Question 19.9: Predicting the Products of Electrolysis Reactions Predict th...
Predicting the Products of Electrolysis Reactions
Predict the half-reaction occurring at the anode and the half-reaction occurring at the cathode for electrolysis for each reaction.
(a) a mixture of molten AlBr_3 and MgBr_2
(b) an aqueous solution of LiI
| (a) In the electrolysis of a molten salt, the anion is oxidized and the cation is reduced. However, this mixture contains two cations. Start by writing the possible oxidation and reduction halfreactions that might occur. Because Br^{-} is the only anion, write the equation for its oxidation, which occurs at the anode.
At the cathode, both the reduction of Al^{3+} and the reduction of Mg^{2+} are possible. The one that actually occurs is the one that occurs more easily. Since the reduction of Al^{3+} has a more positive electrode potential in aqueous solution, this ion is more easily reduced. Therefore, the reduction of Al^{3+} occurs at the cathode. |
Oxidation: 2 Br ^{-}(l) \longrightarrow Br _2(g)+2 e ^{-} Reduction: \underset{\overset{\uparrow }{\text{Reduction that actually occurs (more positive potential)}} }{Al ^{3+}} (l) +3 e^{-} \longrightarrow Al (s) E° = –1.66 V (for aqueous solution)
Mg ^{2+}(l) +2 e^{-} \longrightarrow Mg (s) E° = –2.37 V (for aqueous solution) |
| (b) Since LiI is in an aqueous solution, two different oxidation half-reactions are possible at the anode, the oxidation of I^- and the oxidation of water. Write half-reactions for each including the electrode potential. Remember to use the electrode potential of water under conditions in which [H^+]=10^{-7} M.
Since the oxidation of I^- has the more negative electrode potential, it is the half-reaction to occur at the anode. Similarly, write half-reactions for the two possible reduction half-reactions that might occur at the cathode, the reduction of Li^+ and the reduction of water. Since the reduction of water has the more positive electrode potential (even when considering overvoltage, which would raise the necessary voltage by about 0.4–0.6 V), it is the halfreaction to occur at the cathode. |
Oxidation: \underset{\overset{\uparrow }\text{{Oxidation that actually occurs (more negative potential)}} }{2 I ^{-}} (aq) \longrightarrow I_2 (l)+ 2e^{-} E° = 0.54 V Oxidation: 2 H_2O(l) \longrightarrow O_2 (g)+ 4 H^+ (aq)+ 4 e^{-} E° =0.82 V ([H^+] = 10^{−7} M) Reduction: 2 Li^{+}(aq) +2e^{-} \longrightarrow 2 Li (s) E°= –3.04 V Reduction: \underset{\overset{\uparrow }{\text{Reduction that actually occurs (more positive potential)}} }{2 H_2O(l)} + 2 e^{-} \longrightarrow H_2 (g)+2 OH^-(aq) E = −0.41 V ([OH^−] = 10^{−7} M) |