(a) Compute the voltage at \(25^{\circ} \mathrm{C}\) of an electrochemical cell consisting of pure lead immersed in a \(5 \times 10^{-2} M\) solution of \(\mathrm{Pb}^{2+}\) ions and pure tin in a \(0.25 M\) solution of \(\mathrm{Sn}^{2+}\) ions. (b) Write the spontaneous electrochemical reaction.

For a galvanic cell with a pure lead electrode in a \(\mathrm{Pb}^{2+}\) solution and a pure tin electrode in a \(\mathrm{Sn}^{2+}\) solution, we have the following half-reactions: Oxidation half-reaction (anode): \(\mathrm{Pb}(s) \rightarrow \mathrm{Pb}^{2+}(aq) + 2e^-\) Reduction half-reaction (cathode): \(\mathrm{Sn}^{2+}(aq) + 2e^- \rightarrow \mathrm{Sn}(s)\)

To calculate the standard cell potential, we first need to look up the standard reduction potentials of the two half-reactions: \(\mathrm{Pb}^{2+}(aq) + 2e^- \rightarrow \mathrm{Pb}(s)\), \(E^{\circ}_\mathrm{Pb} = -0.126\,V\) \(\mathrm{Sn}^{2+}(aq) + 2e^-\rightarrow \mathrm{Sn}(s)\), \(E^{\circ}_\mathrm{Sn} = -0.136\,V\) The standard cell potential (\(E^{\circ}_{\text{cell}}\)) is given by the difference between the standard reduction potential of the cathode and the anode: \(E^{\circ}_{\text{cell}} = E^{\circ}_\mathrm{Sn} - E^{\circ}_\mathrm{Pb} = (-0.136 \,\text{V}) - (-0.126\, \text{V}) = -0.010\, \text{V}\)

The Nernst equation allows us to find the non-standard cell potential (\(E_{\text{cell}}\)) at a given temperature and concentration: \(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \dfrac{RT}{nF} \ln \dfrac{[\text{ox}_2][\text{red}_1]}{[\text{ox}_1][\text{red}_2]}\) where \(R=8.314\, \text{J/mol} \cdot \text{K}\) is the gas constant, \(T=298\, \text{K}\) is the absolute temperature, \(n=2\) is the number of electrons transferred, and \(F = 96,485\, \text{C/mol}\) is the Faraday constant. At \(25^{\circ}\mathrm{C}\), we have \([\mathrm{Pb}^{2+}] = 5\times 10^{-2} \,\text{M}\) and \([\mathrm{Sn}^{2+}] = 0.25\, \text{M}\). Plugging this into the Nernst equation: \(E_{\text{cell}} = -0.010\, \text{V} - \dfrac{8.314\, \text{J/mol} \cdot \text{K} \cdot 298\, \text{K}}{2 \cdot 96,485\, \text{C/mol}} \ln \dfrac{(0.25\,\text{M})(0)}{5\times 10^{-2}\, \text{M} \cdot (0)}\) However, since we have zero concentration values in the equation, the concentration term for the Nernst equation will simplify. The lead oxide concentration, \([\text{ox}_2]\), is zero, and the tin electrode is a solid and doesn't appear in the Nernst equation: \(E_{\text{cell}} = -0.010\, \text{V} - \dfrac{8.314\, \text{J/mol} \cdot \text{K} \cdot 298\, \text{K}}{2 \cdot 96,485\, \text{C/mol}} \ln \dfrac{(0.25\,\text{M})}{5\times 10^{-2}\, \text{M}}\) Calculating the result gives: \(E_{\text{cell}} ≈ -0.058\, \text{V}\)

Since \(E_{\text{cell}}\) is negative, the reaction is non-spontaneous in the given direction. Therefore, the spontaneous reaction is the reverse of the overall reaction: Spontaneous electrochemical reaction: \(\mathrm{Pb}(s) + \mathrm{Sn}^{2+}(aq) \rightarrow \mathrm{Pb}^{2+}(aq) + \mathrm{Sn}(s)\)