A voltaic cell is constructed that is based on the following reaction: $$ \mathrm{Sn}^{2+}(a q)+\mathrm{Pb}(s) \longrightarrow \mathrm{Sn}(s)+\mathrm{Pb}^{2+}(a q) $$ (a) If the concentration of \(\mathrm{Sn}^{2+}\) in the cathode half-cell is \(1.00 M\) and the cell generates an emf of \(+0.22 \mathrm{~V},\) what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell? \((\mathbf{b})\) If the anode half-cell contains \(\left[\mathrm{SO}_{4}^{2-}\right]=1.00 M\) in equilibrium with \(\mathrm{PbSO}_{4}(s),\) what is the \(K_{s p}\) of \(\mathrm{PbSO}_{4} ?\)

The Nernst equation relates the concentration of the species involved in the reaction to the cell potential (emf): \(E_{cell} = E_{cell}^{0} - \dfrac{0.0592}{n} \log_{10}{\dfrac{[\mathrm{Pb}^{2+}]}{[\mathrm{Sn}^{2+}]}}\) We need to substitute the given values and calculate \([\mathrm{Pb}^{2+}]\), i.e., the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell.

We have: \(E_{cell}= +0.22\,\mathrm{V}\), \([\mathrm{Sn}^{2+}] = 1.00 \,\mathrm{M}\). The standard cell potential, \(E_{cell}^{0}\), can be found from the standard reduction potentials of \(\mathrm{Sn}^{2+}\) and \(\mathrm{Pb}^{2+}\). Since both ions have a +2 charge, \(n=2\). For the reaction \(\mathrm{Sn}^{2+} + 2e^{-} \longrightarrow \mathrm{Sn}(s)\), the standard reduction potential is \(E^{0}(\mathrm{Sn}) = -0.14\,\mathrm{V}\). For the reaction \(\mathrm{Pb}^{2+} + 2e^{-} \longrightarrow \mathrm{Pb}(s)\), the standard reduction potential is \(E^{0}(\mathrm{Pb}) = -0.13\,\mathrm{V}\). The standard cell potential can be determined as: \(E_{cell}^{0}= E^{0}(\mathrm{Sn}) - E^{0}(\mathrm{Pb}) = -0.14-(-0.13)= -0.01\,\mathrm{V}\). Now, substitute these values into the Nernst equation: \(0.22 = -0.01 - \dfrac{0.0592}{2} \log_{10}{\dfrac{[\mathrm{Pb}^{2+}]}{1}}\)

Now, we can solve for \([\mathrm{Pb}^{2+}]\): \(0.23= \dfrac{0.0592}{2} \log_{10}{[\mathrm{Pb}^{2+}]}\) \(7.776=\log_{10}{[\mathrm{Pb}^{2+}]}\) \(10^{7.776}=[\mathrm{Pb}^{2+}]\) Therefore, \([\mathrm{Pb}^{2+}]=6.0\)x\(10^{7}\,\mathrm{M}\).

The reaction for \(\mathrm{PbSO}_{4}(s)\) dissolution is: \(\mathrm{PbSO}_{4}(s) \longleftrightarrow \mathrm{Pb}^{2+}(a q) + \mathrm{SO}_{4}^{2-}(a q)\) Now we can write the expression for the solubility product constant, \(K_{sp}\): \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{SO}_{4}^{2-}]\)

We have: \([\mathrm{Pb}^{2+}] = 6.0\)x\(10^{7}\,\mathrm{M}\) and \([\mathrm{SO}_{4}^{2-}] = 1.00\,\mathrm{M}\). Now, plug in the values in the expression for \(K_{sp}\): \(K_{sp} = (6.0\times10^{7})(1.00)\) Hence, \(K_{sp} = 6.0\)x\(10^{7}\).