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^-                                         E°_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)                                  E°_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