Question 21.10: Predicting the Electrolysis Products of Aqueous Salt Solutio......

Plan We identify the reacting ions and compare their electrode potentials with those of water, taking the additional 0.4–0.6 V overvoltage into account. The reduction half-reaction with the less negative electrode potential occurs at the cathode, and the oxidation half-reaction with the less positive electrode potential occurs at the anode.

Solution
(a)         K^+(aq)  +  e^−  ⟶  K(s)                         E°  =  −2.93  V

                2H_2O(l)  +  2e^−  ⟶  H_2(g)  +  2OH^−(aq)               E  =  −0.42  V
Despite the overvoltage, which makes E for the reduction of water between –0.8 V and –1.0 V, H_2O is still easier to reduce than K^+, so H_2(g) forms at the cathode.

               2Br^−(aq)  ⟶  Br_2(l)  +  2e^−                       E°  =  1.07  V
               2H_2O(l)  ⟶  O_2(g)  +  4 H^+(aq)  +  4e^−               E  =  0.82  V

Because of the overvoltage, which makes E for the oxidation of water between 1.2 V and 1.4 V, Br^– is easier to oxidize than water, so Br_2(l) forms at the anode (see photo (Fig 21.10)).

(b)         Ag^+(aq)  +  e^−  ⟶  Ag(s)                      E°  =  0.80  V
                2H_2O(l)  +  2e^−  ⟶  H_2(g)  +  2OH^−(aq)              E  =  −0.42  V

As the cation of an inactive metal, Ag^+ is a better oxidizing agent than H_2O, so Ag(s) forms at the cathode. NO_3^− cannot be oxidized, because N is already in its highest (+5) oxidation state. Thus, O_2(g) forms at the anode:
               2H_2O(l)  ⟶  O_2(g)  +  4H^+(aq)  +  4e^−

(c)         Mg^{2+}(aq)  +  2e^−  ⟶  Mg(s)                       E°  =  −2.37  V
               2H_2O(l)  +  2e^−  ⟶  H_2(g)  +  2OH^−(aq)                E  =  −0.42  V

Like K^+ in part (a), Mg^{2+} cannot be reduced in the presence of water, so H_2(g) forms at the cathode. The SO_4^{2−} ion cannot be oxidized because S is in its highest (+6) oxidation state. Thus, H_2O is oxidized, and O_2(g) forms at the anode: