1. Electrochemistry The Study of the Interchange of Chemical and Electrical Energy

2. Galvanic or Voltaic Cells The reactants and products of some oxidation-reduction reactions can be physically separated so that the electron transfer can only occur via a wire. The device or apparatus that is used to convert chemical energy to electrical energy is called a galvanic (or voltaic) cell.

3. Galvanic or Voltaic Cells As the electrical current passes through the wire, it can be used to run a device, such as a motor, light bulb, voltmeter, etc. As a result, the electrochemical reaction can be used to provide useful work.

4. Reaction of Zn with Cu 2+ If a strip of zinc metal is immersed in a solution of copper(II)sulfate, a reaction will spontaneously occur. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) As the reaction proceeds, some of the metallic zinc dissolves into solution, and the blue copper(II) ion plates out as elemental copper.

5. Reaction of Zn with Cu 2+ As each zinc atom dissolves, it provides two electrons to the copper ions.

6. Reaction of Zn with Cu 2+ As the reaction proceeds, a thin black layer of Cu is formed on the zinc surface. The blue color of the Cu 2+ ion fades as it is reduced.

7. Reaction of Zn with Cu 2+ The transfer of electrons occurs directly on the zinc surface. As a result, the movement of electrons cannot be utilized. Construction of a galvanic cell will allow the electron transfer to occur via a wire. In this way, the electrical current can be used to do work.

8. Galvanic Cells In the galvanic cell for the reaction, the oxidizing reagent (Cu 2+ ) and the reducing agent (Zn) are physically separated into two half-cells. The overall net-ionic equation for the reaction is: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

9. Galvanic Cells Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) The half-reactions are: Zn(s)  Zn 2+ (aq) + 2 e - (oxidation) Cu 2+ (aq) + 2 e -  Cu(s) (reduction)

10. Galvanic Cells In galvanic cells, the components of two half-reactions are physically separated into two beakers. The two beakers can then be connected by a wire and a salt bridge so that the electron transfer can occur.

11. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) The half-reactions are: Zn(s)  Zn 2+ (aq) + 2 e - (oxidation) Cu 2+ (aq) + 2 e -  Cu(s) (reduction) One half-cell will contain metallic Zn in a solution of zinc ion, and the other half-cell will contain metallic copper in a solution of copper (II) ion.

12. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Zn(s)  Zn 2+ (aq) + 2 e - (oxidation) Oxidation takes place at the anode, so the strip of zinc metal will serve as the anode. It will be immersed in an aqueous solution of a zinc salt, such as zinc sulfate. The sulfate ions are inert, and are just spectator ions.

13. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Cu 2+ (aq) + 2 e -  Cu(s) (reduction) Reduction takes place at the cathode, so the copper strip will serve as the cathode. It will be immersed in an aqueous solution of a copper (II) salt, such as copper(II)sulfate. The sulfate ion is inert, and will serve as a spectator.

14. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

15. The Salt Bridge The salt bridge, porous cup, or glass frit allows the flow of ions. This is necessary in order to maintain a neutral charge in each half- cell.

16. The Salt Bridge The salt bridge, porous cup, or glass frit allows the flow of ions.

17. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

18. Each combination of half-cells produces a characteristic voltage.

19. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

20. Oxidation at the Anode

21. Reduction at the Cathode

22. Voltage or EMF The voltage of the galvanic cell is also called the cell potential, or the electromotive force ( emf). It is related to the driving force of the reaction. The units of cell potential are volts (V). A volt is exactly 1 joule of work per coulomb of charge transferred.

23. Cell Potential There can be no reduction without oxidation, and thus, each galvanic cell needs two half-cells in order to produce a voltage. Scientists have devised a system to measure the potential (voltage) of any half-cell relative to a standard half cell. The potential of the standard half-cell is set at 0.00 volts.

24. The Hydrogen Half-Cell The half-cell consists of an inert platinum electrode immersed in 1M strong acid. Hydrogen gas is bubbled over the electrode. Pt electrode

25. Standard Reduction Potentials Each half-cell is connected to a standard hydrogen electrode. All cells contain solutions which are 1.00M, and all gases are at a pressure of 1.00 atmospheres. Since the voltage of the hydrogen electrode is set at zero, the voltage of the galvanic cell represents the assigned voltage of the other half- cell.

26. Standard Reduction Potentials The emf of the standard Zn half- cell is 0.76 volts relative to the hydrogen half- cell. In this cell, Zn is being oxidized, and H + reduced.

27. Standard Reduction Potentials The half-reactions are: Zn(s)  Zn 2+ (aq) + 2e - 2H + (aq) + 2e -  H 2 (g) Since the potential of the hydrogen reaction is set at zero, the potential for the oxidation of zinc is the measured value of 0.76 volts.

28. Standard Reduction Potentials All half-reactions are tabulated as reductions. Since the potential for the oxidation of zinc is +0.76 volts, the reduction potential is -.76 volts. Zn 2+ (aq) + 2e -  Zn(s) E o = -0.76V

29. Standard Reduction Potentials In this way, the reduction potentials for all half-cells are obtained relative to the hydrogen electrode. The values of reduction potentials are listed from highest potential to lowest.

31. Cell Potentials When a galvanic cell is constructed, one half- reaction is a reduction reaction, and the other is an oxidation reaction. When a reduction half-reaction is reversed to make it an oxidation reaction, the sign on its cell potential is reversed.

32. Cell Potentials The reaction which will occur spontaneously is the oxidation and reduction that produces the most positive cell potential.

33. Cell Potentials - Problem  Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:  Cr 2 O 7 2- + 14 H + + 6 e -  2 Cr 3+ + 7 H 2 O H 2 O 2 + 2H + + 2 e -  2 H 2 O

34. Cell Potentials - Problem  Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:  Cr 2 O 7 2- + 14 H + + 6 e -  2 Cr 3+ + 7 H 2 O H 2 O 2 + 2H + + 2 e -  2 H 2 O 1. Look up the standard reduction potentials for both half-reactions

36. Cell Potentials - Problem  Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:  Cr 2 O 7 2- + 14 H + + 6 e -  2 Cr 3+ + 7 H 2 O 1.33V H 2 O 2 + 2H + + 2 e -  2 H 2 O 1.78V 2. Reverse one half-reaction so that the net cell potential is the largest positive number.

37. Cell Potentials - Problem  Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:  Cr 2 O 7 2- + 14 H + + 6 e -  2 Cr 3+ + 7 H 2 O 1.33V H 2 O 2 + 2H + + 2 e -  2 H 2 O 1.78V 2. Reverse one half-reaction so that the net cell potential is the largest positive number. reverse

38. Cell Potentials - Problem  Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:  2 Cr 3+ + 7 H 2 O  Cr 2 O 7 2- + 14 H + + 6 e - -1.33V H 2 O 2 + 2H + + 2 e -  2 H 2 O 1.78V 3. Multiply half-reactions so that the electrons lost = electrons gained. Reduction potentials are not multiplied.

39. Cell Potentials - Problem  Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half- cells:  2 Cr 3+ + 7 H 2 O  Cr 2 O 7 2- + 14 H + + 6 e - -1.33V (H 2 O 2 + 2H + + 2 e -  2 H 2 O )3 1.78V 3. Multiply half-reactions so that the electrons lost = electrons gained. Reduction potentials are not multiplied. lost = electrons gained. Reduction potentials are not multiplied.

40. Cell Potentials - Problem  Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:  2 Cr 3+ + 7 H 2 O  Cr 2 O 7 2- + 14 H + + 6 e - -1.33V 3H 2 O 2 + 6H + + 6 e -  6 H 2 O 1.78V 4. Add the two half-reactions and their cell potentials.

41. Cell Potentials - Problem  Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half- cells:  2 Cr 3+ + 7 H 2 O  Cr 2 O 7 2- + 14 H + + 6 e - -1.33V 3H 2 O 2 + 6H + + 6 e -  6 H 2 O 1.78V 3H 2 O 2 + 2 Cr 3+ + 1 H 2 O  Cr 2 O 7 2- + 8 H + 3H 2 O 2 + 2 Cr 3+ + 1 H 2 O  Cr 2 O 7 2- + 8 H + E o =0.45V 8 1

42. Line Notation There is a system of notation, called line notation, used to describe a galvanic cell. For the cell pictured below: The anode is written on the left, the cathode on the right.

43. Line Notation The anode is written on the left, the cathode on the right. A single vertical line represents a phase boundary, and a pair of vertical lines indicate a salt bridge or porous disk.

44. Line Notation The notation for this cell is: Zn(s)|Zn 2+ (aq)||H + (aq)|H 2 (g)|Pt(s)

45. Line Notation The notation for this cell is: Zn(s)|Zn 2+ (aq)||H + (aq)|H 2 (g)|Pt(s) Note that Zn is written on the left because it is the anode.

46. Line Notation The notation for this cell is: Zn(s)|Zn 2+ (aq)||H + (aq)|H 2 (g)|Pt(s) Note that the standard hydrogen electrode is written on the right because it is the cathode.

47. Cell Potential and Free Energy An electrochemical cell produces a voltage as a result of the driving force for electron transfer. As a result, cell potentials are directly related to ∆G for the reaction.

48. Cell Potential and Free Energy For standard conditions, ∆G o = -nFE o where n is the moles of electrons transferred, and F = 96,485 coulumbs/mol e - (Faraday’s constant), and a volt equals 1 joule/coulomb.

49. Cell Potential and Free Energy For standard conditions, ∆G o = -nFE o A positive cell potential (spontaneous reaction) yields a negative value for ∆G.

50. Cell Potential and Concentration Standard conditions dictate that all solutions be 1M, and all gases have a pressure of 1 atm. Cell potential will vary with concentration. As a galvanic cell produces voltage, the concentrations in each half-cell change, and the voltage gradually decreases to zero.

51. Cell Potential and Concentration The cell potential of a non-standard cell can be calculated using the Nernst equation: E = E o -RT ln(Q) n where T is temperature in Kelvins and Q is the reaction quotient.

52. Cell Potential and Concentration For standard temperature (25 o C), the equation becomes: E = E o -0.0591 log(Q) n where n is the number of moles of electrons transferred.

53. Cell Potential and K At equilibrium, the cell potential is 0.0 volts, and Q=K. Using the Nernst Equation at 25 o C: E = E o -0.0591 log(Q) n 0 = E o - 0.0591 log(K) n log K = nE o /0.0591

54. Cell Potential and K log K = nE o /0.0591 Since redox reactions often have very large equilibrium constants, measuring cell potential is often the only way to obtain the value of K.

55. Problem Write the chemical reaction and calculate the equilibrium constant for the following galvanic cell under standard conditions. Write the chemical reaction and calculate the equilibrium constant for the following galvanic cell under standard conditions. Pt(s) | Cu 1+ (aq), Cu 2+ (aq) || Au 3+ (aq) | Au(s) 1. Write the half-reactions. Since the anode is on the left, copper(I) is oxidized to form copper(II). Gold(III) ion is reduced to elemental gold at the cathode.

56. Problem Write the chemical reaction and calculate the equilibrium constant for the following galvanic cell under standard conditions. Write the chemical reaction and calculate the equilibrium constant for the following galvanic cell under standard conditions. Pt(s) | Cu 1+ (aq), Cu 2+ (aq) || Au 3+ (aq) | Au(s) 1. Write the half-reactions. Cu 1+ (aq)  Cu 2+ (aq) + 1 e - Au 3+ (aq) + 3 e -  Au(s)

57. Problem Pt(s) | Cu 1+ (aq), Cu 2+ (aq) || Au 3+ (aq) | Au(s) 2. Look up the reduction potentials. Cu 1+ (aq)  Cu 2+ (aq) + 1 e - E o = -(0.16V) Cu 1+ (aq)  Cu 2+ (aq) + 1 e - E o = -(0.16V) Au 3+ (aq) + 3 e -  Au(s) E o = 1.50V Au 3+ (aq) + 3 e -  Au(s) E o = 1.50V

58. Problem Pt(s) | Cu 1+ (aq), Cu 2+ (aq) || Au 3+ (aq) | Au(s) 3. Combine the reactions. 3[Cu 1+ (aq)  Cu 2+ (aq) + 1 e - ] E o = -(0.16V) 3[Cu 1+ (aq)  Cu 2+ (aq) + 1 e - ] E o = -(0.16V) Au 3+ (aq) + 3 e -  Au(s) E o = 1.50V Au 3+ (aq) + 3 e -  Au(s) E o = 1.50V 3 Cu 1+ (aq) + Au 3+ (aq)  3 Cu 2+ (aq) + Au(s) E o = 1.34V

59. Problem 3 Cu 1+ (aq) + Au 3+ (aq)  3 Cu 2+ (aq) + Au(s) E o = 1.34V 4. Calculate the value of K using E o. log K = nE o /0.0591 log K = (3) (1.34)/.0591 = 68.0 K = 1 x 10 68

60. Concentration Cells Concentration cells are electrochemical cells with half-cells that differ only in the concentration of reactants.

61. Concentration Cells The two solutions would mix if they weren’t physically separated. The transfer of electrons occurs until the concentration in both beakers is the same.

62. Concentration Cells The electrode in the dilute solution dissolves, thus raising the concentration. Electrons flow to the concentrated beaker, where silver ion plates out.

63. Concentration Cells 1.0M CuSO 4 0.1 M CuSO 4

64. Concentration Cells The potential of a concentration cell can be calculated. E cell = -.0591 log [dilute] n [conc.] n [conc.]

65. Applications of Concentration Cells Special electrodes have been developed to determine the concentration of a variety of ions. The pH probe is an electrode that contains dilute hydrochloric acid. A thin glass membrane in the electrode is put in contact with a solution of unknown pH. The difference in potential results in the measurement of pH.

66. pH Electrodes The pH electrode is a half-cell containing a silver wire, silver chloride and dilute hydrochloric acid. The potential depends upon the difference in [H 3 O + ] inside and outside of the electrode.

67. Electrolysis An electrolytic cell uses electricity to produce a non-spontaneous chemical reaction. Examples include the electrolysis of water to produce hydrogen and oxygen.

68. Electrolysis Electrolysis is also used in electroplating, in which a metal such as silver is formed on the surface of a less expensive metal. Typically, an electrical current is passed through a solution of the ion to be deposited. The current is expressed in amperes (amp), indicated with the symbol A.

69. Electrolysis 1 amp = 1 coulomb/second If the current (in A) is multiplied by the time the current flows (in s), the total number of coulombs of charge is obtained. Coulombs of charge = amps x seconds

70. Electrolysis Coulombs can be converted to moles of an element deposited as follows: coulombs  moles e -  moles of element 96,485 C/mol e - divide by n

71. Problem: Electrolysis How many grams of copper will be plated out if a current of 5.00 amps is passed through a Cu 2+ solution for an hour. Assume an excess of Cu 2+. How many grams of copper will be plated out if a current of 5.00 amps is passed through a Cu 2+ solution for an hour. Assume an excess of Cu 2+.

72. Application: Corrosion The rusting of iron involves oxidation of iron at the anode, and reduction of oxygen with water at the cathode.

73. Application: Corrosion To help prevent corrosion of underground fuel tanks or the hulls of ships, a sacrificial anode of a more reactive metal than iron is used.

74. Application - Batteries A battery is a galvanic cell or a group of galvanic cells connected in series. They are “storage devices” for electrochemical energy.

75. Lead Storage Battery Each cell of this battery produces approximately 2 volts. Six are connected in series to make the typical 12V car battery.

76. Lead Storage Battery The anode is made of lead, and the cathode is PbO 2. Anode: Pb + HSO 4 -1  PbSO 4 + H + + 2e - Cathode: PbO 2 + HSO 4 -1 +3 H + + 2e -  PbSO 4 + 2H 2 O PbSO 4 + 2H 2 O

77. Lead Storage Battery Net Reaction: Pb(s) + PbO 2 (s) + 2H + (aq) + 2 HSO 4 -1 (aq) Pb(s) + PbO 2 (s) + 2H + (aq) + 2 HSO 4 -1 (aq)  2 PbSO 4 (s) + 2H 2 O(l)

78. Other Batteries 1.5V Dry Cell Mercury Battery

79. The Potato Clock An electrochemical reaction that depends upon salts and acids in the potato can be used to power a digital clock.