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 \(1.00 \mathrm{M}\) and the cell generates an emf of \(+0.22 \mathrm{~V},\) what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell? (b) If the anode half-cell contains \(\left[\mathrm{SO}_{4}^{2-}\right]=1.00 \mathrm{M}\) in equilibrium with \(\mathrm{PbSO}_{4}(s),\) what is the \(K_{s p}\) of \(\mathrm{PbSO}_{4} ?\)

In order to use the Nernst equation, we must first identify the half reactions for the voltaic cell. For this cell, the half-reactions can be determined as: Anode: Pb(s) -> Pb2+(aq) + 2e- Cathode: Sn2+(aq) + 2e- -> Sn(s) Now that we have the half-reactions, we can proceed to the next step.

By looking up the values in a table of reduction potentials, we can find: E°(Pb2+/Pb) = -0.126 V E°(Sn2+/Sn) = -0.136 V

We can find the standard cell potential by adding the reduction potentials for the anode and cathode: E°(cell) = E°(cathode) - E°(anode) = -0.136 - (-0.126) V = -0.010 V

The Nernst equation relates the cell potential, standard cell potential, concentrations, and temperature: E(cell) = E°(cell) - \(\frac{RT}{nF}\) lnQ Where R is the gas constant, T is the temperature, n is the number of electrons transferred, F is the Faraday's constant, and Q is the reaction quotient. In this case, E(cell) = 0.22 V, E°(cell) = -0.010 V, n = 2 (as per the half-reactions), and Q = \(\frac{[Pb^{2+}]}{[Sn^{2+}]}\). Assuming the temperature is 298 K, we can plug the values into the equation: 0.22 V = -0.010 V - \(\frac{8.314\ \text{J}\ (\text{mol} \ \text{K})^{-1}\times 298\ \text{K}}{2\times 96485\ \text{C}\ \text{mol}^{-1}} \) ln\(\frac{[Pb^{2+}]}{1.00}\) Solve for [Pb²⁺]: [Pb²⁺] ≈ 0.0717 M

For PbSO₄ dissolved in water, the equilibrium reaction can be written as: PbSO4(s) -−−> Pb²⁺ (aq) + SO4²⁻ (aq) The Ksp expression for PbSO₄ is: Ksp = [Pb²⁺] [SO₄²⁻]

Using the determined concentration of Pb²⁺ (0.0717 M) and the provided concentration of SO₄²⁻ (1.00 M), we can calculate the Ksp value for PbSO₄: Ksp = (0.0717)(1.00) = 0.0717 Therefore, the Ksp of PbSO₄ is approximately 0.0717.