1. Chapter 19: Electrochemistry: Voltaic Cells Generate Electricity which can do electrical work. Voltaic or galvanic cells are devices in which electron transfer occurs via an external circuit. Batteries are Practical Voltaic Cells. Reactions are Spontaneous for Practical Voltaic Cells.

2. Voltaic Cell/Battery Batteries are Practical Voltaic Cells. Reactions are Spontaneous for Practical Voltaic Cells. –Commonly Refer to Half Reactions. Oxidation Half-Reaction Reduction Half-Reaction

3. Chapter 21: Voltaic Cells Voltaic cells have: –Anode –Cathode –Salt bridge (used to complete the electrical circuit) –Electrolyte –Switch –Voltmeter

4. Voltaic Cells

5. Electrodes 1. Anode: oxidation 2. Cathode: Reduction Electrons flow from the anode to the cathode. Anode is negative and the cathode is positive.

6. Voltaic Cells

7. Batteries Alkaline Battery

8. Batteries Lead-Acid Battery

9. Cell EMF Standard Reduction Potentials

10. Cell EMF: Potential (E) The flow of electrons from anode to cathode is spontaneous. Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode. Potential difference: difference in electrical potential. Measured in volts. One volt is the potential difference required to impart one joule of energy to a charge of one coulomb:

11. Cell EMF Electromotive force (emf) is the force required to push electrons through the external circuit. Cell potential: E cell is the emf of a cell. For 1M solutions at 25  C (standard conditions), the standard emf (standard cell potential) is called E  cell. E  cell = E  oxidation + E  reduction

12. Cell EMF Standard Reduction Potentials

13. Cell EMF Table of Standard Reduction PotentialsTable of Standard Reduction Potentials Spontaneous: +Spontaneous: + E  Nonspontaneous: - E  Standard reduction potentials, E  red are measured E  oxidation = - E  reduction

14. Problem Using standard reduction potentials (page 846), calculate the standard emf for each of the following reactions. (a) Cl 2 (g) + 2I - (aq)  2Cl - (aq) + I 2 (s)

15. Problem Oxidation Half-Reaction : E  oxidation = - E  reduction 2I -  I 2 + 2e - E  oxidation = - (+0.536 V) = -0.536 V Reduction Half-Reaction Cl 2 (g) + 2e - (aq)  2Cl - (aq) E  reduction = +1.359 volts E  cell = E  oxidation + E  reduction= -0.536 V + 1.359 V

16. Half-Cell Potentials: Cell EMF Oxidizing and Reducing Agents The more positive E  red the stronger the oxidizing agent Target molecule + Oxid Agent  Oxidized target + Reduced agent The more positive E  oxid the stronger the reducing agent Target molecule + Red Agent  Reduced target + Oxid agent

17. Spontaneity of Redox Reactions Consider the displacement of silver by nickel: Ni(s) + 2Ag + (aq)  Ni 2+ (aq) + 2Ag(s) has E  = E  red (Ag + /Ag) - E  red (Ni 2+ /Ni) = (0.80 V) - (-0.28 V) = 1.08 V, which indicates a spontaneous process.

18. EMF and Free-Energy Change  G = -nFE  G  = -nFE   G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant (96,500 Joules/V Mole electrons E is the emf of the cell.

19. Effect of Concentration, Temperature on Cell EMF E  cell calculated under standard state conditions. The Nernst Equation The Nernst equation relates emf to concentrations and temperatures which are not standard state.

20. Effect of Concentration, Temperature on Cell EMF The Nernst Equation This rearranges to give the Nernst equation: At 298 K can supply constants (Note that change from natural logarithm to base- 10 log.) Remember that n is number of moles of electrons.

21. Problem A voltaic cell is constructed that uses the following reaction and operates at 298 K. Zn (s) + Ni 2+ (aq)  Zn 2+ (aq) + Ni (s) (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when [Ni2+] = 3.00 M, [Zn2+] = 0.100 M?

22. Chapter 19: Voltaic Cells Voltaic cells have: –Anode –Cathode –Salt bridge (used to complete the electrical circuit) –Electrolyte –Switch –Voltmeter –HALF REACTIONS ARE USUALLY SPONTANEOUS.

23. Chapter 19: Shorthand Designation for Voltaic Cells Cu(s) + Ag + (aq)  Cu 2+ (aq) + Ag(s) Cu(s)|Cu 2+ (aq)||Ag + (aq)|Ag(s) (anode) (cathode)

24. Problem 19.30-A: Page 875 Calculate E , E, and  G for the following. Mg (s) + Sn 2+ (aq)  Mg 2+ (aq) +Sn (s) [Mg 2+ ] = 0.045 M, [Sn 2+ ] = 0.035 M

25. Chapter 19: Electrolysis Electrolytic cells have: –Anode –Cathode –Electrolyte –Source of DC Voltage –Voltmeter –REDOX REACTIONS NOT NECESSARILY SPONTANEOUS.

26. Chapter 19: Electrolysis ELECTROLYTIC CELL

27. Electrolysis: General Comments Electrolysis reactions are usually nonspontaneous. Focus is on Reduction Half-Reaction Ag + (aq) + e -  Ag (s) E  red =+0.799 V Na + (aq) + e -  Na (l) E  red = -2.71 V Cu 2+ (aq) + 2e -  Cu (s) E  red = +0.153 V Al 3+ (aq) + 3e-  Al (l) E  red = -1.66 V

28. Chapter 19: Electrolysis In voltaic and electrolytic cells: –reduction occurs at the cathode, and –oxidation occurs at the anode. –However, in electrolytic cells, electrons are forced to flow from the anode to cathode. –In electrolytic cells the anode is positive and the cathode is negative. –In galvanic cells the anode is negative and the cathode is positive

29. Electrolysis Electrolysis of Aqueous Solutions

30. Chapter 19: Electrolysis Electrolysis with Active Electrodes Active electrodes: electrodes that take part in electrolysis. Example: electrolytic plating.

31. Electrolysis Electrolysis with Active Electrodes

32. Electrolysis: Stoichiometry Consider the reduction of Cu 2+ to Cu. –Cu 2+ (aq) + 2e -  Cu(s). –2 mol of electrons will plate 1 mol of Cu. –The charge of 1 mol of electrons is 96,500 C (1 F). –Coulomb is quantity of charge passing in electrical circuit in 1 second when current is 1 Ampere.

33. Problem (a) Calculate the mass of Li formed by electrolysis of molten LiCl by a current of 7.5E4 Amperes flowing for a period of twenty-four hours. Assume that the cell is 85% efficient. Li + + e -  Li

34. Stoichiometry of Electrolysis –What current must be supplied to deposit 3.00 g Au from a solution of AuCl 3 in 200.0 s?

35. Cell EMF Standard Reduction Potentials

36. Hall-Herroult Electrolytic Cell Al 2 O 3 + Na 3 AlF 6  Electrolyte At Anode C (s) + 2O 2-  CO 2(g) + 4 electrons At Cathode Al 3+ + 3 electrons  Al(s)

37. Copper Purification Copper Ore (Cu 2 S) Cu 2 S + O 2  2Cu(l) + SO 2 - Blister Copper is impure copper metal. Usually will contain Cu, Zn, Fe, Au, Ag

38. Copper Purification Anode (blister copper) Cu(s)  Cu2+ + 2 electron Cathode (pure copper) Cu 2+ + 2 electrons  Cu Zn, Fe easily oxidize. Are not easily reduced. Au, Ag are not easily oxidized. Easily reduced.

39. Cell EMF Standard Reduction Potentials

40. Problem 19.58: Page 876 A quantity of 0.300 grams of copper was deposited from a CuSO 4 solution by passing a current of 3.00 A through the solution for 304 seconds. Calculate the value of the Faraday constant.

41. Chapter 19: CORROSION Oxidation of Metal Fe (s) + O 2(g)  Fe 2 O 3(s)

42. Cell EMF Standard Reduction Potentials

43. Fig. 19.13a

44. Fig. 19.13b

45. Fig. 19.13c

46. CIA p799

47. Corrosion: 1. Oxidation of Metal Fe (s) + O 2(g)  Fe 2 O 3(s) 2. Factors which facilitate Corrosion Aqueous Solution Electrolyte Acidic or Basis Solution

48. Corrosion Corrosion of Iron

49. Corrosion: Preventing Corrosion can be prevented by coating the iron with paint or another metal. Galvanized iron is coated with a thin layer of zinc. Zinc protects the iron since Zn is the anode and Fe the cathode: Zn 2+ (aq) +2e -  Zn(s), E  red = -0.76 V Fe 2+ (aq) + 2e -  Fe(s), E  red = -0.44 V

50. Corrosion: Preventing To protect underground (Fe) pipelines, a sacrificial anode is added. The water pipe is turned into the cathode and an active metal is used as the anode. Often, Mg is used as the sacrificial anode: Mg 2+ (aq) +2e -  Mg(s), E  red = -2.37 V Fe 2+ (aq) + 2e -  Fe(s), E  red = -0.44 V

51. Corrosion Preventing the Corrosion of Iron

52. Corrosion