1. Hour Exam II Wednesday, March 15 7:00 – 9:00 pm 103 Mumford HallAQG, AQI AllenAQJ BlairAQF 150 Animal Science FisherAQB, AQC KoysAGD PearsonAQA conflict exam 4:30 – 6:30 162 Noyes Review session 7:00 – 9:00, Monday 124 Burrill Hall

2. “minimal” galvanic cells only need reactants Zn H+H+  o = 0.34 V Zn 2+ (aq) + 2e -  Zn(s)  o = -0.76 V 2H + (aq) + 2e -  H 2 (g)  o = 0.00 V reduction b) oxidation Zn   Zn 2+ + 2e -  = 0.76 V 2H + (aq)  0 =.76 V  = very large Q = a) oxidation Cu Cu 2+ (aq) + 2e -  Cu(s) + Zn (s)  Zn 2+ (aq) + H 2 (g) 0

3. Electrolytic Cells  cell > 0  cell < 0  G < 0 spontaneous galvanic cell  G > 0 non-spontaneous electrolytic cell 2 H 2(g)  G = -474 kJ redox reaction: O spontaneous H0 0 + O 2(g)  2 H 2 O (l)  1+  2-

4. Electrolytic Cells 2 H 2 O (l)  G = 474 kJ oxygen half-cell:H2OH2O H 2 O  O 2 a) oxidation b) reduction a) anode b) cathode 2+ 4 H + + 4 e -  O2 O2 reaction  2 H 2(g) + O 2(g)

5. Electrolytic Cells 2 H 2 O (l)  2 H 2(g) + O 2(g)  G = 474 kJ hydrogen half-cell: H 2 O  H 2 H2O H2O reduction reactioncathode + 2e -  H2H2 + OH - 22

6. Electrolytic Cells 2 H 2 O oxidation: anode reduction: cathode 2( ) ____________________________________ 6 H 2 O 2 H 2 O   O2 O2 + 2 H 2 + 4H + + 4 OH - O2O2 + 2 H 2  O2O2 + 4 H + + 4 e - + 2e -  H2H2 + 2 OH -

7. Electrolysis of water oxidation reduction 2H 2 O  O 2 + 4H + + 4e - 4H 2 O + 4e -  2H 2 + 4OH - Pt electrodes battery + - e-e- e-e- anode cathode acid base 1 mol gas 2 mol gas

8. Electrolysis of water 2.5 amp Power source  current = chargemol e - mol product gram product current and time A(C/s) x s x 1mol e - 96,500 C x mol product mol e- x g product mol product 3.2 g O 2 amperes (A) = coulombs/sec (C/s)

9. Electrolysis of water chargemol e - mol product gram product current and time 2 H 2 O  O 2 + 4 H + + 4 e - 2.5 A, 3.2 g O 2 (C/s) x s 2.5 A 1414 32.0 g/mol 3.2 x mol e - C x mol O 2 mol e - mol O 2 x g O 2 = g O 2 1 mol e - 96500 C

10. Electrolysis of water 2 H 2 O  O 2 + 4 H + + 4 e - 2.5 A, 3.2 g O 2 3.2 g O 2 x 2.5 C x s 15440 s x 1 mol O 2 x 32 g O 2 4 mol e - x 1 mol O 2 96500 C = 1 mol e - 38600 C = 38600 C s = 15440 s 1 min x 60 s 1 hr 60 min = 4.3 hr s

11. Electroplating Cu 2+ + 2e -  Cu anodecathode oxidationreduction Cu (s) Cu 2+ (aq)  Cu 2+ (aq) + 2e -  Cu (s)  o = 0.34 V

12. Electroplating anodecathode oxidation reduction Cu (s)  Cu 2+ (aq) + 2e - Cu 2+ (aq) + 2e -  Cu (s) 0.75 A atomic mass of Cu = for 25 min.deposits 0.37 g Cu

13. Electroplating Cu 2+ (aq) + 2e -  Cu (s) 0.75 A for 25 min deposits 0.37 g Cu A (C) s x sx 1 mol e - 96500 C x 1 mol Cu 2 mol e - x g Cu mol Cu

14. Electroplating 0.75(C) s x 25 min 0.37 g Cu = 1.125 x 10 3 C 1.125 x 10 3 C x 1 mol e - 96500 C = 1.17 x 10 -2 mol e - 1.17 x 10 -2 mol e - x 1mol Cu 2mol e - =5.83 x 10 -3 mol Cu = 63.5 g mol 5.8x10 -3 mol Cu atomic mass Cu x 60s min

15. Electrolysis of salt solutions CuBr 2 Cu 2+ Br - +2e -  Cu 2  Br 2 + 2e - 0.153 -1.083 -.934 H 2 O+2e -  H 2 +2OH - -0.83 2H 2 O  O 2 +2H + +4e - -1.23 -1.91 -1.08 -2.06

16. Electrolysis of salt solutions CaI 2 Ca 2+ I-I- +2e -  Ca 2  I 2 + 2e - -2.86 -0.535 -3.40 H 2 O+2e -  H 2 +2OH - -0.83 2H 2 O  O 2 +2H + +4e - -1.23 -1.36 -4.09 -2.06

17. Electrolysis of salt solutions NaCl Na + Cl - + e -  Na 2  Cl 2 + 2e - -2.71 -1.36 -4.07 H 2 O+2e -  H 2 +2OH - -0.83 2H 2 O  O 2 +2H + +4e - -1.23 -2.19 -3.94 -2.06 overvoltage