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Hour Exam II

Hour Exam II. Wednesday, March 15. 7:00 – 9:00 pm. 103 Mumford Hall AQG, AQI Allen AQJ Blair AQF. 150 Animal Science Fisher AQB, AQC Koys AGD Pearson AQA. conflict exam 4:30 – 6:30 162 Noyes. Review session 7:00 – 9:00, Monday 124 Burrill Hall. “minimal” galvanic cells.

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Hour Exam II

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  1. Hour Exam II Wednesday, March 15 7:00 – 9:00 pm 103 Mumford Hall AQG, AQI Allen AQJ Blair AQF 150 Animal Science Fisher AQB, AQC Koys AGD Pearson AQA 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 Cu Cu2+(aq) + 2e- Cu(s) o = 0.34 V a) oxidation Zn Zn2+(aq) + 2e- Zn(s) Zn Zn2+ + 2e-  = 0.76 V o = -0.76 V b) oxidation H+ 2H+(aq) + 2e- H2(g) o = 0.00 V reduction  Zn2+ (aq) + H2(g) 2H+ (aq) + Zn (s) very large 0 = .76 V Q = 0  =

  3. Electrolytic Cells spontaneous cell > 0 G < 0 galvanic cell electrolytic cell cell < 0 G > 0 non-spontaneous  2 H2O(l) G = -474 kJ 2 H2(g) + O2(g) spontaneous redox reaction:  1+ H 0  2- O 0

  4. Electrolytic Cells  2 H2(g) + O2(g) G = 474 kJ 2 H2O(l) oxygen half-cell: H2O  O2 H2O  O2 2 + 4 H+ + 4 e- a) oxidation b) reduction reaction a) anode b) cathode

  5. Electrolytic Cells 2 H2O(l)  2 H2(g) + O2(g) G = 474 kJ H2O  H2 hydrogen half-cell: 2 H2O  H2 + OH- 2 + 2e- reduction reaction cathode

  6. Electrolytic Cells  O2 + 4 H+ + 4 e- anode 2 H2O oxidation: 2( )  H2 + 2 OH- cathode 2 H2O + 2e- reduction: ____________________________________  O2 6 H2O + 2 H2 + 4H+ + 4 OH- O2 2 H2O  + 2 H2

  7. Electrolysis of water - + battery e- Pt electrodes e- anode cathode reduction oxidation 2H2O  O2+ 4H++ 4e- 4H2O + 4e- 2H2 + 4OH- base acid 2 mol gas 1 mol gas

  8. Electrolysis of water 2.5 amp Power source  3.2 g O2 current = amperes (A) = coulombs/sec (C/s) current and time charge mol e- mol product gram product A(C/s) x x mol product mol e- x g product mol product s x 1mol e- 96,500 C

  9. x mol O2 x g O2 mol e- mol O2 1 mol e- 96500 C Electrolysis of water 2 H2O  O2 + 4 H++ 4 e- 2.5 A, 3.2 g O2 current and time charge mol e- mol product gram product x mol e- C (C/s) x s = g O2 1 4 3.2 2.5 A 32.0 g/mol

  10. Electrolysis of water 2 H2O  O2 + 4 H++ 4 e- 2.5 A, 3.2 g O2 38600 C 1 mol O2 x 32 g O2 4 mol e- x 1 mol O2 96500 C = 1 mol e- 3.2 g O2 x 2.5 C x s s s = 15440 s = 38600 C 15440 s x 1 min x 60 s 1 hr 60 min = 4.3 hr

  11. Electroplating anode cathode oxidation reduction  Cu(s) Cu(s) Cu2+(aq) + 2e-  Cu2+(aq) + 2e- Cu2+ + 2e- Cu o = 0.34 V

  12. Electroplating anode cathode reduction oxidation Cu(s) Cu2+(aq) + 2e- Cu2+(aq) + 2e-Cu(s) 0.75 A for 25min. deposits 0.37g Cu atomic mass of Cu =

  13. Electroplating Cu2+(aq) + 2e-Cu(s) 0.75 A for 25min deposits 0.37g Cu A (C) s x 1 mol Cu 2 mol e- x g Cu mol Cu x s x 1 mol e- 96500 C

  14. Electroplating x 25 min x 60s min = 1.125 x 103 C 0.75(C) s x1 mol e- 96500 C 1.125 x 103 C = 1.17 x 10-2 mol e- x 1mol Cu 2mol e- 1.17 x 10-2 mol e- =5.83 x 10-3mol Cu 0.37 g Cu = 63.5 g mol atomic mass Cu 5.8x10-3mol Cu

  15. Electrolysis of salt solutions  Br2 + 2e- 2 CuBr2 +2e- Cu Cu2+ Br- -1.083 0.153 -.934 -1.91 -1.08 2H2O  O2+2H++4e- -1.23 H2O+2e- H2+2OH- -0.83 -2.06

  16. Electrolysis of salt solutions  I2 + 2e- 2 CaI2 +2e- Ca Ca2+ I- -0.535 -2.86 -3.40 -1.36 -4.09 2H2O  O2+2H++4e- -1.23 H2O+2e- H2+2OH- -0.83 -2.06

  17. Electrolysis of salt solutions  Cl2 + 2e- 2 NaCl + e- Na Na+ Cl- -1.36 -2.71 -4.07 -2.19 -3.94 overvoltage 2H2O  O2+2H++4e- -1.23 H2O+2e- H2+2OH- -0.83 -2.06

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