1. 1 © 2009 Brooks/Cole - Cengage ELECTROCHEMISTRY Chapter 20-21

2. 2 © 2009 Brooks/Cole - Cengage TRANSFER REACTIONS Atom/Group transfer HCl + H 2 O  Cl - + H 3 O + Electron transfer Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s)

3. 3 © 2009 Brooks/Cole - Cengage Electron Transfer Reactions Electron transfer reactions are oxidation- reduction or redox reactions.Electron 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.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 ELECTROCHEMISTRY.Therefore, this field of chemistry is often called ELECTROCHEMISTRY.

4. 4 © 2009 Brooks/Cole - Cengage Review of Terminology for Redox Reactions OXIDATION—loss of electron(s) by a species; increase in oxidation number.OXIDATION—loss of electron(s) by a species; increase in oxidation number. REDUCTION—gain of electron(s); decrease in oxidation number.REDUCTION—gain of electron(s); decrease in oxidation number. OXIDIZING AGENT—electron acceptor; species is reduced.OXIDIZING AGENT—electron acceptor; species is reduced. REDUCING AGENT—electron donor; species is oxidized.REDUCING AGENT—electron donor; species is oxidized. OXIDATION—loss of electron(s) by a species; increase in oxidation number.OXIDATION—loss of electron(s) by a species; increase in oxidation number. REDUCTION—gain of electron(s); decrease in oxidation number.REDUCTION—gain of electron(s); decrease in oxidation number. OXIDIZING AGENT—electron acceptor; species is reduced.OXIDIZING AGENT—electron acceptor; species is reduced. REDUCING AGENT—electron donor; species is oxidized.REDUCING AGENT—electron donor; species is oxidized.

5. 5 © 2009 Brooks/Cole - Cengage OXIDATION-REDUCTION REACTIONS Direct Redox Reaction Oxidizing and reducing agents in direct contact. Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) PLAY MOVIE

6. 6 © 2009 Brooks/Cole - Cengage OXIDATION-REDUCTION REACTIONS Indirect Redox Reaction A battery functions by transferring electrons through an external wire from the reducing agent to the oxidizing agent. PLAY MOVIE

7. 7 © 2009 Brooks/Cole - Cengage Why Study Electrochemistry? BatteriesBatteries CorrosionCorrosion Industrial production of chemicals such as Cl 2, NaOH, F 2 and AlIndustrial production of chemicals such as Cl 2, NaOH, F 2 and Al Biological redox reactionsBiological redox reactions The heme group

8. 8 © 2009 Brooks/Cole - Cengage Electrochemical Cells An apparatus that allows a redox reaction to occur by transferring electrons through an external connector.An apparatus that allows a redox reaction to occur by transferring electrons through an external connector. Product favored reaction  voltaic or galvanic cell: chemical change produces electric currentProduct favored reaction  voltaic or galvanic cell: chemical change produces electric current Reactant favored reaction  electrolytic cell: electric current used to cause chemical change.Reactant favored reaction  electrolytic cell: electric current used to cause chemical change. Batteries are voltaic cells

9. 9 © 2009 Brooks/Cole - Cengage Electrochemistry Alessandro Volta, 1745-1827, Italian scientist and inventor. Luigi Galvani, 1737-1798, Italian scientist and inventor.

10. 10 © 2009 Brooks/Cole - Cengage Balancing Equations for Redox Reactions Some redox reactions have equations that must be balanced by special techniques Mn = +7 Fe = +2 Fe = +3 Mn = +2 MnO 4 - + 5 Fe 2+ + 8 H +  Mn 2+ + 5 Fe 3+ + 4 H 2 O PLAY MOVIE

11. 11 © 2009 Brooks/Cole - Cengage Balancing Equations Cu + Ag + --give--> Cu 2+ + Ag

12. 12 © 2009 Brooks/Cole - Cengage Balancing Equations Step 1:Divide the reaction into half-reactions, one for oxidation and the other for reduction. OxCu  Cu 2+ Red Ag +  Ag Step 2:Balance each for mass. Already done in this case. Step 3:Balance each half-reaction for charge by adding electrons. Ox Cu  Cu 2+ + 2e- Red Ag + + e-  Ag

13. 13 © 2009 Brooks/Cole - Cengage Balancing 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 Cu  Cu 2+ + 2e- Oxidizing agent 2 Ag + + 2 e-  2 Ag Step 5:Add half-reactions to give the overall equation. Cu + 2 Ag +  Cu 2+ + 2Ag The equation is now balanced for both charge and mass.

14. 14 © 2009 Brooks/Cole - Cengage Reduction of VO 2 + with Zn

15. 15 © 2009 Brooks/Cole - Cengage Balancing Equations Balance the following in acid solution— VO 2 + + Zn  VO 2+ + Zn 2+ 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 VO 2 +  VO 2+ + H 2 O 2 H + + Add H 2 O on O-deficient side and add H + on other side for H-balance.

16. 16 © 2009 Brooks/Cole - Cengage Balancing Equations 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- Red 2e- + 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

17. 17 © 2009 Brooks/Cole - Cengage Tips on Balancing Equations Never add O 2, O atoms, or O 2- to balance oxygen.Never add O 2, O atoms, or O 2- to balance oxygen. Never add H 2 or H atoms to balance hydrogen.Never add H 2 or H atoms to balance hydrogen. Be sure to write the correct charges on all the ions.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.Check your work at the end to make sure mass and charge are balanced. PRACTICE!PRACTICE!

18. 18 © 2009 Brooks/Cole - Cengage 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. CHEMICAL CHANGE f ELECTRIC CURRENT With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”

19. 19 © 2009 Brooks/Cole - Cengage To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire.To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire. CHEMICAL CHANGE f ELECTRIC CURRENT This is accomplished in a GALVANIC or VOLTAIC cell. A group of such cells is called a battery.

20. 20 © 2009 Brooks/Cole - Cengage Electrons travel thru external wire.Electrons travel thru external wire. Salt bridge allows anions and cations to move between electrode compartments.Salt bridge allows anions and cations to move between electrode compartments. Electrons travel thru external wire.Electrons travel thru external wire. Salt bridge allows anions and cations to move between electrode compartments.Salt bridge allows anions and cations to move between electrode compartments. Zn  Zn 2+ + 2e- Cu 2+ + 2e-  Cu  Anions Cations  OxidationAnodeNegativeOxidationAnodeNegative ReductionCathodePositiveReductionCathodePositive

21. 21 © 2009 Brooks/Cole - Cengage The Cu|Cu 2+ and Ag|Ag + Cell

22. 22 © 2009 Brooks/Cole - Cengage Electrochemical Cell Electrons move from anode to cathode in the wire. Anions & cations move thru the salt bridge. PLAY MOVIE

23. 23 © 2009 Brooks/Cole - Cengage Terms Used for Voltaic Cells See Figure 20.6

24. 24 © 2009 Brooks/Cole - Cengage The Voltaic Pile Drawing done by Volta to show the arrangement of silver and zinc disks to generate an electric current. What voltage does a cell generate?

25. 25 © 2009 Brooks/Cole - Cengage CELL POTENTIAL, E Electrons are “driven” from anode to cathode by an electromotive force or emf.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.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

26. 26 © 2009 Brooks/Cole - Cengage animation

27. 27 © 2009 Brooks/Cole - Cengage 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.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 oThis 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.—a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 ˚C.

28. 28 © 2009 Brooks/Cole - Cengage Calculating Cell Voltage Balanced half-reactions can be added together to get overall, balanced equation.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) 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.

29. 29 © 2009 Brooks/Cole - Cengage CELL POTENTIALS, E o STANDARD HYDROGEN CELL, SHE. Can’t measure 1/2 reaction E o directly. Therefore, measure it relative to a STANDARD HYDROGEN CELL, SHE. 2 H + (aq, 1 M) + 2e- e H 2 (g, 1 atm) E o = 0.0 V

30. 30 © 2009 Brooks/Cole - Cengage Cu/Cu 2+ and H 2 /H + Cell E o = +0.34 V Acceptor of electrons Supplier of electrons Cu 2+ + 2e-  Cu ReductionCathode H 2  2 H + + 2e- OxidationAnode Positive Negative

31. 31 © 2009 Brooks/Cole - Cengage Cu/Cu 2+ and H 2 /H + Cell 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

32. 32 © 2009 Brooks/Cole - Cengage Zn/Cu Electrochemical Cell Zn 2+ (aq) + 2e-  Zn(s) E o = - 0.76 V Cu 2+ (aq) + 2e-  Cu(s)E o = +0.34 V --------------------------------------------------------------- +0.34 V – (- 0.76 V) = 1.1V Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) Cathode, positive, sink for electrons Anode, negative, source of electrons + Note that when a reaction is reversed the sign of E˚ is reversed!

33. 33 © 2009 Brooks/Cole - Cengage

34. 34 Using Standard Potentials, E o Table 21.12 Which is the best oxidizing agent (reduced): O 2, H 2 O 2, or Cl 2 ? _________________ Which is the best reducing agent (oxidized): Hg, Al, or Sn? ____________________

35. 35 © 2009 Brooks/Cole - Cengage Cd Cd 2+ + 2e- Cd  Cd 2+ + 2e-or Cd 2+ + 2e- Cd Cd 2+ + 2e-  Cd Fe  Fe 2+ + 2e- or Fe 2+ + 2e-  Fe E o for a Voltaic Cell All ingredients are present. Which way does reaction proceed?

36. 36 © 2009 Brooks/Cole - Cengage From the table, you see Fe is less likely to be reduced (oxidized)Fe is less likely to be reduced (oxidized) Cd 2+ is more likely to be reducedCd 2+ is more likely to be reduced E o for a Voltaic Cell Overall reaction Fe + Cd 2+  Cd + Fe 2+ E o = E˚ cathode - E˚ anode = (-0.40 V) - (-0.44 V) = +0.04 V Overall reaction Fe + Cd 2+  Cd + Fe 2+ E o = E˚ cathode - E˚ anode = (-0.40 V) - (-0.44 V) = +0.04 V

37. 37 © 2009 Brooks/Cole - Cengage More About Calculating Cell Voltage 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 = (-0.828 V) - (+0.535 V) = -1.363 V Minus E˚ means rxn. occurs in opposite direction

38. 38 © 2009 Brooks/Cole - Cengage BATTERIES Primary, Secondary, and Fuel Cells

39. 39 © 2009 Brooks/Cole - Cengage Dry Cell Battery Anode (-) Zn  Zn 2+ + 2e- Cathode (+) 2 NH 4 + + 2e-  2 NH 3 + H 2 Primary battery — uses redox reactions that cannot be restored by recharge.

40. 40 © 2009 Brooks/Cole - Cengage Nearly same reactions as in common dry cell, but under basic conditions. Alkaline Battery Anode (-): Zn + 2 OH -  ZnO + H 2 O + 2e- Cathode (+): 2 MnO 2 + H 2 O + 2e-  Mn 2 O 3 + 2 OH - PLAY MOVIE

41. 41 © 2009 Brooks/Cole - Cengage Lead Storage Battery Secondary batterySecondary battery Uses redox reactions that can be reversed.Uses redox reactions that can be reversed. Can be restored by rechargingCan be restored by recharging

42. 42 © 2009 Brooks/Cole - Cengage Lead Storage Battery Anode (-) E o = +0.36 V Pb + HSO 4 -  PbSO 4 + H + + 2e- Cathode (+) E o = +1.68 V PbO 2 + HSO 4 - + 3 H + + 2e-  PbSO 4 + 2 H 2 O PLAY MOVIE

43. 43 © 2009 Brooks/Cole - Cengage Ni-Cad Battery Anode (-) Cd + 2 OH -  Cd(OH) 2 + 2e- Cathode (+) NiO(OH) + H 2 O + e-  Ni(OH) 2 + OH - PLAY MOVIE

44. 44 © 2009 Brooks/Cole - Cengage Fuel Cells: H 2 as a Fuel Fuel cell - reactants are supplied continuously from an external source.Fuel cell - reactants are supplied continuously from an external source. Cars can use electricity generated by H 2 /O 2 fuel cells.Cars can use electricity generated by H 2 /O 2 fuel cells. H 2 carried in tanks or generated from hydrocarbons.H 2 carried in tanks or generated from hydrocarbons.

45. 45 © 2009 Brooks/Cole - Cengage See Figure 20.12 Hydrogen—Air Fuel Cell

46. 46 © 2009 Brooks/Cole - Cengage H 2 as a Fuel Comparison of the volumes of substances required to store 4 kg of hydrogen relative to car size. (Energy, p. 264)

47. 47 © 2009 Brooks/Cole - Cengage Storing H 2 as a Fuel One way to store H 2 is to adsorb the gas onto a metal or metal alloy. (Energy, p. 264)

48. 48 © 2009 Brooks/Cole - Cengage Electrolysis Using electrical energy to produce chemical change. Sn 2+ (aq) + 2 Cl - (aq)  Sn(s) + Cl 2 (g) How is this different from a voltaic cell?

49. 49 © 2009 Brooks/Cole - Cengage Electrolysis of Aqueous NaOH Anode (+) 4 OH -  O 2 (g) + 2 H 2 O + 4e- Cathode (-) 4 H 2 O + 4e-  2 H 2 + 4 OH - E o for cell = -1.23 V Electric Energy f Chemical Change AnodeCathode PLAY MOVIE

50. 50 © 2009 Brooks/Cole - Cengage Electrolysis Electric Energy  Chemical Change Electrolysis of molten NaCl. Electrolysis of molten NaCl. Here a battery “pumps” electrons from Cl - to Na +. Here a battery “pumps” electrons from Cl - to Na +. NOTE: Polarity of electrodes is reversed from batteries. NOTE: Polarity of electrodes is reversed from batteries.

51. 51 © 2009 Brooks/Cole - Cengage Electrolysis of Molten NaCl See Figure 20.18

52. 52 © 2009 Brooks/Cole - Cengage Electrolysis of Molten NaCl Anode (+) 2 Cl -  Cl 2 (g) + 2e- Cathode (-) Na + + e-  Na E o for cell (in water) = E˚ c - E˚ a = - 2.71 V – (+1.36 V) = - 2.71 V – (+1.36 V) = - 4.07 V (in water) = - 4.07 V (in water) External energy needed because E o is (-).

53. 53 © 2009 Brooks/Cole - Cengage 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 is more easily reduced than Na +. Also, Cl - is oxidized in preference to H 2 O because of kinetics.

54. 54 © 2009 Brooks/Cole - Cengage Electrolysis of Aqueous NaCl Cells like these are the source of NaOH and Cl 2. In 1995: 25.1 x 10 9 lb Cl 2 and 26.1 x 10 9 lb NaOH Also the source of NaOCl for use in bleach.

55. 55 © 2009 Brooks/Cole - Cengage Electrolysis of Aqueous NaI Anode (+): 2 I -  I 2 (g) + 2e- Cathode (-): 2 H 2 O + 2e-  H 2 + 2 OH - E o for cell = -1.36 V

56. 56 © 2009 Brooks/Cole - Cengage 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 +.

57. 57 © 2009 Brooks/Cole - Cengage Electrolytic Refining of Copper See Figure 22.11 Impure copper is oxidized to Cu 2+ at the anode. The aqueous Cu 2+ ions are reduced to Cu metal at the cathode.

58. 58 © 2009 Brooks/Cole - Cengage Producing Aluminum 2 Al 2 O 3 + 3 C f 4 Al + 3 CO 2 Charles Hall (1863-1914) developed electrolysis process. Founded Alcoa.

59. 59 © 2009 Brooks/Cole - Cengage Michael Faraday 1791-1867 Originated the terms anode, cathode, anion, cation, electrode. Discoverer of electrolysiselectrolysis magnetic props. of mattermagnetic props. of matter electromagnetic inductionelectromagnetic induction benzene and other organic chemicalsbenzene and other organic chemicals Was a popular lecturer.

60. 60 © 2009 Brooks/Cole - Cengage E o and Thermodynamics E o is related to ∆G o, the free energy change for the reaction.E o is related to ∆G o, the free energy change for the reaction. ∆G˚ proportional to –nE˚∆G˚ proportional to –nE˚ ∆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

61. 61 © 2009 Brooks/Cole - Cengage E o and ∆G o ∆G o = - n F E o For a product-favored reaction Reactants f Products Reactants f Products ∆G o 0 E o is positive For a reactant-favored reaction Reactants Products Reactants r Products ∆G o > 0 and so E o 0 and so E o < 0 E o is negative

62. 62 © 2009 Brooks/Cole - Cengage Quantitative Aspects of Electrochemistry Consider electrolysis of aqueous silver ion. Ag + (aq) + e- f Ag(s) 1 mol e- f 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-?

63. 63 © 2009 Brooks/Cole - Cengage But how is charge related to moles of electrons? Quantitative Aspects of Electrochemistry = 96,500 C/mol e- = 1 Faraday

64. 64 © 2009 Brooks/Cole - Cengage Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a)Calc. charge Charge (C) = current (A) x time (t) = (1.5 amps)(15.0 min)(60 s/min) = 1350 C

65. 65 © 2009 Brooks/Cole - Cengage Quantitative Aspects of Electrochemistry Solution (a)Charge = 1350 C (b)Calculate moles of e- used 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? (c)Calc. quantity of Ag

66. 66 © 2009 Brooks/Cole - Cengage Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) f 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

67. 67 © 2009 Brooks/Cole - Cengage Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) f 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