1. Slide 1 Electrochemistry Chapter 17

2. Slide 2 Why Study Electrochemistry? Batteries Batteries Corrosion Corrosion Industrial production of chemicals such as Cl 2, NaOH, F 2 and Al Industrial production of chemicals such as Cl 2, NaOH, F 2 and Al Biological redox reactions,photosynthesis Biological redox reactions,photosynthesis 6CO 2 + 6H 2 O --> C 6 H 12 O 6 + 6O 2 C 6 H 12 O 6 + O 2 --> 6CO 2 + 6H 2 O +Energy

3. Slide 3 Redox Reactions01

4. Slide 4 Redox reaction are those involving the oxidation and reduction of species. LEO – L oss of E lectrons is Oxidation. GER –Gain of Electrons Is Reduction. Oxidation and reduction must occur together. They cannot exist alone. Redox Reactions01

5. Slide 5 Oxidation Half-Reaction: Zn(s)  Zn 2+ (aq) + 2 e –. The Zn loses two electrons to form Zn 2+. Redox Reactions02

6. Slide 6 Redox Reactions03 Reduction Half-Reaction: Cu 2+ (aq) + 2 e –  Cu(s) The Cu 2+ gains two electrons to form copper.

7. Slide 7 Electrochemical Cells spontaneous redox reaction anode oxidation cathode reduction

8. Slide 8 Overall: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Redox Reactions04

9. Slide 9 Electromotive Force (emf) Water only spontaneously flows one way in a waterfall. Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

10. Slide 10 Electromotive Force (emf) The potential difference between the anode and cathode in a cell is called the electromotive force (emf). It is also called the cell potential, and is designated E cell.

11. Slide 11 Cell Potential Cell potential is measured in volts (V). C (Coulomb): The amount of charge transferred when a current of 1 ampere (A) Flows for one second. Volt, a potential difference between two points which results to a current of one ampere through a resistance of one ohm

12. Slide 12 Cell Potentials and Free-Energy Changes for Cell Reactions 1 J = 1 C x 1 V volt SI unit of electric potential joule SI unit of energy coulomb Electric charge 1 coulomb is the amount of charge transferred when a current of 1 ampere flows for 1 second. 1 V = 1 JCJC

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14. Slide 14 Electrochemical Cells01 Electrodes: are usually metal strips/wires connected by an electrically conducting wire. Salt Bridge: is a U-shaped tube that contains a gel permeated with a solution of an inert electrolyte. Anode: is the electrode where oxidation takes place, (-). Cathode: is the electrode where reduction takes place, (+) terminal

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16. Slide 16 Cells Notation02 Anode Half-Cell || Cathode Half-Cell Electrode | Anode Soln || Cathode Soln | Electrode Zn(s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu(s)

17. Slide 17 Electrochemical Cells02 Anode Half-Cell || Cathode Half-Cell Electrode | Anode Soln || Cathode Soln | Electrode Fe(s) | Fe 2+ (aq) || Fe 3+ (aq), | Pt(s)

18. Slide 18 Write the cell notation for: Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s)

19. Slide 19 Electrochemical Cell Potentials The standard half-cell potentials are determined from the difference between two electrodes. The reference point is called the standard hydrogen electrode (S.H.E.) and consists of a platinum electrode in contact with H 2 gas (1 atm) and aqueous H + ions (1 M). The standard hydrogen electrode is assigned an arbitrary value of exactly 0.00 V.

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21. Slide 21 Cu 2+ (aq) + 2e- Cu (s) E 0 = 0.34 V ∆G˚ = –nFE˚

22. Slide 22 Electrochemical Cells06 ---

23. Slide 23 E 0 is for the reaction as written The more positive E 0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0 ∆G˚ = –nFE˚` n = number of moles of electrons in reaction F = 96,500 C/mole

24. Slide 24 Electrochemical Cells03 The standard potential of any galvanic cell is the sum of the standard half-cell potentials for the oxidation and reduction half-cells. E° cell = E° oxidation + E° reduction Standard half-cell potentials are always tabulated as a reduction process. The sign must be changed for the oxidation process.

25. Slide 25 Electrochemical Cells07 When selecting two half-cell reactions the more negative value, or smaller E° will form the oxidation half-cell. Consider the reaction between zinc and silver: Ag + (aq) + e –  Ag(s)E° = 0.80 V Zn 2+ (aq) + 2 e –  Zn(s)E° = – 0.76 V Therefore, zinc forms the oxidation half-cell: Zn(s)  Zn 2+ (aq) + 2 e – E° = – (–0.76 V) E 0 = E° Ox + E ° Red cell 00 E 0 = 0.76 V+ 0.80 V = 1.56 V cell 00

26. Slide 26 Write the cell rection and calculate the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E 0 = -0.40 V Cr 3+ (aq) + 3e - Cr (s) E 0 = -0.74 V Cd is the stronger oxidizer Cd will oxidize Cr 2e - + Cd 2+ (1 M) Cd (s) Cr (s) Cr 3+ (1 M) + 3e - Anode (oxidation): Cathode (reduction): 2Cr (s) + 3Cd 2+ (1 M) 3Cd (s) + 2Cr 3+ (1 M) x 2 x 3 E° cell = E° oxidation + E° reduction E 0 = 0.74 V + (-0.4 V) cell E 0 = 0.34 V cell

27. Slide 27 Spontaneity of a Reaction01 The value of E˚ cell is related to the thermodynamic quantities of ∆G˚ and K. The value of E˚ cell is related to ∆G˚ by: ∆G˚ = –nFE˚ cell The value of K is related to ∆G˚ by: ∆G˚ = –RT ln K F = 96,500 C/mole  G 0 = -RT ln K = -nFE cell 0

28. Slide 28 Spontaneity of Redox Reactions ΔG = -nFE cell  G 0 = -nFE cell 0 n = number of moles of electrons in reaction F = 96,500 J V mol = 96,500 C/mol  G 0 = -RT ln K = -nFE cell 0 E cell 0 = RT nF ln K (8.314 J/K mol)(298 K) n (96,500 J/V mol) ln K = = 0.0257 V n ln K E cell 0 = 0.0592 V n log K E cell 0 0.0257 V x nE0E0 cell exp K = ln K = 2.3035 Log K º ( see slide 4)

29. Slide 29 Spontaneity of Redox Reactions ∆G˚ = –RT ln K = 0.0257 V n ln K E cell 0 ---

30. Slide 30 2e - + Fe 2+ Fe Oxidation: Reduction: What is the equilibrium constant for the following reaction at 25 0 C? Fe 2+ (aq) + 2Ag (s) Fe (s) + 2Ag + (aq) = 0.0257 V n ln K E cell 0 E 0 = (-0.80) +( -0.45) E 0 = -1.25 V 0.0257 V x nE0E0 cell exp K = n = 2 0.0257 V x 2x 2-1.25 V = exp K = 5.67 x 10 -43 E° cell = E° oxidation + E° reduction E 0 = -0.80 V E 0 = -0.45V 2Ag 2Ag + + 2e -

31. Slide 31 Cell Emf Under Nonstandard Conditions  G =  G 0 + RT ln Q  G = -nFE  G 0 = -nFE 0 -nFE = -nFE 0 + RT ln Q E = E 0 - ln Q RT nF Nernst equation At 298 K - 0.0257 V n ln Q E 0 E = - 0.0592 V n log Q E 0 E = If we divide both sides by “-nF” ln Q = 2.3035 Log Q

32. Slide 32 Will the following reaction occur spontaneously at 25 0 C if [Fe 2+ ] = 0.60 M and [Cd 2+ ] = 0.010 M? Fe 2+ (aq) + Cd (s) Fe (s) + Cd 2+ (aq) 2e - + Fe 2+ Fe Cd Cd 2+ + 2e - Oxidation: Reduction: n = 2 E 0 = 0.40 V + (-0.45 V) E 0 = -0.05 V - 0.0257 V n ln Q E 0 E = - 0.0257 V 2 ln -0.05 VE = 0.010 0.60 E = 0.0026 E > 0 Spontaneous E 0 = 0.40 V E 0 = -0.45 V E° cell = E° oxidation + E° reduction  G = -nFE cell ΔG < 0

33. Slide 33 Write Cell Notation and Calculate the Cell Potential for the Following Cell E = 0.088 V - 0.0592 V n log Q E 0 E =

34. Slide 34 Concentration Cells Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. For such a cell,would be 0, but Q would not. E cell  Therefore, as long as the concentrations are different, E will not be 0. - 0.0592V n log Q E 0 E =

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43. Slide 43 Calculate the concentration of H + in the following if the Cell voltage is 0.7 V. Zn(s)|Zn 2+ (aq, 1 M)|| H + (aq, ?)|H 2 (g, 1 atm)|Pt E ° red (Zn/Zn 2+ )=  0.76 V. H + = 0.1 M - 0.0592 V n log Q E 0 E = Nernst equation

44. Slide 44 Nernst Equation Could be Applied to Half Cell Potential A particularly important use of the Nernst equation is in the electrochemical determination of pH. Pt | H 2 (1 atm) | H + (? M) || Reference Cathode E cell = E H 2  H + + E ref The Nernst equation can be applied to the half-reaction: H 2 (g)  2 H + (aq) + 2 e –

45. Slide 45 The Nernst Equation05 For the half-reaction: H 2 (g)  2 H + (aq) + 2 e – E° = 0 V for this reaction (standard hydrogen electrode). According to the problem, P H 2 is 1 atm. - 0.0592 V n log Q E 0 E =

46. Slide 46 The Nernst Equation07 The overall potential is given by: Which rearranges to give an equation for the determination of pH:

47. Slide 47 Replacing Standard Hydrogen Electrode With Glass Electrode (Ag/AgCl wire in dilute HCl)

48. Slide 48 pH meter A higher cell potential indicates a higher pH, therefore we can measure pH by measuring E cell. A glass electrode (Ag/AgCl wire in dilute HCl) with a calomel reference is the most common arrangement. Glass: Ag(s) + Cl – (aq)  AgCl(s) + e – E° = –0.22 V Calomel: Hg 2 Cl 2 (s) + 2 e –  2 Hg(l) + 2 Cl – (aq) E° = 0.28 V

49. Slide 49 pH Electrode09 The glass pH probe is constructed as follows: Ag(s) | AgCl(s) | HCl(aq) | glass | H + (aq) || reference The difference in [H + ] from one side of the glass membrane to the other causes a potential to develop, which adds to the measured E cell.

50. Slide 50 The following cell has a potential of 0.28 V at 25°C: Pt(s) | H 2 (1 atm) | H + (? M) || Pb 2+ (1 M) | Pb(s) What is the pH of the solution at the anode? H 2 (g) + Pb 2+ (aq)  2 H + (aq) + Pb(s) E o ref = ­- 0.13 V ( Please see page 698);

51. Slide 51 Batteries Lead storage battery

52. Batteries Lead Storage Battery 2PbSO 4 (s) + 2H 2 O(l)Pb(s) + PbO 2 (s) + 2H 1+ (aq) + 2HSO 4 1- (aq) PbSO 4 (s) + 2H 2 O(l)PbO 2 (s) + 3H 1+ (aq) + HSO 4 1- (aq) + 2e - PbSO 4 (s) + H 1+ (aq) + 2e - Pb(s) + HSO 4 1- (aq) Overall: Anode: Cathode:

53. Slide 53 Batteries Leclanché cell Dry cell Zn (s) Zn 2+ (aq) + 2e - Anode: Cathode: 2NH 4 (aq) + 2MnO 2 (s) + 2e - Mn 2 O 3 (s) + 2NH 3 (aq) + H 2 O (l) + Zn (s) + 2NH 4 (aq) + 2MnO 2 (s) Zn 2+ (aq) + 2NH 3 (aq) + H 2 O (l) + Mn 2 O 3 (s) 1.5 V but deteriorates to 0.8 V with use Zn (s) + 2NH 4 (aq) + 2MnO 2 (s) Zn 2+ (aq) + 2NH 3 (aq) + H 2 O (l) + Mn 2 O 3 (s)

54. Slide 54 Batteries A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioningfuel cell Anode: Cathode: O 2 (g) + 2H 2 O (l) + 4e - 4OH - (aq) 2H 2 (g) + 4OH - (aq) 4H 2 O (l) + 4e - 2H 2 (g) + O 2 (g) 2H 2 O (l)

55. Slide 55 Figure 7.13: The essentials of a typical fuel cell. © 2003 John Wiley and Sons Publishers

56. Slide 56 Replacing KOH with Proton ExchangeMembrane

57. Slide 57 Proton Exchange Membrane (PEM) Fuel Cell Anode side: 2H 2 => 4H + + 4e - Cathode side: O 2 + 4H + + 4e - => 2H 2 O Net reaction: 2H 2 + O 2 => 2H 2 O

58. Slide 58 Corrosion01 Corrosion is the oxidative deterioration of metal. 25% of steel produced in USA goes to replace steel structures and products destroyed by corrosion. Rusting of iron requires the presence of BOTH oxygen and water. Rusting results from tiny galvanic cells formed by water droplets.

59. Slide 59 Corrosion Corrosion: The oxidative deterioration of a metal.

60. Slide 60 Corrosion 4Fe +2 (aq) + O 2 (g)+ 4H + (aq) 4Fe +3 (aq) + 2H 2 O 2Fe +3 (aq) + 4H 2 O(l) Fe 2 O 3.H 2 O + 6H +

61. Slide 61 …Corrosion Prevention: Galvanizing

62. Slide 62 Corrosion Prevention03 Galvanizing: is the coating of iron with zinc. Zinc is more easily oxidized than iron, which protects and reverses oxidation of the iron. Cathodic Protection: is the protection of a metal from corrosion by connecting it to a metal (a sacrificial anode) that is more easily oxidized. Attaching a magnesium stake to iron will corrode the magnesium instead of the iron

63. Slide 63 Electrolysis01

64. Slide 64 Electrolysis of Water

65. Slide 65 Electrolysis: is the process in which electrical energy is used to drive a nonspontaneous chemical reaction. An electrolytic cell is an apparatus for carrying out electrolysis. Processes in an electrolytic cell are the reverse of those in a galvanic cell. Electrolysis01

66. Slide 66 Electrolysis05 Electrolysis of Water: Requires an electrolyte species, that is less easily oxidized and reduced than water, to carry the current. Anode: Water is oxidized to oxygen gas. 2 H 2 O(l)  O 2 (g) + 4 H + (aq) + 4 e – Cathode: Water is reduced to hydrogen gas. 4 H 2 O(l) + 4 e –  2 H 2 (g) + 4 OH – (aq)

67. Slide 67 Electrolysis Applications01 Manufacture of Sodium (Downs Cell): Bp (NaCl) = 801 o C Bp (NaCl-CaCl 2 ) = 580 o C

68. Slide 68 Electrolysis07 Quantitative Electrolysis: The amount of substance produced at an electrode by electrolysis depends on the quantity of charge passed through the cell. Reduction of 1 mol of sodium ions requires 1 mol of electrons to pass through the system. The charge on 1 mol of electrons is 96,500 coulombs.

69. Slide 69 Electrolysis08 To determine the moles of electrons passed, we measure the current and time that the current flows: Charge (C) = Current (A) x Time (s) Because the charge on 1 mol of e – is 96,500 C, the number of moles of e – passed through the cell is:

70. Slide 70 Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mole e - = 96,500 C

71. Slide 71 How much Ca will be produced in an electrolytic cell of molten CaCl 2 if a current of 0.452 A is passed through the cell for 1.5 hours? Anode: Cathode: Ca 2+ (l) + 2e - Ca (s) 2Cl - (l) Cl 2 (g) + 2e - Ca 2+ (l) + 2Cl - (l) Ca (s) + Cl 2 (g) 2 mole e - = 1 mole Ca 0.452 C s x 1.5 hr x 3600 s hr96,500 C 1 mol e - x 2 mol e - 1 mol Ca x = 0.0126 mol Ca = 0.50 g Ca

72. Slide 72 Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Calculate moles of e- c)Calculate charge 4.38 mol e- 96,500 C/mol e- = 423,000 C 4.38 mol e- 96,500 C/mol e- = 423,000 C

73. Slide 73 Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (s) + H + (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a)454 g Pb = 2.19 mol Pb b)Mol of e- = 4.38 mol c)Charge = 423,000 C About 78 hours d)Calculate time

74. Slide 74 Electrolysis Applications01 Manufacture of Sodium (Downs Cell): Bp (NaCl) = 801 o C Bp (NaCl-CaCl 2 ) = 580 o C

75. Slide 75 Electrolysis Applications02 Manufacture of Cl 2 and NaOH (Chlor–Alkali): Chlorine bleach: Cl 2 + 2 NaOH → NaCl + NaClO + H 2 O

76. Slide 76 Manufacture of Aluminum (Hall–Heroult): Anode: (positive electrode) C (s) + 2O 2- (l) ---> CO 2(g) + 4e- Cathode: (negative electrode) Al 3+ (l) + 3e- ---> Al (l) Overall Reaction: 2Al 2 O 3(l) + 3C (s) ---> 4Al (l) + 3CO 2(g) Bauxite: Al2O3 + SiO2 + TiO2 + Fe2O3 Hot NaOH used to dissolve alumnum Comounds and other materials separated by filration Mixture mp. 1000 C Al2O3 mp. 2045 C

77. Slide 77 Electrorefining and Electroplating:Electroplating

78. Given the following reaction, which is true? 1. Plating Ag onto Cu is a spontaneous process. 2. Plating Cu onto Ag is a spontaneous process. 3. Plating Ag onto Cu is a nonspontaneous process. 4. Plating Cu onto Ag is a nonspontaneous process. 5. Energy will have to be put in for the reaction to proceed. Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) E° = +0.46 V

79. Given the following reaction, which is true? 1. Plating Ag onto Cu is a spontaneous process. 2. Plating Cu onto Ag is a spontaneous process. 3. Plating Ag onto Cu is a nonspontaneous process. 4. Plating Cu onto Ag is a nonspontaneous process. 5. Energy will have to be put in for the reaction to proceed. Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) E° = +0.46 V

80. Based on the standard reduction potentials, which metal would not provide cathodic protection to iron? 1.Magnesium 2.Nickel 3.Sodium 4.Aluminum

81. Slide 81 Electrochemical Cells06 ---

82. Correct Answer: In order to provide cathodic protection, the metal that is oxidized while protecting the cathode must have a more negative standard reduction potential. Here, only Ni has a more positive reduction potential (  0.26 V) than Fe 2+ (  0.44 V) and cannot be used for cathodic protection. 1.Magnesium 2.Nickel 3.Sodium 4.Aluminum

83. Ni 2+ is electrolyzed to Ni by a current of 2.43 amperes. If current flows for 600 s, how much Ni is plated (in grams)? (AW Ni = 58.7 g/mol) 1.0.00148 g 2.0.00297 g 3.0.444 g 4.0.888 g

84. Correct Answer: Fn FWti    mass  C/mol) 96,500(2 g/mol) (58.7s) (600. A2.43 mass    1.0.00148 g 2.0.00297 g 3.0.444 g 4.0.888 g

85. Slide 85 Alkaline Dry-Cell 04 Alkaline Dry-Cell: Modified Leclanché cell which replaces NH 4 Cl with NaOH or KOH. Anode: Zinc metal can on outside of cell. Zn(s) + 2 OH – (aq)  ZnO(s) + H 2 O(l) + 2 e – Cathode: MnO 2 and carbon black paste on graphite. 2 MnO 2 (s) + H 2 O(l) + 2 e –  Mn 2 O 3 (s) + 2 OH – (aq) Electrolyte: NaOH or KOH, and Zn(OH) 2 paste. Cell Potential: 1.5 V but longer lasting, higher power, and more stable current and voltage.

86. Slide 86 Batteries Zn(Hg) + 2OH - (aq) ZnO (s) + H 2 O (l) + 2e - Anode: Cathode: HgO (s) + H 2 O (l) + 2e - Hg (l) + 2OH - (aq) Zn(Hg) + HgO (s) ZnO (s) + Hg (l) Mercury Battery 0.9 V Non-rechargeable button cells for watches, hearing aids, and calculatorsbutton cells

87. 1. N of NO 2 - is reduced, Cr of Cr 2 O 7 2- is oxidized 2. N of NO 2 - is oxidized, Cr of Cr 2 O 7 2- is reduced 3. O of NO 2 - is oxidized, Cr of Cr 2 O 7 2- is reduced 4. Cr 3+ is reduced, N of NO 2 - is oxidized 5. N of NO 3 - is oxidized, Cr 3+ is reduced For the reaction given below, what substance is oxidized and what is reduced? 3 NO 2 - + Cr 2 O 7 2- + 8 H + 2 Cr 3+ + 3 NO 3 - + 4 H 2 O

88. 1. N of NO 2 - is reduced, Cr of Cr 2 O 7 2- is oxidized 2. N of NO 2 - is oxidized, Cr of Cr 2 O 7 2- is reduced 3. O of NO 2 - is oxidized, Cr of Cr 2 O 7 2- is reduced 4. Cr 3+ is reduced, N of NO 2 - is oxidized 5. N of NO 3 - is oxidized, Cr 3+ is reduced For the reaction given below, what substance is oxidized and what is reduced? 3 NO 2 - + Cr 2 O 7 2- + 8 H + 2 Cr 3+ + 3 NO 3 - + 4 H 2 O

89. 1. 1, 1, 2, 1, 1 2. 1, 1, 4, 1, 2 3. 1, 1, 2, 1, 2 4. 4, 1, 2, 1, 2 5. 4, 1, 4, 4, 2 When the following reaction is balanced, what are the coefficients for each substance? __Ag + __O 2 + __H + __Ag + + __H 2 O

90. 1. 1, 1, 2, 1, 1 2. 1, 1, 4, 1, 2 3. 1, 1, 2, 1, 2 4. 4, 1, 2, 1, 2 5. 4, 1, 4, 4, 2 When the following reaction is balanced, what are the coefficients for each substance? __Ag + __O 2 + __H + __Ag + + __H 2 O

91. 1. 1 barr pressure for Cl 2(g) and 1 M solution for Cl – (aq). 2. 1 M solution for Cl 2(g) and for Cl – (aq). 3. 1 atm pressure for Cl 2(g) and for Cl – (aq). 4. 1 atm pressure for Cl 2(g) and 1 M solution for Cl – (aq). 1. 1 barr pressure for Cl 2(g) and 1 M solution for Cl – (aq). 2. 1 M solution for Cl 2(g) and for Cl – (aq). 3. 1 atm pressure for Cl 2(g) and for Cl – (aq). 4. 1 atm pressure for Cl 2(g) and 1 M solution for Cl – (aq).

92. 1. 1 barr pressure for Cl 2(g) and 1 M solution for Cl – (aq). 2. 1 M solution for Cl 2(g) and for Cl – (aq). 3. 1 atm pressure for Cl 2(g) and for Cl – (aq). 4. 1 atm pressure for Cl 2(g) and 1 M solution for Cl – (aq). 1. 1 barr pressure for Cl 2(g) and 1 M solution for Cl – (aq). 2. 1 M solution for Cl 2(g) and for Cl – (aq). 3. 1 atm pressure for Cl 2(g) and for Cl – (aq). 4. 1 atm pressure for Cl 2(g) and 1 M solution for Cl – (aq). 1Bar =.987 atm

93. Which substance is the stronger oxidizing agent? Br 2 O 2 NO 3 - H + Cl 2

94. Which substance is the stronger oxidizing agent? Br 2 O 2 NO 3 - H + Cl 2

95. 1.+0.76 V 2.+1.52 V 3.  0.76 V 4.  1.52 V Calculate the emf of the following cell: Zn(s)|Zn 2+ (aq, 1 M)|| H + (aq, 1 M)|H 2 (g, 1 atm)|Pt E° (Zn/Zn 2+ )=  0.76 V.

96. Correct Answer: Zn is the anode, hydrogen at the Pt wire is the cathode. 1.+0.76 V 2.+1.52 V 3.  0.76 V 4.  1.52 V E° cell = E° red + E°ox E° cell = 0.00 V + (0.76 V) E° cell = +0.76 V

97. Slide 97 The End

98. Slide 98 Batteries06 Nickel–Cadmium Battery: is rechargeable. Anode: Cadmium metal. Cd(s) + 2 OH – (aq)  Cd(OH) 2 (s) + 2 e – Cathode: Nickel(III) compound on nickel metal. NiO(OH) (s) + H 2 O(l) + e –  Ni(OH) 2 (s) + OH – (aq) Electrolyte: Nickel oxyhydroxide, NiO(OH). Cell Potential: 1.30 V

99. Slide 99 Batteries Solid State Lithium Battery 1.5 V to about 3.7 V, Slid Electrolyte: Inorganic ceramic andnorganic ceramic and organic polymer solid-electrolyte materials are reviewed

100. Slide 100 Batteries08 Lithium Ion (Li–ion): The newest rechargeable battery is based on the migration of Li + ions. Anode: Li metal, or Li atom impregnated graphite. Li(s)  Li + + e – Cathode: Metal oxide or sulfide that can accept Li +. MnO 2 (s) + Li + (aq) + e –  LiMnO 2 (s) Electrolyte: Lithium-containing salt such as LiClO 4, in organic solvent. Solid state polymers can also be used. Cell Potential: 3.0 V

101. Slide 101 Batteries07 Nickel–Metal–Hydride (NiMH): Replaces toxic Cd anode with a hydrogen atom impregnated ZrNi 2 metal alloy. Applications of NiMH type batteries includes hybrid vehicles such as the Toyota Prius and consumer electronics.hybrid vehicles ToyotaPrius 1.2 V Anode: Cathod: