#### For an electrochemical cell $\mathrm{Sn(s)\left | Sn^{2+}(aq,1M) \right |\left | Pb^{2+}(aq,1M) \right |Pb(s)}$ the ratio $\mathrm{\frac{\left [ Sn^{2+} \right ]}{\left [ Pb^{2+} \right ]}}$ When this cell attains equilibrium is _____. $\mathrm{( Given :E^{0}_{Sn^{2+} | Sn} =-0.14V,}$ $\mathrm{E^{0}_{Pb^{2+} | Pb }=-0.13V,}$ $\mathrm{\frac{2.303RT}{F}=0.06)}$ Option: 1 Option: 2 1.11 Option: 3 7.15 Option: 4 3.14

As we have learnt,

Nernst equation is given as

$\mathrm{E= E^0_{cell}-\frac{2.303RT}{nF}logQ}$

Now, the chemical reaction occuring in the cell is given as

$\begin{array}{l}{\mathrm{Sn}+\mathrm{Pb}^{2+} \longrightarrow \mathrm{Sn}^{2+}+\mathrm{Pb}} \\ {0=0.01-\frac{0.06}{2} \log \left\{\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\}} \\ {0.01=\frac{0.06}{2} \log \left[\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\}} \\ {\frac{1}{3}=\log \left[\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\} \Rightarrow \frac{\left[\mathrm{Sb}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=10^{1 / 3}=2.1544}\end{array}$

Hence, the option number (1) is correct.