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
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             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) [SO42-] = 1.00 M

 

             PbSO4(s) Pb2+(aq) + SO42-(aq)

             Ksp = [Pb2+][SO42-] = 2 x 10-8 x 1.00 = 2 x 10-8