1. Chapter 20 - Electron Transfer Reactions Objectives: 1. Carry out balancing of redox reactions in acidic or basic solutions; 2. Recall the parts of a basic and commercial voltaic cells; 3. Perform cell potential calculations from standard reduction potentials; 4. Classify oxidizing and reducing agents; 5. Apply the Nerst equation to redox problems; 6. Determine K from Ecell; 7. Perform electrolysis calculations.

2. Introduction NADH + (1/2)O 2 + H + -----> NAD + + H 2 O Medicinal Biochemistry: http://web.indstate.edu/thcme/mwking/home.html

3. Introduction Pyruvate + CoA + NAD + ------> CO 2 + acetyl-CoA + NADH + H +

4. Redox Reactions Cu(s) + 2 Ag + (aq) ---> Cu 2+ (aq) + 2 Ag(s) Electron transfer reactions are ________________ or redox reactions. Redox reactions can result in the generation of an electric current or be caused by imposing an electric current. Therefore, this field of chemistry is often called _____________________.

5. Review OXIDATION – _______________________________ REDUCTION – _______________________________ OXIDIZING AGENT – _______________________________ REDUCING AGENT – _______________________________

6. Redox Reactions Direct Redox Reaction Oxidizing and reducing agents in direct contact. Cu(s) + 2 Ag + (aq) ---> Cu 2+ (aq) + 2 Ag(s)

7. Redox Reactions Indirect Redox Reaction A battery functions by transferring electrons through an external wire from the reducing agent to the oxidizing agent.

8. Electrochemical Cells An apparatus that allows a redox reaction to occur by transferring electrons through an external connector. Product favored reaction ---> ________________cell ----> electric current. Reactant favored reaction ---> ________________ cell ---> electric current used to cause chemical change. Batteries are voltaic cells

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

10. Why study electrochemistry? Batteries Corrosion Industrial production of chemicals such as Cl 2, NaOH, F 2 and Al Biological redox reactions The heme group

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

12. Balancing Redox Equations Cu + Ag + --give--> Cu 2+ + Ag Cu + Ag + --give--> Cu 2+ + Ag

13. Balancing Redox Equations Step 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction. Ox Red Step 2: Balance each for mass. Step 3: Balance each half-reaction for charge by adding electrons. Ox Red Cu + Ag + --give--> Cu 2+ + Ag Cu + Ag + --give--> Cu 2+ + Ag

14. Balancing Redox Equations 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 Oxidizing agent Step 5: Add half-reactions to give the overall equation. The equation is now balanced for both charge and mass.

15. Reduction of VO 2 + with Zn

16. Balance the following in ACID solution: VO 2 + + Zn ---> VO 2+ + Zn 2+ Step 1: Write the half-reactions Ox Red Step 2: Balance each half-reaction for mass. Ox Red Add H 2 O on O-deficient side and add H + on other side for H-balance.

17. Balancing… Step 3: Balance half-reactions for charge. Ox Red Step 4: Multiply by an appropriate factor. Ox Red Step 5: Add balanced half-reactions

18. Tips on Balancing 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!

19. Balance the following in basic solution: MnO 4 - + NO 2 -  MnO 2 + NO 3 - Oxidation half reaction:

20. Balancing… MnO 4 - + NO 2 -  MnO 2 + NO 3 - Reduction half reaction:

21. Balancing… MnO 4 - + NO 2 -  MnO 2 + NO 3 - Oxidation half reaction: Reduction half reaction: Multiply by appropriate factor to cancel e- and add both half-reactions Oxid. X Red. X Sum: Study Exp 11 – Procedure to balance redox reactions – practice and E calculation

22. Chemical Change ---> Electric Current Oxidation: Zn(s) ---> Zn 2+ (aq) + 2e- Reduction: Cu 2+ (aq) + 2e- ---> Cu(s) -------------------------------------------------------- Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s) Electrons are transferred from Zn to Cu 2+, but there is no useful electric current. With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”

23. Chemical Change ---> Electric Current To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire. This is accomplished in a GALVANIC or VOLTAIC cell. A group of such cells is called a _____________.

24. Chemical Change ---> Electric Current Electrons travel thru external wire. _____________ allows anions and cations to move between electrode compartments. Zn --> Zn 2+ + 2e-Cu 2+ + 2e- --> Cu <--Anions Cations--> Oxidation Anode Negative Oxidation Anode Negative Reduction Cathode Positive Reduction Cathode Positive

25. The Cu|Cu 2+ and Ag|Ag + Cell

26. Electrochemical Cell _________ move from anode to cathode in the wire. _________ move from anode to cathode in the wire. _______ & _________move thru the salt bridge. _______ & _________move thru the salt bridge.

27. Terminology Figure 20.6

28. What Voltage does a Cell Generate? Electrons are “driven” from anode to cathode by an _______________________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

29. Cell Potential, E 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 o —a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 ˚C.

30. Calculating Cell Voltage Balanced half-reactions can be added together to get overall, balanced equation. Zn(s) ---> Zn 2+ (aq) + 2e- Zn(s) ---> Zn 2+ (aq) + 2e- Cu 2+ (aq) + 2e- ---> Cu(s) Cu 2+ (aq) + 2e- ---> Cu(s)--------------------------------------------------------- Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s) Zn(s) ---> Zn 2+ (aq) + 2e- Zn(s) ---> Zn 2+ (aq) + 2e- Cu 2+ (aq) + 2e- ---> Cu(s) 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. But we need a reference!

31. Cell Potential: SHE ______________________, SHE. Can’t measure 1/2 reaction E o directly. Therefore, measure it relative to a ______________________, SHE. 2 H + (aq, 1 M) + 2e- H 2 (g, 1 atm) E o = 0.0 V

32. Zn/Zn 2+ half-cell hooked to a SHE. E o for the cell = +0.76 V Zn(s) ---> Zn 2+ (aq) + 2e- Negative electrode Supplier of electrons Acceptor of electrons Positive electrode 2 H + + 2e- --> H 2 Reduction Cathode Zn --> Zn 2+ + 2e- Oxidation Anode

33. Reduction of Protons (H + ) by Zn

34. Overall reaction is reduction of H + by Zn metal. Zn(s) + 2 H + (aq) --> Zn 2+ + H 2 (g) E o = +0.76 V Therefore, E o for Zn ---> Zn 2+ (aq) + 2e- is +0.76 V Zn is a (better) (poorer) reducing agent than H 2.

35. Cu/Cu 2+ and H 2 /H + Cell, E 0 for the cell = + 0.34 V Cu 2+ (aq) + 2e- ---> Cu(s) E o = +0.34 V Acceptor of electrons Supplier of electrons Cu 2+ + 2e- --> Cu Reduction Cathode H 2 --> 2 H + + 2e- Oxidation Anode Positive Negative e-

36. Overall reaction is reduction of Cu 2+ by H 2 gas. Cu 2+ (aq) + H 2 (g) ---> Cu(s) + 2 H + (aq) Measured E o = +0.34 V Therefore, E o for Cu 2+ + 2e- ---> Cu is + 0.34 V

37. Zn/Cu Electrochemical Cell Zn(s) ---> Zn 2+ (aq) + 2e-E o = +0.76 V Zn(s) ---> Zn 2+ (aq) + 2e-E o = +0.76 V Cu 2+ (aq) + 2e- ---> Cu(s)E o = +0.34 V Cu 2+ (aq) + 2e- ---> Cu(s)E o = +0.34 V --------------------------------------------------------------- --------------------------------------------------------------- Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s) Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s) Cathode, positive, sink for electrons Anode, negative, source of electrons +

38. Uses of E o values Organize half-reactions by relative ability to act as oxidizing agents. Cu 2+ (aq) + 2e- ---> Cu(s)E o = +0.34 V Zn 2+ (aq) + 2e- ---> Zn(s) E o = –0.76 V Note that when a reaction is reversed the sign of E˚ is reversed! Cu 2+ is better oxidazing agent than Zn 2+ ; Cu 2+ will be reduced and Zn will be oxidized Cu 2+ reaction (reduction) will occur at the cathode Zn reaction (oxidation) will occur at the anode. E˚net = E˚cathode - E˚anode = (0.43) – (-0.76) = 1.1 V

39. Std. Reduction Potentials Organize half- reactions by relative ability to act as oxidizing agents. (All references are written as reduction processes). Table 20.1 Use this to predict direction of redox reactions and cell potentials.

40. Potential Ladder for Reduction Half- Reactions Best oxidizing agents Best reducing agents Figure 20.14

41. Using Standard Potentials, E o Which is the best oxidizing agent: O 2, H 2 O 2, or Cl 2 ? Which is the best reducing agent: Hg, Al, or Sn? E o (V) Cu 2+ + 2e- Cu+0.34 2 H + + 2e- H 2 0.00 Zn 2+ + 2e- Zn-0.76 oxidizing ability of agent reducing ability of agent

42. Standard Reduction Potentials Any substance on the right will reduce any substance higher than it on the left. Zn can reduce H + and Cu +. H 2 can reduce Cu 2+ but not Zn 2+ Cu cannot reduce H + or Zn 2+.

43. Standard Reduction Potentials Cu 2+ + 2e- --> Cu +0.34 + 2 H + 2e- --> H 2 0.00 Zn 2+ + 2e- --> Zn -0.76 Northwest-southeast rule: product-favored reactions occur between reducing agent at southeast corner oxidizing agent at northwest corner Any substance on the right will reduce any substance higher than it on the left. Ox. agent Red. agent

44. Using Standard Reduction Potentials In which direction do the following reactions go? Cu(s) + 2 Ag + (aq) ---> Cu 2+ (aq) + 2 Ag(s) 2 Fe 2+ (aq) + Sn 2+ (aq) ---> 2 Fe 3+ (aq) + Sn(s) What is E o net for the overall reaction?

45. Calculating Cell Potential E˚ net = “distance” from “top” half-reaction (cathode) to “bottom” half-reaction (anode) E˚ net = E˚ cathode - E˚ anode E o net for Cu/Ag + reaction = +0.46 V

46. E o for a Cell 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?

47. E o for a Cell From the table, you see 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 o = E˚ cathode - E˚ anode = = Fe/Fe 2+ // Cd 2+ /Cd

48. More about E o for a Cell Assume I - ion can reduce water. 2 H 2 O + 2e- ---> H 2 + 2 OH - Cathode 2 I - ---> I 2 + 2e- Anode ------------------------------------------------- 2 I - + 2 H 2 O --> I 2 + 2 OH - + H 2 2 H 2 O + 2e- ---> H 2 + 2 OH - Cathode 2 I - ---> I 2 + 2e- Anode ------------------------------------------------- 2 I - + 2 H 2 O --> I 2 + 2 OH - + H 2 Assuming reaction occurs as written, E˚ net = E˚ cathode - E˚ anode = _________ E˚ means rxn. occurs in ___________ direction It is ____________ favored.

49. E o at non-standard conditions The NERNST EQUATION E = potential under nonstandard conditions R = gas constant (8.314472 J/Kmol) T = temperature (K) n = no. of electrons exchanged F = Faraday constant (9.6485338 x 10 4 C/mol) ln = “natural log” Q = reaction quotent (concentration of products/concentration of reactants to the appropriate power) E = E o - (RT/nF) lnQ E = E o - 0.0257/n lnQ If [P] and [R] = 1 mol/L, then E = E˚ If [R] > [P], then E is ______________ than E˚ If [R] < [P], then E is ______________ than E˚ One ___________ is the quantity of electric charge carried by one mole of electrons.

50. A voltaic cell is set up at 25 o C with the following half-cells: Al 3+ (0.0010 M)/Al and Ni 2+ (0.50 M)/Ni. Write an equation for the reaction that occurs when the cell generates an electric current. a)Determine which substance is oxidized (decide which is the better reducing agent). Al is best reducing agent. Then Al is oxidized and Ni 2+ is reduced. Ox (Anode): Red (Cathode): b) Add the half-reactions to determine the net ionic equation. Net eq: c) Calculate E o and use Nernst eq. to calculate E. E o = E o cathode – E o anode E = E o – 0.0257/n ln Q

51. Calculate the cell potential, at 25 °C, based upon the overall reaction: 3 Cu 2+ (aq) + 2 Al(s)  3 Cu(s) + 2 Al 3+ (aq) if [Cu 2+ ] = 0.75 M and [Al 3+ ] = 0.0010 M. The standard reduction potentials are as follows: Cu 2+ (aq) + 2 e- → Cu(s)E° = +0.34 V Al 3+ (aq) + 3 e- → Al(s)E° = -1.66 V E o = E o cathode – E o anode Cathode Anode E = E o – 0.0257/n ln Q

52. E o and Thermodynamics E o is related to ∆G o, the free energy change for the reaction (energy released by the cell); under standard conditions: ∆G o = -nFE o where F = Faraday constant = 9.6485 x 10 4 J/Vmol of e - (or 9.6485 x 10 4 coulombs/mol) and n is the number of moles of electrons transferred  E = q + w The maximum work done by an electrochemical system (ideally) is proportional to the potential difference (volts) and the quantity of charge (coulombs): W max = nFE E is the cell voltage nF is the quantity of electric charge transferred from anode to cathode.

53. Calculate  G o from E o Cu(s) + 2 Ag + (aq) ---> Cu 2+ (aq) + 2 Ag(s) E o net for Cu/Ag + reaction = +0.46 V 1J = 1C * 1V 1000 J = 1kJ  G o = -nFE o

54. E o and the Equilibrium Constant When E cell = 0, the reactants and products are at equilibrium, Q = K E = 0 = E o – 0.0257/n ln K then ln K = n E o / 0.0257 (at 25 o C) For: Cu(s) + 2 Ag + (aq) ---> Cu 2+ (aq) + 2 Ag(s) E o net for Cu/Ag + reaction = +0.46 V

55. E o and the Equilibrium Constant ∆G o = - n F E o For a product-favored reaction Reactants ----> Products ∆G o 0 E o is positive For a reactant-favored reaction Reactants <---- Products ∆G o > 0 and so E o < 0 E o is negative E o = -  G 0 nF

56. Primary batteries Uses redox reactions that cannot be restored by recharge. * Indicate which reaction goes in the anode which in the cathode. Dry cell battery: _____________ Zn ---> Zn 2+ + 2e- _____________ 2 NH 4 + + 2e- ---> 2 NH 3 + H 2

57. Alkaline batteries Nearly same reactions as in common dry cell, but under basic conditions. _______________ Zn + 2 OH - ---> ZnO + H 2 O + 2e- _______________ 2 MnO 2 + H 2 O + 2e- ---> Mn 2 O 3 + 2 OH -

58. Secondary batteries Uses redox reactions that can be reversed. Can be restored by recharging.

59. Lead storage batteries ___________ E o = +0.36 V Pb + HSO 4 - ---> PbSO 4 + H + + 2e- Pb + HSO 4 - ---> PbSO 4 + H + + 2e- ___________ E o = +1.68 V PbO 2 + HSO 4 - + 3 H + + 2e- ---> PbSO 4 + 2 H 2 O

60. Ni-Cd battery ______________ Cd + 2 OH - ---> Cd(OH) 2 + 2e- ______________ NiO(OH) + H 2 O + e- ---> Ni(OH) 2 + OH -

61. Fuel Cell: H 2 as fuel Reactants are supplied continuously from an external source. Cars can use electricity generated by H 2 /O 2 fuel cells. H 2 carried in tanks or generated from hydrocarbons. Used in space rockets.

62. Fuel Cell: H 2 as fuel Cathode (red) O 2 (g) + 2 H 2 O (l) + 4e-  4 OH - Anode (ox) 2H 2 (g)  4 H + (aq) + 4e- Temperature of 70-140 o C and produce ~ 0.9 V. The two halves are separated by a proton exchange membrane (PEM). Protons combine with hydroxide ions forming water. The net reaction is then: 2 H 2 + O 2  2 H 2 O

63. Electrolysis Electric Energy ----> Chemical Change AnodeCathode 2 H 2 O  2 H 2 + O 2 ________________ 4 OH - ---> O 2 (g) + 2 H 2 O + 4e- ________________ 4 H 2 O + 4e- ---> 2 H 2 + 4 OH - E o for cell = -1.23 V

64. Electrolysis of Molten NaCl Electrolysis of molten NaCl. Here a battery “pumps” electrons from Cl - to Na +. NOTE: Polarity of electrodes is reversed from batteries. ________________ 2 Cl - ---> Cl 2 (g) + 2e- ________________ Na + + e- ---> Na

65. Electrolysis of Molten NaCl E o for cell (in water) =E˚c - E˚a = - 2.71 V – (+1.36 V) = - 4.07 V (in water) External energy needed because E o is (-).

66. Electrolysis of Aqueous NaCl Anode (+) 2 Cl - ---> Cl 2 (g) + 2e- Cathode (-) 2 H 2 O + 2e- ---> H 2 + 2 OH - E o for cell = -2.19 V Note that H 2 O (-0.8277) is more easily reduced than Na + (-2.71). Also, Cl - (1.36) is oxidized in preference to H 2 O (1.33) because of kinetics.

67. Electrolysis of Aqueous CuCl 2 Anode (+) 2 Cl - ---> Cl 2 (g) + 2e- Cathode (-) Cu 2+ + 2e- ---> Cu E o for cell = -1.02 V Note that Cu is more easily reduced than either H 2 O or Na +.

68. Electrolytic Refining of Copper Impure copper is oxidized to Cu 2+ at the _________. The aqueous Cu 2+ ions are reduced to Cu metal at the _______________.

69. Electrolysis of Aqueous SnCl 2 Sn 2+ (aq) + 2 Cl - (aq) ---> Sn(s) + Cl 2 (g) E o cell = E o cathode-E o anode =

70. Al production Charles Hall (1863-1914) developed electrolysis process. Founded Alcoa. 2 Al 2 O 3 + 3 C ---> 4 Al + 3 CO 2

71. Counting electrons The number of e- consumed or produced in an electron transfer reaction is obtained by measuring the current flowing in the circuit in a given time. The current flowing is the amount of charge (coulombs, C) per unit time, the unit is the ampere (A). 1 A = 1 C/s then 1C = A *s 1 F = 9.6485338 x 10 4 C/mol e- 1 mol e- = 96,500 C

72. 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? a)Calculate the charge Charge (C) = current (A) x time (t) b) Calculate moles of e- used c) Calculate the mass

73. 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? a)Calculate moles of Pb b)Calculate moles of e- c)Calculate charge (C): d) Calculate time Time (sec) = Charge (C) I (amps)

74. End of Chapter Go over all the contents of your textbook. Practice with examples and with problems at the end of the chapter. Practice with OWL tutor. Practice with the quiz on CD of Chemistry Now. Work on your OWL assignment for Chapter 20.