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