}}_{\text{cell}}=0$
${\mathrm{E}}_{\text{cell}}^0=0.01\text{ V}$
$\text{Sn}+\text{P}{\text{b}}^{\text{2+}}\to \text{S}{\text{n}}^{\text{2+}}+\text{Pb}$
${\mathrm{E}}_{\text{cell}}={\mathrm{E}}_{\text{cell}}^0-\frac{0.06}{\mathrm{n}}\log \mathrm{Q}$
$0=0.01-\frac{0.06}{2}\log \frac{\left[\text{S}{\text{n}}^{2+}\right]}{\left[\text{P}{\text{b}}^{2+}\right]}$
$0.01=\frac{0.06}{2}\log \frac{\left[\text{S}{\text{n}}^{2+}\right]}{\left[\text{P}{\text{b}}^{2+}\right]}$
$\frac{1}{3}=\log \frac{\left[\text{S}{\text{n}}^{2+}\right]}{\left[\text{P}{\text{b}}^{2+}\right]}$
$\frac{\left[{\text{Sn}}^{2+}\right]}{\left[{\text{Pb}}^{2+}\right]}={10}^{\frac{1}{3}}=2.14$