Lecture 41 - Electrochemistry V

Review Galvanic Cells: Reaction is spontaneous E o cell > 0 The “product” is an electrical current Some can be reversed (recharged)

Electrolytic Cells l Reaction is not spontaneous l E o cell < 0 l Used for producing a chemical product

Electrolysis of Water We want: 2 H 2 O (l)  2 H 2(g) + O 2(g) 2 H 2 O (l)  O 2(g) + 4 H + (aq) + 4 e - E o = V 2 H 2 O (l) + 2 e -  H 2(g) + 2 OH - (aq) E o = V E o cell = V i.e. non-spontaneous!

Electrolysis of Water H 2 O (l) + salt

Electrolysis of Water H 2 O + 2 e -  H 2(g) + 2 OH - H 2 O (l) + salt cathode anode Note reversal of polarity! battery 2 H 2 O  O 2(g) + 4 H e -

- + anode cathode CuCl 2(aq) Electrowinning of Metals

at the cathode: Cu +2  Cu (s) ?? or H 2 O  H 2(g) ??

Rule of thumb… The cathode reaction with the most positive reduction potential will occur preferentially

Cu +2 (aq) + 2 e - Ž Cu (s) E o = V 2 H 2 O (l) + 2 e - Ž H 2(g) + 2 OH - E o = V

at the anode: Cl -  Cl 2(g) ?? or H 2 O  O 2(g) ??

Same rule of thumb… The anode reaction with the most positive oxidation potential will occur preferentially

Same rule of thumb… 2 Cl - (aq) Ž Cl 2(g) + 2 e - E o = V 2 H 2 O (l) Ž O 2(g) + 4 H e - E o = V Thus, we predict water gets oxidized at the anode.

Overpotential E required > E o cell for electrolysis In this case, Cl - is oxidized

- + anode cathode CuSO 4(aq) 2 Cl -  Cl 2(g) + 2 e - Cu +2 (aq) + 2 e -  Cu (s) Electrowinning of Metals

Overall Reaction: Cu +2 (aq) + 2 Cl - (aq) Ž Cu (s) + Cl 2(g) E o cell = V

Electrorefining of Metals - + Impure Cu Pure Cu Cu

Making Fluorine, F 2(g) Electrolysis of HF  2 KF (l) oxidation: 2 F -  F 2(g) + 2 e - reduction: 2 H e -  H 2(g)

Making Chlorine, Cl 2(g) Electrolysis of brine (NaCl (aq) ) oxidation: 2 Cl - (aq)  Cl 2(g) + 2 e - reduction: 2 H 2 O (l) + 2 e -  H 2(g) + 2 OH - (aq)

Overall, 2 NaCl (aq) + 2 H 2 O (l)  Cl 2(g) + H 2(g) + 2 NaOH (aq) disinfection, plastics (PVC) cleaners, paper, etc

Manufacture of Aluminum Why not just electrowin Al +3 (aq) ? Al +3 (aq) + 3 e -  Al (s) E o = V 2 H 2 O (l) + 2 e -  H 2(g) + 2 OH - (aq) E o = V (reduction of water is favored)

Manufacture of Aluminum Bauxite = impure Al 2 O 3(s) m.p. = 2045 o C Solution? 1.Lower the melting point 2.Perform electrolysis

Manufacture of Aluminum Al 2 O 3(s) + AlF 3(s) + 3 NaF (s) Na 3 AlF 6, cryolite m.p.  900 o C

Manufacture of Aluminum reduction: AlF x O y 3-x-2y + n e -  Al (l) + ? oxidation: AlF x O y 3-x-2y  O 2(g) + n e - + ? sinks!oxidizes C electrodes!