1. Jeffrey Mack California State University, Sacramento Chapter 20 Principles of Chemical Reactivity: Electron Transfer Reactions

2. oxidation- reductionredoxElectron transfer reactions are oxidation- reduction or redox reactions. Redox reactions can result in the generation of an electric current or be caused by imposing an electric current. ELECTROCHEMISTRYTherefore, this field of chemistry is often called ELECTROCHEMISTRY. Electron Transfer Reactions

3. OXIDATIONOXIDATION: loss of electron(s) by a species; increase in oxidation number. REDUCTIONREDUCTION: gain of electron(s); decrease in oxidation number. OXIDIZING AGENTOXIDIZING AGENT: electron acceptor; species is reduced. REDUCING AGENTREDUCING AGENT: electron donor; species is oxidized. Review of Terminology for Redox Reactions

4. Applications: Batteries Corrosion Industrial production of chemicals such as Cl 2, NaOH, F 2 and Al Biological electron transfer reactions The heme group Electrochemistry

5. An apparatus that allows an electron transfer reaction to occur via an external connector. Voltaic or galvanic cells: Reactions that produce a chemical current that are product favored Electrolytic cell: Reactions which require an electric current to cause chemical change. (reactant favored) Batteries are voltaic cells Electrochemical Cells

6. Alessandro Volta, 1745- 1827, Italian scientist and inventor. Luigi Galvani, 1737-1798, Italian scientist and inventor. Electrochemistry Pioneers

7. Cu(s) + 2Ag + (aq)  Cu 2+ (aq) + 2Ag(s) Oxidation/Reduction Reactions

8. Direct Redox Reaction Oxidizing and reducing agents in direct contact. Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) Oxidation-Reduction Reactions: “Redox”

9. Indirect Redox Reaction A battery functions by transferring electrons through an external wire from the reducing agent to the oxidizing agent. Oxidation-Reduction Reactions: “Redox”

10. In any Oxidation Reduction reaction: One species is Oxidized and one species is Reduced. Neither can occur alone. That which is reduced is the Oxidizing Agent. That which is Oxidized is the Reducing Agent. The total number of electrons lost in oxidation must equal the total number of electrons gained in reduction. Redox reactions must therefore be balanced for mass and charge. Oxidation-Reduction Reactions: “Redox”

11. Step 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction. OxCu(s)  Cu 2+ (aq) RedAg + (aq)  Ag(s) Step 2: Balance each for mass. Already done in this case. Step 3: Balance each half-reaction for charge by adding electrons. Ox Cu(s)  Cu 2+ (aq) + 2e  Red Ag + (aq) + e   Ag(s) Balancing Redox Equations

12. Step 4: Multiply each half-reaction by a factor so that the reducing agent supplies as many electrons as the oxidizing agent requires. Reducing agent:Cu(s)  Cu 2+ (aq) + 2e  Oxidizing agent: 2Ag + (aq) + 2e   2Ag(s) Step 5: Add half-reactions to give the overall equation. Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2Ag(s) The equation is now balanced for both charge and mass. Balancing Redox Equations

13. Some redox reactions have equations that must be balanced by special techniques Mn = +7Fe = +2Fe = +3Mn = +2 MnO 4 - + 5 Fe 2+ + 8 H +  Mn 2+ + 5 Fe 3+ + 4 H 2 O Balancing Equations for Redox Reactions

14. Reduction of VO 2 + by Zn

15. Balance the following in acid solution: VO 2 + + Zn  VO 2+ + Zn 2+ Step 1:Write the half-reactions OxZn  Zn 2+ RedVO 2 +  VO 2+ Step 2:Balance each half-reaction for mass. OxZn  Zn 2+ Red 2 H + + VO 2 +  VO 2+ + H 2 O Add H 2 O on O-deficient side and add H + on other side for H- balance. Balancing Redox Equations

16. Step 3:Balance half-reactions for charge. OxZn  Zn 2+ + 2e- Rede - + 2 H + + VO 2 +  VO 2+ + H 2 O Step 4:Multiply by an appropriate factor. OxZn  Zn 2+ + 2e- Red2e - + 4 H + + 2 VO 2 +  2 VO 2+ + 2 H 2 O Step 5:Add balanced half-reactions Zn + 4 H + + 2 VO 2 +  Zn 2+ + 2 VO 2+ + 2 H 2 O Balancing Redox Equations

17. Never add O 2, O atoms, or O 2- to balance oxygen. Never add H 2 or H atoms to balance hydrogen. Be sure to write the correct charges on all the ions. Check your work at the end to make sure mass and charge are balanced. PRACTICE! Tips on Balancing Redox Reaction Equations

18. Oxidation: 2e- Oxidation: Zn(s)  Zn 2+ (aq) + 2e- Reduction: 2e- Reduction: Cu 2+ (aq) + 2e-  Cu(s) -------------------------------------------------------- Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” Chemical Change & Electric Current

19. To obtain a current that can do work, the oxidizing and reducing agents are separated so that electron transfer occurs via an external wire. GALVANICVOLTAIC This is accomplished in a GALVANIC or VOLTAIC cell. battery A group of such cells is called a battery. Chemical Change & Electric Current

20. Anode The electrode where oxidation occurs The electrode where mass is lost The electrode that attracts anions Cathode The electrode where reduction occurs The electrode where mass is gained The electrode that attracts cations Chemical Change & Electric Current

21. Electrons travel thru external wire. Salt bridgeSalt bridge allows anions and cations to move between electrode compartments. Zn  Zn 2+ + 2e- Cu 2+ + 2e-  Cu  Anions Cations  OxidationAnodeNegative ReductionCathodePositive

22. The Cu|Cu 2+ & Ag|Ag + Cell

23. Electrons move from anode to cathode in the wire. Anions & cations move thru the salt bridge. Electrochemical Cell

24. Terminology for Voltaic Cells

25. An arrangement of silver and zinc disks used to generate an electric current is show in this drawing by Alessandro Volta. The Voltaic Pile

26. Electrons are “driven” from anode to cathode by an electromotive force or emf. For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 °C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M. Zn and Zn 2+, anode Cu and Cu 2+, cathode 1.10 V 1.0 M Cell Potential, E

27. For Zn/Cu cell, potential is +1.10 V at 25 °C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M. This is the STANDARD CELL POTENTIAL, E° E° is a quantitative measure of the tendency of reactants to proceed to products at standard state conditions: 1 atm, 1M solutions @ 25 °C. Cell potential in a voltaic cell is the electrical potential energy difference between the cathode and anode. Cell potential depends on the ease of reduction at the cathode verses that of the anode. Cell Potential, E

28. Balanced half-reactions can be added together to get overall, balanced equation. Zn(s)  Zn 2+ (aq) + 2e- Cu 2+ (aq) + 2e-  Cu(s) -------------------------------------------- Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) If we know E o for each half-reaction, we could get E o for net reaction. Calculating Cell Voltage

29. Individual oxidation/reduction half reaction E o values cannot be measured directly. They are measured relative to a standard reference cell. The Standard Hydrogen Electrode (SHE) 2 H + (aq, 1 M) + 2e   H 2 (g, 1 atm) The potential of the cell is set to an E° = 0.0 V Cell Potentials, E°

30. Zn/Zn 2+ half-cell hooked to a SHE. E° for the cell = +0.76 V Negative electrode Electron donor Electron acceptor Positive electrode 2 H + + 2e-  H 2 ReductionCathode Zn  Zn 2+ + 2e- OxidationAnode

31. Reduction of H + by Zn

32. Overall reaction is reduction of H + by Zn metal. Zn(s) + 2 H + (aq)  Zn 2+ + H 2 (g) E° = +0.76 V Therefore, E° for Zn  Zn 2+ (aq) + 2e- is +0.76 V stronger Zn is a stronger reducing agent than H 2. Reduction of H + by Zn

33. E° = +0.34 V Electron acceptor Electron donor Cu 2+ + 2e-  Cu ReductionCathode H 2  2 H + + 2e- OxidationAnode Positive electrode Negative electrode Cu/Cu 2+ and H 2 /H + Cell

34. Overall reaction is reduction of Cu 2+ by H 2 gas. Cu 2+ (aq) + H 2 (g)  Cu(s) + 2 H + (aq) Measured E° = +0.34 V +0.34V Therefore, E° for Cu 2+ (aq)+ 2e-  Cu(s) is +0.34V Cu/Cu 2+ and H 2 /H + Cell

35. Zn(s)  Zn 2+ (aq) + 2e-E° = +0.76 V Cu 2+ (aq) + 2e-  Cu(s)E° = +0.34 V --------------------------------------------------------------- Cu 2+ (aq) + Zn(s) f Zn 2+ (aq) + Cu(s) E° (calc’d) = +1.10 V Cathode, positive, sink for electrons Anode, negative, source of electrons + Zn/Cu Electrochemical Cell

36. Half-reactions are organized by relative ability to act as oxidizing agents (Reduction Potential) Cu 2+ (aq) + 2e-  Cu(s)E° = +0.34 V Zn 2+ (aq) + 2e-  Zn(s) E° = –0.76 V Note that when a reaction is reversed the sign of E˚ is reversed! E° Values

37. Table 20.1 Table 20.1 tabulates the standard reduction potentials for half reactions from highest (most positive) to lowest (most negative) The greater the reduction potential (more positive) the greater the tendency to undergo reduction. One can use these values to predict the direction and cell potentials of redox reactions. E° Values

39. Stronger oxidizing agents Stronger reducing agents Potential Ladder for Reduction Half- Reactions

40. Which species is the strongest oxidizing agent: O 2, H 2 O 2, or Cl 2 ? _________________ Which is the strongest reducing agent: Hg, Al, or Sn? ____________________ Using Standard Potentials, E°

41. H 2 O 2 > Cl 2 > O 2Which species is the strongest oxidizing agent: O 2, H 2 O 2, or Cl 2 ? H 2 O 2 > Cl 2 > O 2 Which is the strongest reducing agent: Hg, Al, or Sn? ____________________ Using Standard Potentials, E°

42. H 2 O 2 > Cl 2 > O 2Which species is the strongest oxidizing agent: O 2, H 2 O 2, or Cl 2 ? H 2 O 2 > Cl 2 > O 2 Al > Sn > HgWhich is the strongest reducing agent: Hg, Al, or Sn? Al > Sn > Hg Using Standard Potentials, E°

43. 2 E°E° (V) Cu 2+ + 2e- Cu+0.34 2 H + + 2e- H0.00 Zn 2+ + 2e- Zn-0.76 oxidizing ability of ion reducing ability of element Standard Redox Potentials, E°

44. A substance on the right will reduce any substance higher than it on the left. Zn can reduce H + and Cu 2+. H 2 can reduce Cu 2+ but not Zn 2+ Cu cannot reduce H + or Zn 2+. Standard Redox Potentials, E°

45. Cu 2+ + 2e-  Cu Or Cu  Cu 2+ + 2 e- H 2  2 H + + 2 e- or 2 H + + 2e-  H 2 CathodePositiveAnodeNegative Electrons  1 M Cu(NO 3 ) 2 1 M H 3 O + Cu(s) H 2 (g) Salt Bridge KNO 3 (aq) Cu(s) | Cu 2+ (aq) || H + (aq) | H 2 (g)

46. Galvanic Cells The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Cell Diagram Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+ ] = 1 M and [Zn 2+ ] = 1 M Zn (s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (s) anode cathode salt bridge phase boundary

47. Cu 2+ + 2e-  Cu +0.34 + 2 H + 2e-  H 2 0.00 Zn 2+ + 2e-  Zn -0.76 Northwest-southeast rule: product-favored when… The reducing agent is the species at southeast corner The oxidizing agent is the species at northwest corner Any substance on the right will reduce any substance higher than it on the left. Ox. agent Red. agent Standard Redox Potentials, E°Standard Redox Potentials, E°

48. In which direction do the following reactions go? Using Standard Potentials, E°Using Standard Potentials, E°

49. In which direction do the following reactions go? 1)Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s)

50. Using Standard Potentials, E°Using Standard Potentials, E° In which direction do the following reactions go? 1)Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) Goes right as writtenGoes right as written

51. Using Standard Potentials, E°Using Standard Potentials, E° In which direction do the following reactions go? 1)Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) Goes right as writtenGoes right as written 2)2 Fe 2+ (aq) + Sn 2+ (aq)  2 Fe 3+ (aq) + Sn(s)

52. Using Standard Potentials, E°Using Standard Potentials, E° In which direction do the following reactions go? 1)Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) Goes right as writtenGoes right as written 2)2 Fe 2+ (aq) + Sn 2+ (aq)  2 Fe 3+ (aq) + Sn(s) Goes LEFT opposite to direction writtenGoes LEFT opposite to direction written

53. Northwest-southeast rule: reducing agent at southeast corner = ANODE oxidizing agent at northwest corner = CATHODE Cu 2+ + 2e-  Cu +0.34 + 2 H + 2e-  H 2 0.00 Zn 2+ + 2e-  Zn -0.76 ANODE CATHODE Standard Redox Potentials, E°Standard Redox Potentials, E°

54. (cathode) (anode) E° net = “distance” from “top” half-reaction (cathode) to “bottom” half-reaction (anode) E° net = E 0 cathode + E° anode E 0 ox + E 0 red E° net for Cu/Ag + reaction = +0.459 V Standard Redox Potentials, E°Standard Redox Potentials, E° 1) ANODE Cu(s)  Cu 2+ (aq) + 2 eE 0 ox =-0.340V 2)Cathode 2 Ag + (aq) + 2e  2 Ag(s) E 0 red =0.799 NET: Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s)

55. Cd  Cd 2+ + 2e- or Cd 2+ + 2e-  Cd Fe  Fe 2+ + 2e- or Fe 2+ + 2e-  Fe All ingredients are present. Which way does reaction proceed? E° for a Voltaic Cell

56. From the table, one sees that Fe is a better reducing agent than Cd Cd 2+ is a better oxidizing agent than Fe 2+ Overall reaction Fe + Cd 2+  Cd + Fe 2+ E° = E° cathode(red) + E° anode(ox) = (-0.40 V) + 0.44 V = +0.04 V E° for a Voltaic Cell

57. Predicting Reaction Direction Consider the following electrochemical reaction: Cathode 2H 2 O (l) + 2e-  H 2 (g) + 2OH - (aq)Cathode Anode 2I - (aq)  I 2 (aq) + 2e-(aq)Anode ------------------------------------------------- 2I - (aq)+ 2 H 2 O(l)  I 2 (aq) + 2OH - (aq) + H 2 (g) Will the reaction occur as written? E° net = E° cathode, red + E° anode, ox = (  0.828 V) - 0.535 V =  1.363 V E° < 0 indicates the rxn. occurs in rev. direction

58. The NERNST EQUATION E = potential under nonstandard conditions n = no. of electrons exchanged If [P] and [R] = 1 mol/L, then E = E° If [R] > [P], then E is GREATER THAN than E° If [R] < [P], then E is LESS THAN than E° E at Non-Standard Conditions

59. 59 The Effect of Concentration on Cell Emf  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 =

60. 60 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+ 2Fe Cd Cd 2+ + 2e - Oxidation: Reduction: n = 2 E 0 = -0.44 +0.40 E 0 = -0.04 V E 0 = E Fe /Fe + E Cd /Cd 00 2+ - 0.0257 V n ln Q E 0 E = - 0.0257 V 2 ln -0.04 VE = 0.010 0.60 E = 0.013 E > 0Spontaneous

61. 61 Concentration Cells Galvanic cell from two half-cells composed of the same material but differing in ion concentrations.

62. Batteries: Primary, Secondary, and Fuel Cells

63. Anode (-) Zn(s)  Zn 2+ (aq) + 2e- Cathode (+) 2NH 4 + (aq) + 2e-  2NH 3 (aq) + H 2 (g) Primary batteries use redox reactions that cannot be reversed by recharge. (non- rechargeable) Dry Cell Battery

64. Similar reactions as in common dry cells, but under basic (alkaline) conditions. Anode (-): Zn(s) + 2 OH - (aq)  ZnO(s) + H 2 O(l) + 2e- Cathode (+): MnO 2 (s) + H 2 O(l) + 2e-  Mn 2 O 3 (s)+ 2OH - (aq) Alkaline Battery

65. Pb storage batteries are rechargeable, they are classified as “secondary batteries”. These utilize redox reactions that are reversible. The components can be restored by recharging Lead Storage Battery

66. Anode (-): E° = +0.36 V Pb(s) + HSO 4 - (aq)  PbSO 4 (aq)+ H + (aq) + 2e- Cathode (+): E° = +1.68 V PbO 2 (s) + HSO 4 - (aq) + 3H + (aq) + 2e-  PbSO 4 (aq) + 2H 2 O(l) Lead Storage Battery

67. Anode (-): Cd(s) + 2OH - (aq)  Cd(OH) 2 (s)+ 2e- Cathode (+): NiO(OH) + H 2 O(l) + e-  Ni(OH) 2 (s) + OH - (aq) Ni-Cad Battery

68. In a “fuel cell”, reactants are supplied continuously from an external source. Electricity is generated by H 2 /O 2 fuel cells. H 2 (g) is carried in tanks or is generated from hydrocarbon decomposition. O 2 is supplied from the atomsphere. Fuel Cells: H 2 as a Fuel

69. Hydrogen - Air Fuel Cell

70. Mass equivalents between different sources of hydrogen gas. H 2 as a Fuel

71. H 2 (g) can be adsorbed onto a metal or metal alloy. Gentle heating releases the gas. Storing H 2 as a Fuel

72. A non-spontaneous reaction may be brought about using electrical energy to produce chemical change.  Sn 2+ (aq) + 2 Cl - (aq)  Sn(s) + Cl 2 (g) Some applications of electrolysis: Electroplating, separation of elements and separation of ions.Electrolysis

73. Electrolysis uses direct current (DC) The Cathode is the negative terminal, the source of electrons. Anode is the positive terminal, electrons move to ward it. Anions flow to the anode, cations to the cathode. Electrons are lost by the anions (oxidation). Electrons are taken on by the cations (reductoin). Sometimes and over voltage (E > E total ) is needed to initiate a reaction.Electrolysis

74. Anode (+) 4OH - (aq)  O 2 (g) + 2H 2 O(l) + 4e- Cathode (-) 4H 2 O + 4e-  2H 2 (aq) + 4OH - (aq) E o = -1.23 V Electric Energy & Chemical Change AnodeCathode Electrolysis of Aqueous NaOH

75. Electrolysis of molten NaCl. Here a battery moves electrons from Cl - to Na +. The polarity of the electrodes are reversed from galvanic cells. Electrolysis: Electric Energy & Chemical Change

76. Electrolysis of Molten NaCl

77. Anode (+): 2 Cl -  Cl 2 (g) + 2e- Cathode (-): Na + + e-  Na E° for cell (in water) = E° cat - E° an = - 2.71 V – (+1.36 V) = - 4.07 V (in water) The reaction will work when a voltage of 4.07 V or greater is applied. Electrolysis of Molten NaCl

78. Anode (+):  2Cl - (aq)  Cl 2 (g) + 2e- Cathode (-):  2H 2 O(l)+ 2e-  H 2 (g) + 2OH - (aq) E° = -2.19 V The reaction works because water is more easily reduced than Na +. Cl - is oxidized over water H 2 O as it is favored by kinetics. Electrolysis of Aqueous NaCl

79. NaOH(s) and Cl 2 (g) are commercially produced in electrolytic cells like below. (25.1  10 9 lb Cl 2 and 26.1  10 9 lb NaOH in 1995) Also the source of NaOCl for use in bleach. Electrolysis of Aqueous NaCl

80. Anode (+): 2I - (aq)  I 2 (aq) + 2e- Cathode (-): 2H 2 O + 2e-  H 2 (g) + 2 OH - (aq) E° = -1.36 V Electrolysis of Aqueous NaI

81. Anode (+): 2Cl - (aq)  Cl 2 (g) + 2e- Cathode (-): Cu 2+ (aq)+ 2e-  Cu(s) E° = -1.02 V Cu is favored to be reduced over either H 2 O or Na +. Electrolysis of Aqueous CuCl

82. Impure copper is oxidized to Cu 2+ at the anode. The aqueous Cu 2+ ions are reduced to Cu metal at the cathode. Electrolytic Refining of Copper

83. 2Al 2 O 3 (s) + 3C(graphite)  4Al(a) + 3CO 2 (g) The Al electrolysis process was developed by Charles Hall (1863-1914), founder of Alcoa. Aluminum Production

84. Coined the terms “anode”, “cathode”, “anion”, “cation” & “electrode”. Father of  Electrolysis  paramagnetism  electromagnetic induction  benzene and other organic  chemicals Also a noted chemical lecturer. Michael Faraday: 1791-1867

85. Since E° can predict product or reactant favored reactoins, it must be related to the Gibb’s free energy, ∆G°, for the reaction. ∆G° =  nFE° F = Faraday constant = 9.6485 x 10 4 J/Vmol of e - = 9.6485 x 10 4 coulombs/mol n is the number of moles of electrons transferred E° & Thermodynamics

86. ∆G° =  n F E° For a product-favored reaction: Reactants  Products ∆G° 0 E° is positive For a reactant-favored reaction Reactants  Products ∆G° > 0 and so E° < 0 E° is negative E° and ∆G°

87. Consider electrolysis of aqueous silver ion. Ag + (aq) + e-  Ag(s) 1 mol e-  mol Ag Moles of e- transferred is related to the moles of Ag(s) formed via stoichiometry. Current flow and time yield moles of e- Quantitative Aspects of Electrochemistry

88. How does charge related to moles of electrons? Quantitative Aspects of Electrochemistry

89. Problem: A 1.50 amp current is passed through a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Quantitative Aspects of Electrochemistry

90. Problem: A 1.50 amp current is passed through a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Quantitative Aspects of Electrochemistry

91. Problem: A 1.50 amp current is passed through a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Quantitative Aspects of Electrochemistry

92. Problem: A 1.50 amp current is passed through a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Quantitative Aspects of Electrochemistry

93. Problem: At the anode of a lead storage battery is, the following reaction occurs: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e-

94. Quantitative Aspects of Electrochemistry Problem: At the anode of a lead storage battery is, the following reaction occurs: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e- If the battery has 454g of Pb available, how long will it last if it running at 1.50 amps of current.

95. Problem: At the anode of a lead storage battery is, the following reaction occurs: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e- If the battery has 454g of Pb available, how long will it last if it running at 1.50 amps of current. Quantitative Aspects of Electrochemistry

96. Problem: At the anode of a lead storage battery is, the following reaction occurs: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e- If the battery has 454g of Pb available, how long will it last if it running at 1.50 amps of current.

97. Quantitative Aspects of Electrochemistry Problem: At the anode of a lead storage battery is, the following reaction occurs: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e- If the battery has 454g of Pb available, how long will it last if it running at 1.50 amps of current.

98. Quantitative Aspects of Electrochemistry Problem: At the anode of a lead storage battery is, the following reaction occurs: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e- If the battery has 454g of Pb available, how long will it last if it running at 1.50 amps of current.

99. Quantitative Aspects of Electrochemistry Problem: At the anode of a lead storage battery is, the following reaction occurs: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e- If the battery has 454g of Pb available, how long will it last if it running at 1.50 amps of current.

100. Quantitative Aspects of Electrochemistry Problem: At the anode of a lead storage battery is, the following reaction occurs: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e- If the battery has 454g of Pb available, how long will it last if it running at 1.50 amps of current.

101. Quantitative Aspects of Electrochemistry Problem: At the anode of a lead storage battery is, the following reaction occurs: Pb(s) + HSO 4 - (aq)  PbSO 4 (s) + H + (aq) + 2e- If the battery has 454g of Pb available, how long will it last if it running at 1.50 amps of current. 78.3 hrs

102. 102 Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mol e - = 96,500 C

103. 103 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 mol 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