Electrochemistry

 

 26) AuBr4-(aq) + 3e- Au(s) + 4Br-(aq)                     E°red = -0.858 V

       Eu3+(aq) + e- Eu2+(aq)                                        E°red = -0.43 V
             
       IO-(aq) + H2O(l) + 2e- I-(aq) + 2OH-(aq)          E°red = +0.49 V

       Sn2+(aq) + 2e- Sn(s)                                            E°red = -0.14 V

 

       (a) Because E°red( IO-) > E°red( AuBr4-), IO-(aq) will be reduced and Au(s) will be oxidized.

 

            Au(s) + 4Br-(aq) AuBr4-(aq) + 3e-

            IO-(aq) + H2O(l) + 2e- I-(aq) + 2OH-(aq)

 

                                             2Au(s) + 8Br-(aq) 2AuBr4-(aq) + 6e-                                         red = -0.858 V

                                  3IO-(aq) + 3H2O(l) + 6e- 3I-(aq) + 6OH-(aq)                             E°red = 0.49 V
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           2Au(s) + 8Br-(aq) + 3IO-(aq) + 3H2O(l) 2AuBr4-(aq) + 3I-(aq) + 6OH-(aq)     E°red = 1.35 V

 

            E° = E°red(reduction) - E°red(oxidation) = 0.49 V - (-0.858 V) = 1.35 V

             

       (b) Because E°red( Sn2+) > E°red( Eu3+), Sn2+(aq) will be reduced and Eu2+(aq) will be oxidized.

 

             Sn2+(aq) + 2e- Sn(s)

             Eu2+(aq) Eu3+(aq) + e-

 

                       Sn2+(aq) + 2e- Sn(s)                                  red = -0.14 V

                              2Eu2+(aq) 2Eu3+(aq) + 2e-         E°red = -0.43 V
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            Sn2+(aq) + 2Eu2+(aq) Sn(s) + 2Eu3+(aq)      E°red = 0.29 V

 

            E° = E°red(reduction) - E°red(oxidation) = -0.14 - (-0.43 V) = 0.29 V