1. Intersection 15: The End 12/12/06

2. The Final Exam Questions 1.Electron configurations, atomic structure, periodic trends, atomic orbitals. 2.Quantitative calculations. 3.Lewis & VSEPR Structures and polarity. 4.Thermochemistry 5.Thermochemistry 6.Thermochemistry 7.Acid/Base chemistry 8.Acid/Base chemistry 9.Oxidation number 10.Redox reactions 11.Redox reactions 12.Electrochemistry

3. Research at the University of Michigan A unique opportunity in your life Women in Science and Engineering (WISE) –http://www.wise.umich.edu/http://www.wise.umich.edu/ Undergraduate Research Opportunities Program –http://www.lsa.umich.edu/urop/http://www.lsa.umich.edu/urop/ Your future class professors Dr. Gottfried and Dr. Banaszak Holl Send specific request with information about yourself and interests (CV). Use contacts. Blanket emails not great approach. For Help Finding Research Labs

4. Outline Vitamin C note Electrochemistry –Nernst –Ampere –Electrolytic cells –Environmental –Hybrid cars

5. Vitamin C Which fruit keeps best over a long period of time? Fruits known to sailors –Orange, lemon ~1400 –Limes 1638 –Shaddock (grapefruit) 1700 Used lemon or lime juice preserved in brandy

6. Picture from: www.corrosion-doctors.org/ Biographies/images/ www.corrosion-doctors.org/ Biographies/images/ Nernst M

7. Is potential always the same? Standard conditions: 1 atm, 25 o C, 1 M What will influence the potential of a cell? M

8. Mathematical Relationships: Nernst The Nernst Equation: E o = standard potential of the cell R = Universal gas constant = 8.3145 J/mol*K T = temperature in Kelvin n = number of electrons transferred F = Faraday’s constant = 96,483.4 C/mol Q = reaction quotient (concentration of anode divided by the concentration of the cathode) E = E o - RT ln Q nF Cu +2 + Zn(s) → Zn +2 + Cu(s) Q = M

9. Applying the Nernst Equation This cell is operating at 25 o C with 1.00x10 -5 M Zn 2+ and 0.100M Cu 2+ ? Predict if the voltage will be higher or lower than the standard potential E = E o - RT ln Q nF Cu +2 + Zn(s) → Zn +2 + Cu(s) M

10. E o = standard potential of the cell R = Universal gas constant = 8.3145 J/mol*K T = temperature in Kelvin n = number of electrons transferred F = Faraday’s constant = 96,483.4 C/mol Q = reaction quotient (concentration of anode divided by the concentration of the cathode) E = E o - RT ln Q nF Zn +2 + 2e - -> Zn -0.76 V Cu +2 + 2e - -> Cu 0.34 V 25 o C + 273 = K n = 1.00x10 -5 M Zn 2+ and 0.100M Cu 2 Cu +2 + Zn(s) → Zn +2 + Cu(s) Q = [Zn +2 ]/[Cu +2 ] M

11. Were your predictions correct? M

12. Picture from: musee-ampere.univ-lyon1.fr/ musee-ampere.univ-lyon1.fr/ Ampere A

13. The Units of Electrochemistry Coulomb –1 coulomb equals 2.998 x 10 9 electrostatic units (eu) –eu is amount of charge needed to repel an identical charge 1 cm away with unit force –Charge on one electron is -1.602 x 10 -19 coulomb Problem:An aluminum ion has a +3 charge. What is this value in coulombs? magnitude of charge is same at that of e-, opposite sign 3 x 1.602 x10 -19 = 4.806 x 10 -19 coulomb Key Point: electrons or ions charges can be measured in coloumbs A

14. The Units of Electrochemistry Ampere –Amount of current flowing when 1 coulomb passes a given point in 1 second –Units of Amperes are Coulombs per second –Current (I) x time (C/s x s) gives an amount of charge. Problem:How much current is flowing in a wire in which 5.0 x 10 16 electrons are flowing per second? The charge transferred each second = (5.0 x 10 16 electrons/sec) x (1.602 x 10 -19 coulomb/electron) = 8.0 x 10 -3 coulombs/sec = amps A

15. The Units of Electrochemistry Ampere –Amount of current flowing when 1 coulomb passes a given point in 1 second –Units of Amperes are Coulombs per second –Current (I) x time (C/s x s) gives an amount of charge. –We can express electron or ION flow in amps! Problem:If 1 mol Al +3 ions passes a given point in one hour, what is the current flow? 1 mol Al +3 ions6.022 x 10 23 Al +3 ions4.806 x 10 -19 coulomb1 hour Hour1 mol Al +3 ions1 Al +3 ion3600 sec = 80 C/s = 80 A A

16. From: Moore, Stanitski, and Jurs Chemistry: The Molecular Science 2 nd Edition//

17. Batteries Primary Cells non-reversible, non-rechargeable electrochemical cell "dry" cell & alkaline cell 1.5 v/cell mercury cell 1.34 v/cell fuel cell 1.23v/cell Secondary Cells reversible, rechargeable electrochemical cell Lead-acid (automobile battery) 2 v/cell NiCad 1.25 v/cell Lithium batteries A

18. “Flash Light” Batteries Primary Cells "Dry" Cell Zn (s) + 2 MnO 2(s) + 2 NH 4 +  Zn +2 (aq) + 2 MnO(OH) (s) + 2 NH 3 Alkaline Cell Zn (s) + 2 MnO 2(s)  ZnO (s) + Mn 2 O 3 (s) A

19. Leclanche “Dry” Cell A

20. Mercury Battery Primary Cells Zn (s) + HgO (s)  ZnO (aq) + Hg (l) A

21. Fuel Cells anode H 2(g) + 2 OH - (aq) → 2 H 2 O (l) + 2 e - cathode O 2(g) + 2 H 2 O (l) + 4 e - → 4 OH - (aq) Picture from: http://www.bpa.gov/Energy/N/projects/fuel_cell/education/fuelcellcar/ A

22. Lead-Acid (Automobile Battery) Pb (s) + PbO 2(s) + 2 H 2 SO 4 2 PbSO 4(s) + 2 H 2 O 2 v/cell thus 12 volt battery = 6-2 volt cells Secondary Cell A

23. Nickel-Cadmium (Ni-Cad) Secondary Cell Cd (s) + 2 Ni(OH) 3(s) ↔ Cd(OH) 2(s) + Ni(OH) 2(s) A

24. Lithium Batteries Lithium batters (developed in 1970; used in watches, pacemakers, etc.) The 4-volt lithium battery, which has up to 33 percent higher energy density and 60 less weight than a nickel-metal hydride battery of the same size, has made possible the miniaturization of the current generation of electronic devices M

25. Lithium chemistry What is the ½ reaction involving lithium? Is this a reduction or oxidation reaction? Does it take place at the anode or cathode? M

26. The other ½ of the battery: Co +3 + e - → Co +2 (+1.92 V) What is the standard potential of the cell? [Li + + e - → Li (-3.045 V)] A lithium battery has a potential listed at 4V. Does this differ from the number you calculated? Why? M

27. Alternative Anodes Which elements would you predict to have oxidation potentials similar to lithium? (no peeking at the table of standard reduction potentials!) Is what way is lithium preferable to these elements? M

28. Alternative Cathodes Co +3 + e - → Co +2 Disadvantage: Use of cobalt oxide can lead to thermal runaway. Increased temperature and pressure cause case of battery to break and fumes are released. Lithium is exposed to oxygen and hydrogen in the air. What are the alternatives and why aren’t they being used in conjunction with lithium to build a battery with higher potential? M

29. Table of Standard Reduction Potentials EoEo EoEo F 2 + 2e - --> 2F - 2.87 Fe 3+ + 3e - ---> Fe-0.04 Co 3+ + e - --> Co 2+ 1.80 Pb 2+ + 2e - ---> Pb-0.13 Cl 2 + 2e - ---> 2Cl - 1.36 Ni 2+ + 2e - ---> Ni-0.25 O 2 + 4H + + 4e - -> 2H 2 O1.23 Co 2+ + 2e - ---> Co-0.29 Hg 2+ + 2e - ---> Hg0.85 Cr 3+ + e - ---> Cr 2 -0.40 Ag + + e - ---> Ag0.80 Fe 2+ + 2e - ---> Fe-0.41 I 2 + 2e - ---> 2I - 0.54 Zn 2+ + 2e - ---> Zn-0.76 Cu + + e - ---> Cu0.52 Mn 2+ + 2e - ---> Mn-1.18 2H + + 2e - ---> H 2 0.00 Al 3+ + 3e - ---> Al-1.66 M

30. Other commercially viable batteries Besides cobalt, two other reduction reactions used in lithium batteries are shown below. What is the potential of these batteries? Fe +2 + 2e - → Fe (-0.41 V) ½ I 2 + e - → I - (0.54 V) M

31. Other cathode reactions being explored Al and Mg are also being explored as materials for the cathode. What are the advantages and disadvantages of these materials? M

32. A

33. Corrosion: A Case of Environmental Electrochemistry Facts about formation of rust –Iron does not rust in dry air: moisture must be present –Iron does not rust in air-free water: O 2 must be present –The loss of iron and the deposition of rust often occur at different places on the same object –Iron rusts more under acidic conditions (low pH) –Iron rusts more quickly in contact with ionic solutions –Iron rusts more quickly in contact with a less active metal (such as Cu) and more slowly in contact with a more reactive metal (such as Zn) A

34. Balance the Redox Reaction Fe (s) + O 2 (g)  Fe +2 (aq) + H 2 O (l) in acid

35. Corrosion: A Case of Environmental Electrochemistry O 2(g) + 4 H + (aq) + 4 e - => 2 H 2 O (l) E o = 1.23 V Fe (s) => Fe +2 (aq) + 2 e - E o = 0.44 V 2 Fe (s) + O 2(g) + 4 H + (aq) => 2 H 2 O (l) + Fe +2 (aq) E o = 1.67 V 2Fe 2+ (aq) + ½O 2 (g) + (2+n)H 2 O(l)  Fe 2 O 3. nH 2 O(s) + 4H + (aq) A

36. Corrosion: A Case of Environmental Electrochemistry Sacrificial cathode A

38. Evaluations Please fill out both an evaluation for Dr. Banaszak Holl and Dr. Gottfried and put them in the correct envelopes! Constructive comments very helpful! Thanks for a great semester!