1. The Study of the Interchange of Chemical and Electrical Energy
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 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. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Reaction of Zn with Cu2+
If a strip of zinc metal is immersed in a solution of copper(II)sulfate, a reaction will spontaneously occur.
Zn(s) + Cu2+(aq)  Zn2+(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 Cu2+
As each zinc atom dissolves, it provides two electrons to the copper ions.

6. Reaction of Zn with Cu2+
As the reaction proceeds, a thin black layer of Cu is formed on the zinc surface. The blue color of the Cu2+ ion fades as it is reduced.

7. Reaction of Zn with Cu2+
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. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Galvanic Cells
In the galvanic cell for the reaction, the oxidizing reagent (Cu2+) and the reducing agent (Zn) are physically separated into two half-cells.
The overall net-ionic equation for the reaction is:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

9. Galvanic Cells Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
The half-reactions are:
Zn(s)  Zn2+(aq) + 2 e- (oxidation)
Cu2+(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) + Cu2+(aq)  Zn2+(aq) + Cu(s)
The half-reactions are:
Zn(s)  Zn2+(aq) + 2 e- (oxidation)
Cu2+(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) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Zn(s)  Zn2+(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) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Cu2+(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) + Cu2+(aq)  Zn2+(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) + Cu2+(aq)  Zn2+(aq) + Cu(s)

18. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Each combination of half-cells produces a characteristic voltage.

19. Zn(s) + Cu2+(aq)  Zn2+(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

27. 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.

28. Standard Reduction Potentials
The half-reactions are: Zn(s)  Zn2+(aq) + 2e- 2H+(aq) + 2e-  H2(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.

29. Standard Reduction Potentials
All half-reactions are tabulated as reductions. Since the potential for the oxidation of zinc is volts, the reduction potential is -.76 volts. Zn2+(aq) + 2e- Zn(s) Eo = -0.76V

30. 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.

32. 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.

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

34. Cell Potentials - Problem
Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:
Cr2O72-   +  14 H+  +  6 e-   2 Cr3+  +  7 H2O H2O2  + 2H+  +  2 e-    2 H2O

35. Cell Potentials - Problem
Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:
Cr2O72-   +  14 H+  +  6 e-   2 Cr3+  +  7 H2O H2O2  + 2H+  +  2 e-    2 H2O
1. Look up the standard reduction potentials for both half-reactions.

37. Cell Potentials - Problem
Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:
Cr2O72- + 14 H+  + 6 e- 2 Cr3+  + 7 H2O V H2O2  + 2H+  +  2 e-    2 H2O V
2. Reverse one half-reaction so that the net cell potential is the largest positive number.

38. Cell Potentials - Problem
Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:
Cr2O72- + 14 H+  + 6 e- 2 Cr3+  + 7 H2O V H2O2  + 2H+  +  2 e-    2 H2O V
2. Reverse one half-reaction so that the net cell potential is the largest positive number.
reverse

39. Cell Potentials - Problem
Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:
2 Cr3+  + 7 H2O  Cr2O72- + 14 H+  + 6 e V H2O2  + 2H+  +  2 e-    2 H2O V
3. Multiply half-reactions so that the electrons 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 Cr3+  + 7 H2O  Cr2O72- + 14 H+  + 6 e V (H2O2  + 2H+  +  2 e-    2 H2O ) V
3. Multiply half-reactions so that the electrons
lost = electrons gained.  Reduction potentials are not multiplied.

41. Cell Potentials - Problem
Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:
2 Cr3+  + 7 H2O  Cr2O72- + 14 H+  + 6 e V 3H2O2  + 6H+  +  6 e-    6 H2O V
4. Add the two half-reactions and their cell potentials.

42. Cell Potentials - Problem
Determine the balanced reaction and the standard cell potential for a galvanic cell with the following half-cells:
2 Cr3+  + 7 H2O  Cr2O72- + 14 H+  + 6 e V 3H2O2  + 6H+  +  6 e-    6 H2O V
3H2O2  + 2 Cr3+ + 1 H2O  Cr2O72- + 8 H+ 
Eo=0.45V 1 8

43. 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.

44. 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.

45. Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)
Line Notation
The notation for this cell is:
Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)

46. Line Notation Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)
Note that Zn is written on the left because it is the anode.
The notation for this cell is:
Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)

47. Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)
Line Notation
Note that the standard hydrogen electrode is written on the right because it is the cathode.
The notation for this cell is:
Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)

48. 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.

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

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

51. 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.

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

53. Cell Potential and Concentration
For standard temperature (25oC), the equation becomes:
E = Eo log(Q) n
where n is the number of moles of electrons transferred.

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

55. 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.

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

57. Concentration Cells
0.1 M CuSO4
1.0M CuSO4

58. Concentration Cells
The potential of a concentration cell can be calculated.
Ecell = log [dilute]
n [conc.]

59. 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.

60. 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 [H3O+] inside and outside of the electrode.

61. Cell Potential and K
At equilibrium, the cell potential is 0.0 volts, and Q=K. Using the Nernst Equation at 25oC:
E = Eo log(Q) n
0 = Eo log(K)
log K = nEo/0.0591

62. Cell Potential and K log K = nEo/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.

63. Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(aq) | Au(s)
Problem
Write the chemical reaction and calculate the equilibrium constant for the following galvanic cell under standard conditions.
Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(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.

64. Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(aq) | Au(s)
Problem
Write the chemical reaction and calculate the equilibrium constant for the following galvanic cell under standard conditions.
Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(aq) | Au(s)
1. Write the half-reactions.
Cu1+(aq)  Cu2+(aq) + 1 e-
Au3+(aq) + 3 e-  Au(s)

65. Problem Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(aq) | Au(s)
2. Look up the reduction potentials.
Cu1+(aq)  Cu2+(aq) + 1 e- Eo = -(0.16V)
Au3+(aq) + 3 e-  Au(s) Eo = V

66. Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(aq) | Au(s)
Problem
Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(aq) | Au(s)
3. Combine the reactions.
3[Cu1+(aq)  Cu2+(aq) + 1 e-] Eo = -(0.16V)
Au3+(aq) + 3 e-  Au(s) Eo = V
3 Cu1+(aq) + Au3+(aq)  3 Cu2+(aq) + Au(s)
Eo = 1.34V

67. 3 Cu1+(aq) + Au3+(aq)  3 Cu2+(aq) + Au(s)
Problem
3 Cu1+(aq) + Au3+(aq)  3 Cu2+(aq) + Au(s)
Eo = 1.34V
4. Calculate the value of K using Eo.
log K = nEo/0.0591
log K = (3) (1.34)/.0591 = 68.0
K = 1 x 1068

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

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

70. Lead Storage Battery
The anode is made of lead, and the cathode is PbO2.
Anode: Pb + HSO4-1  PbSO4 + H e-
Cathode: PbO2 + HSO H+ + 2e- 
PbSO4 + 2H2O

71. Lead Storage Battery Net Reaction:
Pb(s) + PbO2(s) + 2H+(aq) + 2 HSO4-1(aq)
 2 PbSO4(s) + 2H2O(l)

72. Other Batteries
1.5V Dry Cell
Mercury Battery

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

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

75. 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.

76. Coulombs of charge = amps x seconds
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

77. Electrolysis Coulombs of charge = amps x seconds
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

78. Problem: Electrolysis
How many grams of copper will be plated out if a current of 5.00 amps is passed through a Cu2+ solution for an hour.  Assume an excess of Cu2+.  

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

80. 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.