1. Identify anode and cathode from standard reduction potentials

Given: $E^\circ_{Sn^{2+}/Sn} = -0.14\,V, \qquad E^\circ_{Pb^{2+}/Pb} = -0.13\,V$

Since $-0.13\,V$ is greater than $-0.14\,V$, $Pb^{2+}$ gets reduced more easily.

• Cathode: $Pb^{2+} + 2e^- \to Pb$
• Anode: $Sn \to Sn^{2+} + 2e^-$

So the overall cell reaction is: $Sn(s) + Pb^{2+}(aq) \to Sn^{2+}(aq) + Pb(s)$

2. Calculate standard cell potential

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ $E^\circ_{cell} = (-0.13) - (-0.14) = 0.01\,V$

3. Write Nernst equation

For the reaction $Sn(s) + Pb^{2+}(aq) \to Sn^{2+}(aq) + Pb(s)$

the reaction quotient is $Q = \frac{[Sn^{2+}]}{[Pb^{2+}]}$

Also, number of electrons transferred is $n=2$.

Thus, $E_{cell} = E^\circ_{cell} - \frac{0.06}{n}\log Q$ $E_{cell} = 0.01 - \frac{0.06}{2}\log\left(\frac{[Sn^{2+}]}{[Pb^{2+}]}\right)$

4. At equilibrium, cell emf becomes zero

At equilibrium: $E_{cell}=0$

So, $0 = 0.01 - 0.03\log\left(\frac{[Sn^{2+}]}{[Pb^{2+}]}\right)$

Therefore, $0.03\log\left(\frac{[Sn^{2+}]}{[Pb^{2+}]}\right) = 0.01$ $\log\left(\frac{[Sn^{2+}]}{[Pb^{2+}]}\right) = \frac{0.01}{0.03} = \frac13$

Hence, $\frac{[Sn^{2+}]}{[Pb^{2+}]} = 10^{1/3} \approx 2.15$

5. Final answer

$\boxed{\frac{[Sn^{2+}]}{[Pb^{2+}]} \approx 2.15}$