For an electrochemical cell Sn(s)|Sn$$^{2+}$$(aq, 1M)||Pb$$^{2+}$$(aq, 1M)|Pb(s) the ratio $$\frac{[Sn^{2+}]}{[Pb^{2+}]}$$ when this cell attains equilibrium is __________.
(Given: E$$^0_{Sn^{2+}|Sn} = -0.14$$ V, E$$^0_{Pb^{2+}|Pb} = -0.13$$ V, $$\frac{2.303RT}{F} = 0.06$$)

We have the cell

$$\text{Sn(s)}\;|\;\text{Sn}^{2+}(1\,\text{M})\;||\;\text{Pb}^{2+}(1\,\text{M})\;|\;\text{Pb(s)}$$

The standard reduction potentials are given as

$$E_{\,\text{Sn}^{2+}/\text{Sn}}^{0}= -0.14\ \text{V},\qquad E_{\,\text{Pb}^{2+}/\text{Pb}}^{0}= -0.13\ \text{V}.$$

The electrode with the higher (more positive) reduction potential acts as the cathode. Hence

$$\text{Cathode (reduction)}:\quad \text{Pb}^{2+}+2e^- \rightarrow \text{Pb},\qquad E_{\text{red}}^{0}= -0.13\ \text{V}$$

$$\text{Anode (oxidation)}:\quad \text{Sn}\rightarrow \text{Sn}^{2+}+2e^-,\qquad E_{\text{ox}}^{0}= +0.14\ \text{V}$$

Using $$E_{\text{cell}}^{0}=E_{\text{cathode}}^{0}-E_{\text{anode}}^{0}$$ we get

$$E_{\text{cell}}^{0}=(-0.13\ \text{V})-(-0.14\ \text{V})=+0.01\ \text{V}.$$

The overall cell reaction is

$$\text{Sn(s)}+\text{Pb}^{2+}(aq)\rightleftharpoons \text{Sn}^{2+}(aq)+\text{Pb(s)}.$$

For this reaction the reaction quotient is

$$Q=\frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]}.$$

The Nernst equation (stated for an $$n$$-electron process) is

$$E_{\text{cell}} = E_{\text{cell}}^{0}-\frac{0.06}{n}\log Q,$$

where $$n=2$$ for the transfer of two electrons.

At equilibrium $$E_{\text{cell}}=0$$, so

$$0 = E_{\text{cell}}^{0}-\frac{0.06}{2}\log Q_{\text{eq}}.$$

Rearranging,

$$\frac{0.06}{2}\log Q_{\text{eq}} = E_{\text{cell}}^{0}.$$

Substituting $$E_{\text{cell}}^{0}=0.01\ \text{V}$$,

$$\log Q_{\text{eq}} = \frac{2\times 0.01}{0.06} = \frac{0.02}{0.06} = \frac13 = 0.333\ldots$$

So

$$Q_{\text{eq}} = 10^{0.333\ldots} \approx 2.15.$$

Because $$Q_{\text{eq}} = \dfrac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]}$$, we have

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

Hence, the correct answer is Option 2.15.