Electrochemistry

 94) [Cu⁺] = 2.5 M          [I^-] = 3.5 M          T = 298 K

       (a)

                        2Cu(s) 2Cu⁺(aq) + 2e             E°_red = 0.521 V

                    I₂(s) + 2e^- 2I^-(aq)                         E°_red = 0.536 V
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             2Cu(s) + I₂(s) 2Cu⁺(aq) + 2I^-(aq)      E° = 0.015 V

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

             E° = 0.536 V - 0.521 V = 0.015 V

             E = E° - 0.0592 V/n log Q

             E = 0.015 V - 0.0592 V/2 x log([Cu⁺]²[I^-]²)

             E = -0.041 V

       (b) Because the cell emf is negative, the cell would be spontaneous in the reverse direction, therefore I₂(s) is the anode.

       (c) No, because under standard condtions, Cu(s) is being oxidized and is the anode.

       (d) [Cu⁺] = 1.4 M          [I^-] = ?          E = 0

             E = E° - 0.0592 V/n log Q

             0 = 0.015 V - 0.0592 V/2 x log([Cu⁺]²[I^-]²)

             [I^-] = 1.3 M