Electrochemistry

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 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. 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. Reaction of Zn with Cu2+

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. 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 +0.76 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 1.33V H2O2  + 2H+  +  2 e-  2 H2O 1.78V 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 1.33V H2O2  + 2H+  +  2 e-  2 H2O 1.78V 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- -1.33V H2O2  + 2H+  +  2 e-  2 H2O 1.78V 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- -1.33V (H2O2  + 2H+  +  2 e-  2 H2O )3 1.78V 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- -1.33V 3H2O2  + 6H+  +  6 e-  6 H2O 1.78V 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- -1.33V 3H2O2  + 6H+  +  6 e-  6 H2O 1.78V 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. Line Notation The notation for this cell is: Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)

46. Line Notation 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. 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.