1. Chapter 17 Electrochemistry Why does LEO say GER?

2. Chapter 17 Table of Contents Copyright © Cengage Learning. All rights reserved 2 17.0Balancing Oxidation–Reduction Equations I 17.1 Galvanic Cells 17.2 Standard Reduction Potentials 17.3Cell Potential, Electrical Work, and Free Energy 17.4Dependence of Cell Potential on Concentration 17.5Batteries 17.6Corrosion 17.7Electrolysis 17.8Commercial Eletrolytic Processes

3. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 3 To understand what oxidation – reduction reactions (REDOX) To assign the oxidation state to elements and ions To identify the oxidizing agent and the reducing agent To write balanced REDOX half-reactions Objectives

4. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 4 Reactions in which one or more electrons are transferred. Redox Reactions

5. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 5 Reaction of Sodium and Chlorine

6. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 6 1.Oxidation state of an atom in an element = 0 2.Oxidation state of monatomic ion = charge of the ion 3.Oxygen =  2 in covalent compounds (except in peroxides where it =  1) 4.Hydrogen = +1 in covalent compounds 5.Fluorine =  1 in compounds 6.Sum of oxidation states = 0 in compounds 7.Sum of oxidation states = charge of the ion in ions Rules for Assigning Oxidation States (Gangsta Switch)

7. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 7 Exercise Find the oxidation states for each of the elements in each of the following compounds: (Start with what you know) K 2 Cr 2 O 7 CO 3 2- MnO 2 PCl 5 SF 4 K = +1; Cr = +6; O = –2 C = +4; O = –2 Mn = +4; O = –2 P = +5; Cl = –1 S = +4; F = –1

8. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 8 Transfer of electrons Transfer may occur to form ions Oxidation – increase in oxidation state (loss of electrons); reducing agent Reduction – decrease in oxidation state (gain of electrons); oxidizing agent Leo says Ger! Redox Characteristics

9. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 9 Concept Check Use the oxidation states to determine which of the following are oxidation-reduction reactions? Identify the oxidizing agent and the reducing agent. a)Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g)

10. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 10 Concept Check Use the oxidation states to determine which of the following are oxidation-reduction reactions? Identify the oxidizing agent and the reducing agent. a)2CuCl(aq)  CuCl 2 (aq) + Cu(s) (1/2 Rxs)

11. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 11 To understand what oxidation – reduction reactions (REDOX) To assign the oxidation state to elements and ions To identify the oxidizing agent and the reducing agent Page 830 #13, 14(every other), 15a-d also write the half reactions Objectives Review

12. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 12 To balance half reactions that occur in acid solutions Objectives Continued

13. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 13 Review of Terms Oxidation – reduction (redox) reaction – involves a transfer of electrons from the reducing agent to the oxidizing agent Oxidation – loss of electrons (LEO) Reduction – gain of electrons (GER) Reducing agent – electron donor (loser) Oxidizing agent – electron acceptor (gainer)

14. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 14 Half–Reactions Write the half reactions for the following: 8H + + MnO 4 - + 5Fe 2+  Mn 2+ + 5Fe 3+ + 4H 2 O Oxidation: 5Fe 2+  5Fe 3+ + 5e - Reduction: 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O

15. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 15 Write the half reactions for the following: Cr 2 O 7 2- (aq) + SO 3 2- (aq)  Cr 3+ (aq) + SO 4 2- (aq) 6e - + Cr 2 O 7 2- (aq)  2Cr 3+ (aq) SO 3 2- (aq)  + SO 4 2- (aq) + 2e - Is this balanced? Rules:

16. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 16 1.Write separate equations for the oxidation and reduction half – reactions. 2.For each half – reaction: A.Balance all the elements except H and O. B.Balance O using H 2 O. C.Balance H using H +. D.Balance the charge using electrons. The Half–Reaction Method for Balancing Equations for Oxidation– Reduction Reactions Occurring in Acidic Solution

17. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 17 3.If necessary, multiply one or both balanced half – reactions by an integer to equalize the number of electrons transferred in the two half – reactions. 4.Add the half – reactions, and cancel identical species. 5.Check that the elements and charges are balanced. The Half–Reaction Method for Balancing Equations for Oxidation– Reduction Reactions Occurring in Acidic Solution

18. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 18 Cr 2 O 7 2- (aq) + SO 3 2- (aq)  Cr 3+ (aq) + SO 4 2- (aq) 6e - + Cr 2 O 7 2- (aq)  2Cr 3+ (aq) SO 3 2- (aq)  + SO 4 2- (aq) + 2e -  Separate into half-reactions.  Balance elements except H and O.  Balance O with H 2 O and H with H +

19. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 19 6e - + Cr 2 O 7 2- (aq)  2Cr 3+ (aq) SO 3 2- (aq)  + SO 4 2- (aq) + 2e - How can we balance the oxygen atoms? Method of Half Reactions (continued)

20. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 20 6e - + Cr 2 O 7 2- (aq)  2Cr 3+ (aq) + 7H 2 O H 2 O +SO 3 2- (aq)  + SO 4 2- (aq) + 2e - How can we balance the hydrogen atoms? Method of Half Reactions (continued)

21. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 21 This reaction occurs in an acidic solution. 14H + + 6e - + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O H 2 O +SO 3 2-  SO 4 2- + 2e - + 2H + How can we balance the electrons? Method of Half Reactions (continued)

22. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 22 14H + + 6e - + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O 3[H 2 O +SO 3 2-  SO 4 2- + 2e - + 2H + ] Cancel to get the Final Balanced Equation: Cr 2 O 7 2- + 3SO 3 2- + 8H +  2Cr 3+ + 3SO 4 2- + 4H 2 O WOW!! Method of Half Reactions (continued)

23. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 23 Exercise Balance the following oxidation – reduction reaction that occurs in acidic solution. Br – (aq) + MnO 4 – (aq)  Br 2 (l)+ Mn 2+ (aq) 10Br – (aq) + 16H + (aq) + 2MnO 4 – (aq)  5Br 2 (l)+ 2Mn 2+ (aq) + 8H 2 O(l)

24. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 24 1.Use the half–reaction method as specified for acidic solutions to obtain the final balanced equation as if H + ions were present. 2.To both sides of the equation, add a number of OH – ions that is equal to the number of H + ions. (We want to eliminate H + by forming H 2 O.) The Half–Reaction Method for Balancing Equations for Oxidation– Reduction Reactions Occurring in Basic Solution

25. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 25 3.Form H 2 O on the side containing both H + and OH – ions, and eliminate the number of H 2 O molecules that appear on both sides of the equation. 4.Check that elements and charges are balanced. The Half–Reaction Method for Balancing Equations for Oxidation– Reduction Reactions Occurring in Basic Solution

26. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 26 To balance half reactions that occur in acid solutions Work Session: page 830 # 16 a-c acid Objectives Review

27. Section 17.1 Atomic MassesGalvanic Cells Return to TOC Copyright © Cengage Learning. All rights reserved 27 To understand what a galvanic cell is and how it works To calculate cell potential for a given reaction Objectives

28. Section 17.1 Atomic MassesGalvanic Cells Return to TOC Copyright © Cengage Learning. All rights reserved 28 Galvanic Cell Device in which chemical energy is changed to electrical energy. Uses a spontaneous redox reaction to produce a current that can be used to do work.

29. Section 17.1 Atomic MassesGalvanic Cells Return to TOC Copyright © Cengage Learning. All rights reserved 29 A Galvanic Cell

30. Section 17.1 Atomic MassesGalvanic Cells Return to TOC Copyright © Cengage Learning. All rights reserved 30 Galvanic Cell Oxidation occurs at the anode. (O and A vowels) Reduction occurs at the cathode. (consonants) Salt bridge or porous disk – devices that allow ions to flow without extensive mixing of the solutions.  Salt bridge – contains a strong electrolyte held in a Jello–like matrix.  Porous disk – contains tiny passages that allow hindered flow of ions.

31. Section 17.1 Atomic MassesGalvanic Cells Return to TOC Copyright © Cengage Learning. All rights reserved 31 Voltaic Cell: Anode Reaction (losing e-s LEO)

32. Section 17.1 Atomic MassesGalvanic Cells Return to TOC Copyright © Cengage Learning. All rights reserved 32 Voltaic Cell: Cathode Reaction (gaining e-s GER)

33. Section 17.1 Atomic MassesGalvanic Cells Return to TOC Copyright © Cengage Learning. All rights reserved 33 Cell Potential A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment. The “pull”, or driving force, on the electrons is called the cell potential ( ), or the electromotive force (emf) of the cell.  Unit of electrical potential is the volt (V).  1 joule of work per coulomb of charge transferred.

34. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 34 Electrochemical Half-Reactions in a Galvanic Cell

35. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 35 Galvanic Cell All half-reactions are given as reduction processes in standard tables.  Table 17.1, 1 M, 1atm, 25°C When a half-reaction is reversed, the sign of E ° is reversed. Won’t be a factor in the calculation* When a half-reaction is multiplied by an integer, E ° remains the same. A galvanic cell runs spontaneously in the direction that gives a positive value for E ° cell. E ° cell = E °(cathode) - E °(anode) *Use numbers directly from Table 17.1**

36. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 36 Example: Fe 3+ (aq) + Cu(s) → Cu 2+ (aq) + Fe 2+ (aq) Write the Half-Reactions and list the E from 17.1  Fe 3+ + e – → Fe 2+ E ° = 0.77 V  Cu 2+ + 2e – → Cu E ° = 0.34 V Balance the overall reaction to determine Anode/Cathode Cu → Cu 2+ + 2e – – E ° = – 0.34 V (LEO Anode)  2Fe 3+ + 2e – → 2Fe 2+ E ° = 0.77 V (GER Cathode)  Cu + 2Fe 3+ → Cu 2+ + 2Fe 2+  Calculate E ° cell = E °(cathode) - E °(anode)  E ° cell = 0.77 – 0.34 = 0.43 V (+ is spontaneous)

37. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 37 Example: Al 3+ (aq) + Mg(s) → Al(s) + Mg 2+ (aq) Write the Half-Reactions and list the E from 17.1  Al 3+ + 3e – → Al E ° = -1.66 V  Mg 2+ + 2e – → Mg E ° = -2.37 V Balance the overall reaction to determine Anode/Cathode 3[Mg → Mg 2+ + 2e – ] – E ° = 2.37 V (LEO Anode)  2[Al 3+ + 3e – → Al] E ° = -1.66 V (GER Cathode)  3Mg + 2Al 3+ → 3Mg 2+ + 2Al  Calculate E ° cell = E °(cathode) - E °(anode)  E ° cell = -1.66 –(- 2.37) = 0.71 V (+ is spontaneous)

38. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 38 MnO 4 - (aq) + H + (aq) + ClO 3 - (aq) → ClO 4 - (aq) + Mn 2+ (aq) + H 2 O Write the Half-Reactions and list the E from 17.1 ClO 4 - + 2H + + 2e -  ClO 3 - + H 2 OE ° = 1.19 V MnO 4 - + 5e- + 8H +  Mn +2 + 4H 2 O E ° = 1.51 V Balance the overall cell reaction E ° = V (LEO Anode)  E ° = V (GER Cath)  Calculate E ° cell = E °(cathode) - E °(anode)  E ° cell = - = V (spontaneous??)

39. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 39 Line Notation Used to describe electrochemical cells. Anode components are listed on the left. Cathode components are listed on the right. Separated by double vertical lines. The concentration of aqueous solutions should be specified in the notation when known. Example: Mg(s)|Mg 2+ (aq)||Al 3+ (aq)|Al(s)  Mg → Mg 2+ + 2e – (anode)  Al 3+ + 3e – → Al(cathode) ABC Order…

40. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 40 Line Notation Page 831 # 25 and 27 also show the line notation Also 26 and 28 with line notation Which way does the Potato Clock work? Cu(s)|Cu 2+ (aq)||Zn 2+ (aq)|Zn(s) Or Zn(s)|Zn +2 (aq)||Cu 2+ (aq)|Cu(s)

41. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 41 Concept Check Sketch a cell using the following solutions and electrodes. Include:  The potential of the cell  The direction of electron flow  Labels on the anode and the cathode a)Ag electrode in 1.0 M Ag + (aq) and Cu electrode in 1.0 M Cu 2+ (aq)

42. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 42 Concept Check Sketch a cell using the following solutions and electrodes. Include:  The potential of the cell  The direction of electron flow  Labels on the anode and the cathode b)Zn electrode in 1.0 M Zn 2+ (aq) and Cu electrode in 1.0 M Cu 2+ (aq)

43. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 43 Concept Check Consider the cell from part b. What would happen to the potential if you increase the [Cu 2+ ]? Explain. The cell potential should increase.

44. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 44 Concept Check Page 831 # 29 sketch the 2 cells, pickup #27 Quiz #30?? end

45. Section 17.3 Cell Potential, Electrical Work, and Free Energy Return to TOC Copyright © Cengage Learning. All rights reserved 45 To understand that the work of a cell is always less that the calculated potential To predict the spontaneity of a galvanic cell Objectives

46. Section 17.3 Cell Potential, Electrical Work, and Free Energy Return to TOC Copyright © Cengage Learning. All rights reserved 46 Work Work is never the maximum possible if any current is flowing. In any real, spontaneous process some energy is always wasted – the actual work realized is always less than the calculated maximum.

47. Section 17.3 Cell Potential, Electrical Work, and Free Energy Return to TOC Copyright © Cengage Learning. All rights reserved 47 Spontaneity Prediction Use the data from Table 17.1 to determine if 1 M HNO 3 will dissolve gold metal to form a 1 M Au +3 solution. In other words, calculate the E ° for this galvanic cell. Remember, + E ° is spontaneous! E ° = -0.54 V

48. Section 17.3 Cell Potential, Electrical Work, and Free Energy Return to TOC Copyright © Cengage Learning. All rights reserved 48 To understand that the work of a cell is always less that the calculated potential To predict the spontaneity of a galvanic cell Work Session: 37, just calculate E °, no G Objectives Review

49. Section 17.4 Dependence of Cell Potential on Concentration Return to TOC Copyright © Cengage Learning. All rights reserved 49 To understand that the Nernst Equation is used to correct E ° for nonstandard conditions- namely different concentrations To calculate the cell potential using the Nernst Equation **have same metal, diff conc—add diff metal, diff conc **specify cathode/anode**smaller M is product, forming ion until [] = [] Objectives (Nernst Eq Excluded from AP Test 2014)

50. Section 17.4 Dependence of Cell Potential on Concentration Return to TOC Copyright © Cengage Learning. All rights reserved 50 Concentration Cell Remember, E ° is reported for 1 M solutions at 25 ° C e-s flow to the higher concentration, forming more Ag + on left, Ag metal on right until the concentrations are equal. Lower concentration is the product side.

51. Section 17.4 Dependence of Cell Potential on Concentration Return to TOC Copyright © Cengage Learning. All rights reserved 51 Nernst Equation At 25°C: Q is the reaction Quotient: Q = [products] A [reactants] B n = number of e-s transferred What is the value of log(Q) when the concentrations are both 1 M?

52. Section 17.4 Dependence of Cell Potential on Concentration Return to TOC Copyright © Cengage Learning. All rights reserved 52 Nernst Equation At 25°C: At equilibrium, the Reaction Quotient = K. At equilibrium, the cell no longer has any chemical driving force to move electrons, making the battery unable to do work. E = 0

53. Section 17.4 Dependence of Cell Potential on Concentration Return to TOC Copyright © Cengage Learning. All rights reserved 53 Concept Check Explain the difference between E and E °. When is E equal to zero? W hen the cell is in equilibrium ("dead" battery). When is E ° equal to zero? E  is equal to zero for a concentration cell.

54. Section 17.4 Dependence of Cell Potential on Concentration Return to TOC Copyright © Cengage Learning. All rights reserved 54 Exercise A concentration cell is constructed using two nickel electrodes with Ni 2+ concentrations of 1.0 M and 1.00 x 10 -4 M in the two half-cells. Calculate the potential of this cell at 25°C. 1)Calculate E ° 2)Balance Eq to get n 3)Use Nernst Eq 0.118 V

55. Section 17.4 Dependence of Cell Potential on Concentration Return to TOC Copyright © Cengage Learning. All rights reserved 55 Concept Check You make a galvanic cell at 25°C containing:  A nickel electrode in 1.0 M Ni 2+ (aq)  A silver electrode in 1.0 M Ag + (aq) Sketch this cell, labeling the anode and cathode, showing the direction of the electron flow, and calculate the cell potential. 1.03 V

56. Section 17.4 Dependence of Cell Potential on Concentration Return to TOC Copyright © Cengage Learning. All rights reserved 56 To understand that the Nernst Equation is used to correct E ° for nonstandard conditions- namely different concentrations To calculate the cell potential using the Nernst Equation Work Session: 53 a-d Quiz 54 a-d?? Objectives

57. Section 17.5 Batteries Return to TOC Copyright © Cengage Learning. All rights reserved 57 One of the Six Cells in a 12–V Lead Storage Battery

58. Section 17.5 Batteries Return to TOC Copyright © Cengage Learning. All rights reserved 58 A Common Dry Cell Battery

59. Section 17.5 Batteries Return to TOC Copyright © Cengage Learning. All rights reserved 59 A Mercury Battery

60. Section 17.5 Batteries Return to TOC Copyright © Cengage Learning. All rights reserved 60 Schematic of the Hydrogen-Oxygen Fuel Cell

61. Section 17.6 Corrosion Return to TOC Copyright © Cengage Learning. All rights reserved 61 Process of returning metals to their natural state – the ores from which they were originally obtained. Involves oxidation of the metal.

62. Section 17.6 Corrosion Return to TOC Copyright © Cengage Learning. All rights reserved 62 The Electrochemical Corrosion of Iron

63. Section 17.6 Corrosion Return to TOC Copyright © Cengage Learning. All rights reserved 63 Corrosion Prevention Application of a coating (like paint or metal plating)  Galvanizing Alloying Cathodic Protection  Protects steel in buried fuel tanks and pipelines.

64. Section 17.6 Corrosion Return to TOC Copyright © Cengage Learning. All rights reserved 64 Cathodic Protection

65. Section 17.7 Electrolysis Return to TOC Copyright © Cengage Learning. All rights reserved 65 Forcing a current through a cell to produce a chemical change for which the cell potential is negative.

66. Section 17.7 Electrolysis Return to TOC Copyright © Cengage Learning. All rights reserved 66 Stoichiometry of Electrolysis How much chemical change occurs with the flow of a given current for a specified time? current and time  quantity of charge  moles of electrons  moles of analyte  grams of analyte

67. Section 17.7 Electrolysis Return to TOC Copyright © Cengage Learning. All rights reserved 67 Stoichiometry of Electrolysis current and time  quantity of charge Coulombs of charge = amps (C/s) × seconds (s) quantity of charge  moles of electrons

68. Section 17.7 Electrolysis Return to TOC Copyright © Cengage Learning. All rights reserved 68 Concept Check An unknown metal (M) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate 0.0719 g of the metal from a solution containing M(NO 3 ) 3. What is the metal? gold (Au)

69. Section 17.7 Electrolysis Return to TOC Copyright © Cengage Learning. All rights reserved 69 Concept Check Consider a solution containing 0.10 M of each of the following: Pb 2+, Cu 2+, Sn 2+, Ni 2+, and Zn 2+. Predict the order in which the metals plate out as the voltage is turned up from zero. Cu 2+, Pb 2+, Sn 2+, Ni 2+, Zn 2+ Do the metals form on the cathode or the anode? Explain.

70. Section 17.8 Commercial Electrolytic Processes Return to TOC Copyright © Cengage Learning. All rights reserved 70 Production of aluminum Purification of metals Metal plating Electrolysis of sodium chloride Production of chlorine and sodium hydroxide

71. Section 17.8 Commercial Electrolytic Processes Return to TOC Copyright © Cengage Learning. All rights reserved 71 Producing Aluminum by the Hall-Heroult Process

72. Section 17.8 Commercial Electrolytic Processes Return to TOC Copyright © Cengage Learning. All rights reserved 72 Electroplating a Spoon

73. Section 17.8 Commercial Electrolytic Processes Return to TOC Copyright © Cengage Learning. All rights reserved 73 The Downs Cell for the Electrolysis of Molten Sodium Chloride

74. Section 17.8 Commercial Electrolytic Processes Return to TOC Copyright © Cengage Learning. All rights reserved 74 The Mercury Cell for Production of Chlorine and Sodium Hydroxide

75. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 75 The Half–Reaction Method for Balancing Equations for Oxidation– Reduction Reactions Occurring in Acidic Solution

76. Section 17.0 Balancing Oxidation–Reduction Equations Return to TOC Copyright © Cengage Learning. All rights reserved 76 The Half–Reaction Method for Balancing Equations for Oxidation– Reduction Reactions Occurring in Basic Solution

77. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 77 Concept Check Order the following from strongest to weakest oxidizing agent and justify. Of those you cannot order, explain why. Fe NaF - Na + Cl 2

78. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 78 Description of a Galvanic Cell The cell potential is always positive for a galvanic cell where E ° cell = E °(cathode)–E °(anode) and the balanced cell reaction. The direction of electron flow, obtained by inspecting the half–reactions and using the direction that gives a positive E ° cell.

79. Section 17.2 The MoleStandard Reduction Potentials Return to TOC Copyright © Cengage Learning. All rights reserved 79 Description of a Galvanic Cell Designation of the anode and cathode. The nature of each electrode and the ions present in each compartment. A chemically inert conductor is required if none of the substances participating in the half–reaction is a conducting solid.

80. Section 17.3 Cell Potential, Electrical Work, and Free Energy Return to TOC Copyright © Cengage Learning. All rights reserved 80 Maximum Cell Potential Directly related to the free energy difference between the reactants and the products in the cell.  ΔG° = –nFE °  F = 96,485 C/mol e –