2. CHEM 160 General Chemistry II Lecture Presentation Electrochemistry December 1, 2004 Chapter 20

3. Electrochemistry Electrochemistry  deals with interconversion between chemical and electrical energy

4. Electrochemistry Electrochemistry  deals with the interconversion between chemical and electrical energy  involves redox reactions

5. Electrochemistry Electrochemistry  deals with interconversion between chemical and electrical energy  involves redox reactions electron transfer reactions Oh No! They’re back!

6. Redox reactions (quick review) Oxidation Reduction Reducing agent Oxidizing agent

7. Redox reactions (quick review) Oxidation  loss of electrons Reduction Reducing agent Oxidizing agent

8. Redox reactions (quick review) Oxidation  loss of electrons Reduction  gain of electrons Reducing agent Oxidizing agent

9. Redox reactions (quick review) Oxidation  loss of electrons Reduction  gain of electrons Reducing agent  donates the electrons and is oxidized Oxidizing agent

10. Redox reactions (quick review) Oxidation  loss of electrons Reduction  gain of electrons Reducing agent  donates the electrons and is oxidized Oxidizing agent  accepts electrons and is reduced

11. Redox Reactions Direct redox reaction

12. Redox Reactions Direct redox reaction  Oxidizing and reducing agents are mixed together

13. CuSO 4 (aq) (Cu 2+ ) Zn rod Direct Redox Reaction

14. CuSO 4 (aq) (Cu 2+ ) Zn rod Deposit of Cu metal forms Direct Redox Reaction

15. Redox Reactions Direct redox reaction  Oxidizing and reducing agents are mixed together Indirect redox reaction  Oxidizing and reducing agents are separated but connected electrically Example –Zn and Cu 2+ can be reacted indirectly  Basis for electrochemistry –Electrochemical cell

16. e-e- e-e- Zn 2+ Cu 2 + Cu cathode Zn anode Zn  Zn 2+ + 2e - Cu 2+ + 2e -  Cu Electrochemical Cells Voltaic Cell Salt bridge

17. Electrochemical Cells

18. Voltaic Cell  cell in which a spontaneous redox reaction generates electricity  chemical energy  electrical energy

19. Electrochemical Cells

20. Voltaic Cell Electrochemical Cells

21. Electrolytic Cell  electrochemical cell in which an electric current drives a nonspontaneous redox reaction  electrical energy  chemical energy

22. Cell Potential

23. Cell Potential (electromotive force), E cell (V)  electrical potential difference between the two electrodes or half-cells Depends on specific half-reactions, concentrations, and temperature Under standard state conditions ([solutes] = 1 M, P solutes = 1 atm), emf = standard cell potential, E  cell 1 V = 1 J/C  driving force of the redox reaction

24. voltmeter e-e- e-e- Zn 2+ Cu 2 + Cu Zn Zn  Zn 2+ + 2e - Cu 2+ + 2e -  Cu Cell Potential

25. high electrical potential low electrical potential Cell Potential

26. E cell = E cathode - E anode = E redn - E ox E° cell = E° cathode - E° anode = E° redn - E° ox (E cathode and E anode are reduction potentials by definition.)

27. Cell Potential E° cell = E° cathode - E° anode = E° redn - E° ox  E cell can be measured Absolute E cathode and E anode values cannot Reference electrode  has arbitrarily assigned E  used to measure relative E cathode and E anode for half- cell reactions Standard hydrogen electrode (S.H.E.)  conventional reference electrode

28. Standard Hydrogen Electrode E  = 0 V (by definition; arbitrarily selected) 2H + + 2e -  H 2

29. e-e- e-e- Cu 2+ + 2e -  Cu Cu H 2  2H + + 2e - 1 M Cu 2+ 1 M H + Pt H 2 (1 atm)

31. Example 1 A voltaic cell is made by connecting a standard Cu/Cu 2+ electrode to a S.H.E. The cell potential is 0.34 V. The Cu electrode is the cathode. What is the standard reduction potential of the Cu/Cu 2+ electrode?

33. e-e- e-e- Zn  Zn 2+ + 2e - Zn 2H + + 2e -  H 2 1 M Zn 2+ 1 M H + Pt H 2 (1 atm)

34. Example 2 A voltaic cell is made by connecting a standard Zn/Zn 2+ electrode to a S.H.E. The cell potential is 0.76 V. The Zn electrode is the anode of the cell. What is the standard reduction potential of the Zn/Zn 2+ electrode?

35. Standard Electrode Potentials Standard Reduction Potentials, E°  E° cell measured relative to S.H.E. (0 V) electrode of interest = cathode  If E° < 0 V: Oxidizing agent is harder to reduce than H +  If E° > 0 V: Oxidizing agent is easier to reduce than H +

36. Standard Reduction Potentials Reduction Half-Reaction E  (V) F 2 (g) + 2e -  2F - (aq) 2.87 Au 3+ (aq) + 3e -  Au(s) 1.50 Cl 2 (g) + 2 e -  2Cl - (aq) 1.36 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O 1.33 O 2 (g) + 4H + + 4e -  2H 2 O(l) 1.23 Ag + (aq) + e -  Ag(s) 0.80 Fe 3+ (aq) + e -  Fe 2+ (aq) 0.77 Cu 2+ (aq) + 2e -  Cu(s) 0.34 Sn 4+ (aq) + 2e -  Sn 2+ (aq) 0.15 2H + (aq) + 2e -  H 2 (g) 0.00 Sn 2+ (aq) + 2e -  Sn(s) -0.14 Ni 2+ (aq) + 2e -  Ni(s) -0.23 Fe 2+ (aq) + 2e -  Fe(s) -0.44 Zn 2+ (aq) + 2e -  Zn(s) -0.76 Al 3+ (aq) + 3e -  Al(s) -1.66 Mg 2+ (aq) + 2e -  Mg(s) -2.37 Li + (aq) + e -  Li(s) -3.04

37. Uses of Standard Reduction Potentials Compare strengths of reducing/oxidizing agents.  the more - E°, stronger the red. agent  the more + E°, stronger the ox. agent

38. Standard Reduction Potentials Reduction Half-Reaction E  (V) F 2 (g) + 2e -  2F - (aq) 2.87 Au 3+ (aq) + 3e -  Au(s) 1.50 Cl 2 (g) + 2 e -  2Cl - (aq) 1.36 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O 1.33 O 2 (g) + 4H + + 4e -  2H 2 O(l) 1.23 Ag + (aq) + e -  Ag(s) 0.80 Fe 3+ (aq) + e -  Fe 2+ (aq) 0.77 Cu 2+ (aq) + 2e -  Cu(s) 0.34 Sn 4+ (aq) + 2e -  Sn 2+ (aq) 0.15 2H + (aq) + 2e -  H 2 (g) 0.00 Sn 2+ (aq) + 2e -  Sn(s) -0.14 Ni 2+ (aq) + 2e -  Ni(s) -0.23 Fe 2+ (aq) + 2e -  Fe(s) -0.44 Zn 2+ (aq) + 2e -  Zn(s) -0.76 Al 3+ (aq) + 3e -  Al(s) -1.66 Mg 2+ (aq) + 2e -  Mg(s) -2.37 Li + (aq) + e -  Li(s) -3.04 Ox. agent strength increases Red. agent strength increases

39. Uses of Standard Reduction Potentials Determine if oxidizing and reducing agent react spontaneously  diagonal rule ox. agent red. agent spontaneous

40. Uses of Standard Reduction Potentials Determine if oxidizing and reducing agent react spontaneously Cathode (reduction) E° redn (cathode) more + Anode (oxidation) E° redn (V) E° redn (anode) more - Spontaneous rxn if E° cathode > E° anode

41. Standard Reduction Potentials Reduction Half-Reaction E  (V) F 2 (g) + 2e -  2F - (aq) 2.87 Au 3+ (aq) + 3e -  Au(s) 1.50 Cl 2 (g) + 2 e -  2Cl - (aq) 1.36 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O 1.33 O 2 (g) + 4H + + 4e -  2H 2 O(l) 1.23 Ag + (aq) + e -  Ag(s) 0.80 Fe 3+ (aq) + e -  Fe 2+ (aq) 0.77 Cu 2+ (aq) + 2e -  Cu(s) 0.34 Sn 4+ (aq) + 2e -  Sn 2+ (aq) 0.15 2H + (aq) + 2e -  H 2 (g) 0.00 Sn 2+ (aq) + 2e -  Sn(s) -0.14 Ni 2+ (aq) + 2e -  Ni(s) -0.23 Fe 2+ (aq) + 2e -  Fe(s) -0.44 Zn 2+ (aq) + 2e -  Zn(s) -0.76 Al 3+ (aq) + 3e -  Al(s) -1.66 Mg 2+ (aq) + 2e -  Mg(s) -2.37 Li + (aq) + e -  Li(s) -3.04

42. Uses of Standard Reduction Potentials Calculate E° cell  E° cell = E° cathode - E° anode Greater E° cell, greater the driving force  E° cell > 0 : spontaneous redox reactions  E° cell < 0 : nonspontaeous redox reactions

43. Example 3 A voltaic cell consists of a Ag electrode in 1.0 M AgNO 3 and a Cu electrode in 1 M Cu(NO 3 ) 2. Calculate E° cell for the spontaneous cell reaction at 25°C.

44. Standard Reduction Potentials Reduction Half-Reaction E  (V) F 2 (g) + 2e -  2F - (aq) 2.87 Au 3+ (aq) + 3e -  Au(s) 1.50 Cl 2 (g) + 2 e -  2Cl - (aq) 1.36 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O 1.33 O 2 (g) + 4H + + 4e -  2H 2 O(l) 1.23 Ag + (aq) + e -  Ag(s) 0.80 Fe 3+ (aq) + e -  Fe 2+ (aq) 0.77 Cu 2+ (aq) + 2e -  Cu(s) 0.34 Sn 4+ (aq) + 2e -  Sn 2+ (aq) 0.15 2H + (aq) + 2e -  H 2 (g) 0.00 Sn 2+ (aq) + 2e -  Sn(s) -0.14 Ni 2+ (aq) + 2e -  Ni(s) -0.23 Fe 2+ (aq) + 2e -  Fe(s) -0.44 Zn 2+ (aq) + 2e -  Zn(s) -0.76 Al 3+ (aq) + 3e -  Al(s) -1.66 Mg 2+ (aq) + 2e -  Mg(s) -2.37 Li + (aq) + e -  Li(s) -3.04

45. Example 4 A voltaic cell consists of a Ni electrode in 1.0 M Ni(NO 3 ) 2 and an Fe electrode in 1 M Fe(NO 3 ) 2. Calculate E° cell for the spontaneous cell reaction at 25°C.

46. Standard Reduction Potentials Reduction Half-Reaction E  (V) F 2 (g) + 2e -  2F - (aq) 2.87 Au 3+ (aq) + 3e -  Au(s) 1.50 Cl 2 (g) + 2 e -  2Cl - (aq) 1.36 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O 1.33 O 2 (g) + 4H + + 4e -  2H 2 O(l) 1.23 Ag + (aq) + e -  Ag(s) 0.80 Fe 3+ (aq) + e -  Fe 2+ (aq) 0.77 Cu 2+ (aq) + 2e -  Cu(s) 0.34 Sn 4+ (aq) + 2e -  Sn 2+ (aq) 0.15 2H + (aq) + 2e -  H 2 (g) 0.00 Sn 2+ (aq) + 2e -  Sn(s) -0.14 Ni 2+ (aq) + 2e -  Ni(s) -0.23 Fe 2+ (aq) + 2e -  Fe(s) -0.44 Zn 2+ (aq) + 2e -  Zn(s) -0.76 Al 3+ (aq) + 3e -  Al(s) -1.66 Mg 2+ (aq) + 2e -  Mg(s) -2.37 Li + (aq) + e -  Li(s) -3.04

47. Cell Potential Is there a relationship between E cell and  G for a redox reaction?

48. Cell Potential Relationship between E cell and  G:   G = -nFE cell F = Faraday constant = 96500 C/mol e-’s, n = # e - ’s transferred redox rxn.

49. Cell Potential Relationship between E cell and  G:   G = -nFE cell F = Faraday constant = 96500 C/mol e-’s, n = # e - ’s transferred redox rxn. 1 J = CV  G 0 = spontaneous

50. Equilibrium Constants from E cell Relationship between E cell and  G:   G = -nFE cell F = Faraday constant = 96500 C/mol e - ’s, n = # e - ’s transferred redox rxn 1 J = CV  G 0 = spontaneous Under standard state conditions:   G° = -nFE° cell

51. Example 5 Calculate  G° at room temperature for the reaction between Sn 4+ (aq) and Fe(s).

52. Standard Reduction Potentials Reduction Half-Reaction E  (V) F 2 (g) + 2e -  2F - (aq) 2.87 Au 3+ (aq) + 3e -  Au(s) 1.50 Cl 2 (g) + 2 e -  2Cl - (aq) 1.36 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O 1.33 O 2 (g) + 4H + + 4e -  2H 2 O(l) 1.23 Ag + (aq) + e -  Ag(s) 0.80 Fe 3+ (aq) + e -  Fe 2+ (aq) 0.77 Cu 2+ (aq) + 2e -  Cu(s) 0.34 Sn 4+ (aq) + 2e -  Sn 2+ (aq) 0.15 2H + (aq) + 2e -  H 2 (g) 0.00 Sn 2+ (aq) + 2e -  Sn(s) -0.14 Ni 2+ (aq) + 2e -  Ni(s) -0.23 Fe 2+ (aq) + 2e -  Fe(s) -0.44 Zn 2+ (aq) + 2e -  Zn(s) -0.76 Al 3+ (aq) + 3e -  Al(s) -1.66 Mg 2+ (aq) + 2e -  Mg(s) -2.37 Li + (aq) + e -  Li(s) -3.04 Ox. agent strength increases Red. agent strength increases

53. Equilibrium Constants from E cell Relationship between E cell and  G:   G = -nFE cell F = Faraday constant = 96500 C/mol e - ’s, n = # e - ’s transferred redox rxn 1 J = CV  G 0 = spontaneous Under standard state conditions:   G° = -nFE° cell

54. Equilibrium Constants from E cell Relationship between E cell and  G:   G = -nFE cell F = Faraday constant = 96500 C/mol e - ’s, n = # e - ’s transferred redox rxn 1 J = CV  G 0 = spontaneous Under standard state conditions:   G° = -nFE° cell and   G° = -RTlnK so -nFE° cell = -RTlnK

55.  H°  S° Calorimetric Data  G° Electrochemical Data Composition Data E° cell Equilibrium constants K

56. Example 5 Calculate E° cell,  G°, and K for the voltaic cell that uses the reaction between Ag and Cl 2 under standard state conditions at 25°C.

57. Standard Reduction Potentials Reduction Half-Reaction E  (V) F 2 (g) + 2e -  2F - (aq) 2.87 Au 3+ (aq) + 3e -  Au(s) 1.50 Cl 2 (g) + 2 e -  2Cl - (aq) 1.36 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O 1.33 O 2 (g) + 4H + + 4e -  2H 2 O(l) 1.23 Ag + (aq) + e -  Ag(s) 0.80 Fe 3+ (aq) + e -  Fe 2+ (aq) 0.77 Cu 2+ (aq) + 2e -  Cu(s) 0.34 Sn 4+ (aq) + 2e -  Sn 2+ (aq) 0.15 2H + (aq) + 2e -  H 2 (g) 0.00 Sn 2+ (aq) + 2e -  Sn(s) -0.14 Ni 2+ (aq) + 2e -  Ni(s) -0.23 Fe 2+ (aq) + 2e -  Fe(s) -0.44 Zn 2+ (aq) + 2e -  Zn(s) -0.76 Al 3+ (aq) + 3e -  Al(s) -1.66 Mg 2+ (aq) + 2e -  Mg(s) -2.37 Li + (aq) + e -  Li(s) -3.04 Ox. agent strength increases Red. agent strength increases

58. The Nernst Equation  G depends on concentrations   G =  G° + RTlnQ and  G = -nFE cell and  G° = -nFE° cell thus -nFE cell = -nFE° cell + RTlnQ or E cell = E° cell - (RT/nF)lnQ (Nernst eqn.)

59. The Nernst Equation E cell = E° cell - (RT/nF)lnQ (Nernst eqn.)  At 298 K (25°C), RT/F = 0.0257 V so E cell = E° cell - (0.0257/n)lnQ or E cell = E° cell - (0.0592/n)logQ

60. Example 7 Calculate the voltage produced by the voltaic cell using the reaction between Zn(s) and Cu 2+ (aq) if [Zn 2+ ] = 0.001 M and [Cu 2+ ] = 1.3 M. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

61. Example 7 Calculate the voltage produced by the galvanic cell which uses the reaction below if [Ag + ] = 0.001 M and [Cu 2+ ] = 1.3 M. 2Ag + (aq) + Cu(s)  2Ag(s) + Cu 2+ (aq)

62. Standard Reduction Potentials Reduction Half-Reaction E  (V) F 2 (g) + 2e -  2F - (aq) 2.87 Au 3+ (aq) + 3e -  Au(s) 1.50 Cl 2 (g) + 2 e -  2Cl - (aq) 1.36 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O 1.33 O 2 (g) + 4H + + 4e -  2H 2 O(l) 1.23 Ag + (aq) + e -  Ag(s) 0.80 Fe 3+ (aq) + e -  Fe 2+ (aq) 0.77 Cu 2+ (aq) + 2e -  Cu(s) 0.34 Sn 4+ (aq) + 2e -  Sn 2+ (aq) 0.15 2H + (aq) + 2e -  H 2 (g) 0.00 Sn 2+ (aq) + 2e -  Sn(s) -0.14 Ni 2+ (aq) + 2e -  Ni(s) -0.23 Fe 2+ (aq) + 2e -  Fe(s) -0.44 Zn 2+ (aq) + 2e -  Zn(s) -0.76 Al 3+ (aq) + 3e -  Al(s) -1.66 Mg 2+ (aq) + 2e -  Mg(s) -2.37 Li + (aq) + e -  Li(s) -3.04 Ox. agent strength increases Red. agent strength increases

63. Commercial Voltaic Cells Battery  commercial voltaic cell used as portable source of electrical energy types  primary cell Nonrechargeable Example: Alkaline battery  secondary cell Rechargeable Example: Lead storage battery

64. How Does a Battery Work cathode (+) anode (-) Electrolyte Paste Seal/cap Assume a generalized battery

65. Battery cathode (+): Reduction occurs here anode (-): oxidation occurs here e - flow Electrolyte paste: ion migration occurs here Placing the battery into a flashlight, etc., and turning the power on completes the circuit and allows electron flow to occur

66. How Does a Battery Work Battery reaction when producing electricity (spontaneous): Cathode: O 1 + e -  R 1 Anode: R 2  O 2 + e - Overall: O 1 + R 2  R 1 + O 2 Recharging a secondary cell  Redox reaction must be reversed, i.e., current is reversed (nonspontaneous) Recharge: O 2 + R 1  R 2 + O 1  Performed using electrical energy from an external power source

67. Batteries Read the textbook to fill in the details on specific batteries.  Alkaline battery  Lead storage battery  Nicad battery  Fuel cell

68. Alkaline Dry Cell

69. Brass rod Plated steel (-) Anode: Mixture of Zn and KOH(aq) Plated steel (+) Cathode: Mixture of MnO 2 and C (graphite) Paper or fabric Separator Insulators

70. Alkaline Dry Cell Half-reactions

71. Alkaline Dry Cell Half-reactions anode: Zn(s) + 2OH - (aq) --> ZnO(s) + H 2 O(l) + 2e -

72. Alkaline Dry Cell Half-reactions anode: Zn(s) + 2OH - (aq) --> ZnO(s) + H 2 O(l) + 2e - cathode: 2MnO 2 (s) + H 2 O(l) + 2e - --> Mn 2 O 3 (s) + 2OH - (aq)

73. Alkaline Dry Cell Half-reactions anode: Zn(s) + 2OH - (aq) --> ZnO(s) + H 2 O(l) + 2e - cathode: 2MnO 2 (s) + H 2 O(l) + 2e - --> Mn 2 O 3 (s) + 2OH - (aq) overall: Zn(s) + 2MnO 2 (s) --> Mn 2 O 3 (s) + ZnO(s) E cell = 1.54 V

74. Lead Storage Battery (cathode) (anode) 6 x 2V = 12 V

75. Lead Storage Battery Half-reactions

76. Lead Storage Battery Half-reactions anode: Pb(s) + SO 4 2- (aq) --> PbSO 4 (s) + 2e -

77. Lead Storage Battery Half-reactions anode: Pb(s) + SO 4 2- (aq) --> PbSO 4 (s) + 2e - cathode: PbO 2 (s) + 4H + (aq) + SO 4 2- (aq) + 2e - --> PbSO 4 (s) + 2H 2 O(l)

78. Lead Storage Battery Half-reactions anode: Pb(s) + SO 4 2- (aq) --> PbSO 4 (s) + 2e - cathode: PbO 2 (s) + 4H + (aq) + SO 4 2- (aq) + 2e - --> PbSO 4 (s) + 2H 2 O(l) overall: Pb(s) + PbO 2 (s) + 2H 2 SO 4 (aq) --> 2PbSO 4 (s) + 2H 2 O(l)

79. Lead Storage Battery Half-reactions anode: Pb(s) + SO 4 2- (aq) --> PbSO 4 (s) + 2e - cathode: PbO 2 (s) + 4H + (aq) + SO 4 2- (aq) + 2e - --> PbSO 4 (s) + 2H 2 O(l) overall: Pb(s) + PbO 2 (s) + 2H 2 SO 4 (aq) --> 2PbSO 4 (s) + 2H 2 O(l) Cell reaction reversed during recharging. 2PbSO 4 (s) + 2H 2 O(l) --> Pb(s) + PbO 2 (s) + 2H 2 SO 4 (aq)

80. Lead Storage Battery Half-reactions anode: Pb(s) + HSO 4 2- (aq) --> PbSO 4 (s) + H + + 2e - cathode: PbO 2 (s) + 3H + (aq) + HSO 4 2- (aq) + 2e - --> PbSO 4 (s) + 2H 2 O(l) overall: Pb(s) + PbO 2 (s) + 2H + + 2HSO 4 - (aq) --> 2PbSO 4 (s) + 2H 2 O(l) Cell reaction reversed during recharging.

81. Lead Storage Battery Half-reactions during recharging (nonspontaneous) cathode: PbSO 4 (s) + H + + 2e - --> Pb(s) + HSO 4 2- (aq) anode: PbSO 4 (s) + 2H 2 O(l) --> PbO 2 (s) + 3H + (aq) + HSO 4 2- (aq) + 2e - overall: 2PbSO 4 (s) + 2H 2 O(l) --> PbO 2 (s) + Pb(s) + 2H + + 2HSO 4 - (aq) Cell converted into electrolytic cell via application of external electrical energy.

82. Fuel Cells Voltaic-like cell that operates with continuous supply of energetic reactants (fuel) to the electrodes  utilize combustion reactions  do not store chemical energy Not self-contained since reactants must be supplied to the electrodes  Example: Hydrogen-Oxygen fuel cell

83. Hydrogen-Oxygen Fuel Cell

84. Half-reactions

85. Hydrogen-Oxygen Fuel Cell Half-reactions anode: 2H 2 (g) + 4OH - (aq) --> 4H 2 O(l) + 4e -

86. Hydrogen-Oxygen Fuel Cell Half-reactions anode: 2H 2 (g) + 4OH - (aq) --> 4H 2 O(l) + 4e - cathode: O 2 (g) + 2H 2 O(l) + 4e - --> 4OH - (aq)

87. Hydrogen-Oxygen Fuel Cell Half-reactions anode: 2H 2 (g) + 4OH - (aq) --> 4H 2 O(l) + 4e - cathode: O 2 (g) + 2H 2 O(l) + 4e - --> 4OH - (aq) overall: 2H 2 (g) + O 2 (g) --> 2H 2 O(l)

89. Corrosion Corrosion  deterioration of metals by a spontaneous redox reaction Attacked by species in environment –Metal becomes a “voltaic” cell Metal is often lost to a solution as an ion Rusting of Iron

90. Corrosion of Iron

91. Half-reactions anode: Fe(s)  Fe 2+ (aq) + 2e - cathode: O 2 (g) + 4H + (aq) + 4e -  2H 2 O(l) overall: 2Fe(s) + O 2 (g) + 4H + (aq)  2Fe 2+ (aq) + 2H 2 O(l) E cell > 0 (E cell = 0.8 to 1.2 V), so process is spontaneous!

92. Corrosion of Iron Rust formation: 4Fe 2+ (aq) + O 2 (g) + 4H + (aq)  4Fe 3+ (aq) + 2H 2 O(l) 2Fe 3+ (aq) + 4H 2 O(l)  Fe 2 O 3  H 2 O(s) + 6H + (aq)

93. Prevention of Corrosion Cover the Fe surface with a protective coating  Paint  Passivation surface atoms made inactive via oxidation 2Fe(s) + 2Na 2 CrO 4 (aq) + 2H 2 O(l) --> Fe 2 O 3 (s) + Cr 2 O 3 (s) + 4NaOH(aq)  Other metal Tin Zn –Galvanized iron

94. Prevention of Corrosion Cathodic Protection  metal to be protected is brought into contact with a more easily oxidized metal  “sacrificial” metal becomes the anode “Corrodes” preferentially over the iron Iron serves only as the cathode

95. Standard Electrode Potentials Half-reaction E° F 2 (g) + 2e - -> 2F - (aq)+2.87 V Ag + (aq) + e - -> Ag(s)+0.80 V Cu 2+ (aq) + 2e - -> Cu(s)+0.34 V 2H + (aq) + 2e - -> H 2 (g)0 V Ni 2+ (aq) + 2e - -> Ni(s)-0.25 V Fe 2+ (aq) + 2e - -> Fe(s)-0.44 V Zn 2+ (aq) + 2e - -> Zn(s)-0.76 V Al 3+ (aq) + 3e - -> Al(s)-1.66 V Mg 2+ (aq) + 2e - ->Mg(s)-2.38 V Metals more easily oxidized than Fe have more negative E°’s

96. Cathodic Protection galvanized steel (Fe)

97. Cathodic Protection (cathode) (electrolyte) (anode)

98. Electrolysis Electrolysis  process in which electrical energy drives a nonspontaneous redox reaction electrical energy is converted into chemical energy Electrolytic cell  electrochemical cell in which an electric current drives a nonspontaneous redox reaction

99. Electrolysis Same principles apply to both electrolytic and voltaic cells  oxidation occurs at the anode  reduction occurs at the cathode  electrons flow from anode to cathode in the external circuit In an electrolytic cell, an external power source pumps the electrons through the external circuit

100. Electrolysis of Molten NaCl

101. Quantitative Aspects of Electrochemical Cells For any half-reaction, the amount of a substance oxidized or reduced at an electrode is proportional to the number of electrons passed through the cell  Faraday’s law of electrolysis  Examples Na + + 1e -  Na Al 3+ + 3e -  Al  Number of electrons passing through cell is measured by determining the quantity of charge (coulombs) that has passed 1 C = 1 A x 1 s 1 F = 1 mole e - = 96500 C

102. Steps for Quantitative Electrolysis Calculations current (A) and time (s), A x s charge in coulombs (C) Number of moles of e - moles of substance oxidized or reduced mass of substance oxidized or reduced

103. Example 8 What mass of copper metal can be produced by a 3.00 A current flowing through a copper(II) sulfate (CuSO 4 ) solution for 5.00 hours?

104. Example 9 An aqueous solution of an iron salt is electrolyzed by passing a current of 2.50 A for 3.50 hours. As a result, 6.1 g of iron metal are formed at the cathode. Calculate the charge on the iron ions in the solution.