1. Electrolysis of ionic compounds Putting electrical energy in to force the nonspontaneous change

2. electrolysis Opposite process to electrochemical (voltaic) cells Electrical energy is put into the reaction The nonspontaneous reactions occur Useful industrial process for the production of “reduced metals” electroplating

3. Molten ionic compounds Molten = melted (NOT dissolved!) Only species present are the cation and anion The cation (metal ion) will be reduced into a neutral metal atom at the cathode – Ex: Fe 2+ + 2e - → Fe The anion (nonmetal ion) will be oxidized to neutral at the anode – Ex: 2 Cl - → Cl 2 + 2e -

4. Aqueous ionic compounds Besides the cation and anion, water is present as well Water can be reduced instead of the metal ion at the cathode: – Ex: 2H 2 O (l) + 2e - → H 2(g) + 2 OH - (aq) E o = -0.83V Water can be oxidized instead of the nonmetal ion at the anode: – Ex: 2H 2 O (l) → O 2(g) + 4e - + 4 H + (aq) E o = 1.23V

5. In an aqueous solution of an ionic compound, the half reaction with the greater (more positive) reduction potential (E o ) will be reduced. Ex: electrolysis of aqueous CoBr 2 Present and able to be reduced: Co 2+, H 2 O Co 2+ + 2e - → Co E o = -0.28V 2H 2 O (l) + 2e - → H 2(g) + 2 OH - (aq) E o = -0.83V Co 2+ will be reduced instead of water

6. In an aqueous solution of an ionic compound, the half reaction with the greater (more positive) reduction potential (E o ) will be reduced. Ex: electrolysis of aqueous BaBr 2 Present and able to be reduced: Ba 2+, H 2 O Ba 2+ + 2e - → Ba E o = -2.90V 2H 2 O (l) + 2e - → H 2(g) + 2 OH - (aq) E o = -0.83V H 2 O will be reduced instead of Ba 2+

7. In an aqueous solution of an ionic compound, the half reaction with the lower (more negative) reduction potential (E o ) will be oxidized. Ex: electrolysis of aqueous CoBr 2 Present and able to be oxidized: Br 1-, H 2 O 2 Br 1- → Br 2 + 2e - E o = 1.07V 2H 2 O (l) → O 2(g) + 4e - + 4 H + (aq) E o = 1.23V Br 1- will be oxidized instead of water

8. In an aqueous solution of an ionic compound, the half reaction with the lower (more negative) reduction potential (E o ) will be oxidized. Ex: electrolysis of aqueous CoF 2 Present and able to be oxidized: F 1-, H 2 O 2 F 1- → F 2 + 2e - E o = 2.87V 2H 2 O (l) → O 2(g) + 4e - + 4 H + (aq) E o = 1.23V H 2 O will be oxidized instead of F 1-