Electrochemistry

 56) Sn²⁺(aq) + Pb(s) Sn(s) + Pb²⁺(aq)

       (a) [Sn²⁺] = 1.00 M          [Pb²⁺] = ?          E = 0.22 V

                Sn²⁺(aq) + 2e^- Sn(s)                        E°_red = -0.136 V

                               Pb(s) Pb²⁺(aq) + 2e^-         E°_red = -0.126 V
             __________________________________________________

             Sn²⁺(aq) + Pb(s) Sn(s) + Pb²⁺(aq)     E° = -0.010 V

             E° = E°_red(reduction) - E°_red(oxidation)

             E° = -0.136 V - (-0.126 V) = -0.010 V

             E = E° - 0.0592 V/n log Q

             0.22 V = -0.010 V - 0.0592 V/2 x log([Pb²⁺]/1.00 m)

             [Pb²⁺] = 2 x 10^-8 M

       (b) [SO₄^2-] = 1.00 M

             PbSO4(s) Pb²⁺(aq) + SO₄^2-(aq)

             K_sp = [Pb²⁺][SO₄^2-] = 2 x 10^-8 x 1.00 = 2 x 10^-8