Chapter 20

1. Chapter 20 Electrochemistry and Oxidation – Reduction

2. Mg + HCl → MgCl2 + H2

3. Mg + HCl → MgCl2 + H2 • Redox Reactions: • Reduction is the decrease of an oxidation number by the gain of electrons. • Oxidation is the increase of an oxidation number by the loss of electrons.

4. Mg + HCl → MgCl2 + H2

5. LEOthe lion saysGER Lose Electrons Oxidation Gain Electrons Reduction

6. Mg + HCl → MgCl2 + H2 • Half-Reactions: Oxidation ½ reaction: Mg → Mg2+ + 2e- Reduction ½ reaction: 2H+ + 2e-→ H2 Electron flow

7. Half-Cells • A half-cell represents a half-reaction from a redox reaction. • A half-cell is made by placing a piece of reactant in an electrolyte solution. • The anode is the half-cell where oxidation occurs. • The cathode is the half-cell where reduction occurs.

8. Mg + HCl → MgCl2 + H2 • Half-Reactions: Oxidation ½ reaction: Mg → Mg2+ + 2e- Reduction ½ reaction: 2H+ + 2e-→ H2 • Half-Cells: A 1 1 1 A A 1 1 1

9. Mg + HCl → MgCl2 + H2 Oxidation (Anode) Reduction (Cathode) Mg → Mg2+ + 2e- 2H+ + 2e-→ H2 A A A A 1 1 1 1 1 1 A A a A a a a a

10. Mg + HCl → MgCl2 + H2 Oxidation (Anode) Reduction (Cathode) Mg → Mg2+ + 2e- 2H+ + 2e- → H2

11. Mg + HCl → MgCl2 + H2 This is a battery. A battery is a type of galvanic cell (voltaic cell). A galvanic cell is a device in which a chemical reaction produces electrical energy.

12. Mg + HCl → MgCl2 + H2 2.37V Oxidation (Anode) Reduction (Cathode) Mg → Mg2+ + 2e- 2H+ + 2e- → H2

13. Standard Reduction PotentialsFigure 20.1 Page 650

14. Mg + HCl → MgCl2 + H2

15. E°cell = E°ox + E°red • E°cell = electromotive force (emf) = voltage = cell potential

16. Understanding Reduction Potentials • Standard conditions: • 1M for ion concentrations • 1atm for gas pressures • 25°C

17. The standard hydrogen electrode

18. Measuring Potential for a Copper Half-Cell

19. Measuring Potential for a Copper Half-Cell

22. Zn + CuSO4(aq)  ZnSO4(aq)+ Cu

23. Zn + CuSO4(aq)  ZnSO4(aq)+ Cu • Write the half-reactions • Draw the half-cells • Label the anode and the cathode • Label the direction of electron flow • Draw a salt bridge and label the direction of ion flow within it. • Describe what you would observe at the anode as the reaction proceeds. • Describe what you would observe at the cathode as the reaction proceeds. • Calculate the cell potential (E°cell)

24. Zn + CuSO4(aq)  ZnSO4(aq)+ Cu A

25. Zn + CuSO4(aq)  ZnSO4(aq)+ Cu

27. E°cell and spontaneity • +E°cell = spontaneous cell reaction • -E°cell = nonspontaneous cell reaction

28. What is the potential for the cell Zn | Zn2+(1.0 M) || Cu2+(1.0 M) | Cu

29. E°cell = +1.10V The reaction is spontaneous Zinc reacts with a solution of copper (II) ions

30. What is the potential for the cell Ag | Ag+(1.0 M) || Li+(1.0 M) | Li • From the table of standard reduction potentials (Appendix H), you find: Ag → Ag+ + e- E°ox = -0.7991V Li+ + e-→ Li E°red = -3.09V E°cell = -3.8891V E°cell = -3.89V • The reaction is nonspontaneous • Silver will not react with a solution of lithium ions

32. A E°red = – 0.440V

33. Write the cell reaction and calculate the potential of this galvanic cell.

34. Co + 2Fe3+→ Co2+ + 2Fe2+ E°cell = +1.048V

35. Write the cell reaction and calculate the potential of a cell consisting of a standard bromine electrode as the anode and a standard chlorine electrode as the cathode. 2Br- + Cl2 → Br2 + 2Cl- E°cell = +0.2943V

36. Write the cell reaction and calculate the potential of a cell consisting of a standard bromine electrode as the anode and a standard chlorine electrode as the cathode.

37. Free Energy Change and Cell Potential ∆G = ‒nFEcell n = number of moles of electrons F = faraday = 96,485 J V-1 mol-1 ≈ 96,500 J V-1 mol-1≈ 96.5 kJV-1 mol-1

38. Free Energy Change and Cell Potential ∆G = ‒nFEcell How do ∆G and Ecell relate?

39. Calculate the cell potential and the ∆G for the reaction:Cd + Pb2+ → Cd2+ + Pb A

40. Calculate the cell potential and the ∆G for the reaction:Cd + Pb2+ → Cd2+ + Pb E°cell = +0.277V ∆G = -53.5 kJ

41. Calculate the cell potential and the ∆G for the reaction:Sn4+ + 2Fe2+ → Sn2+ + 2Fe3+

42. Calculate the cell potential and the ∆G for the reaction:Sn4+ + 2Fe2+ → Sn2+ + 2Fe3+ E°cell = -0.62V ∆G = +120kJ

43. The Effect of Concentration on Cell Potential Ecell vs. E°cell The Nernst Equation (Page 657)

44. Concentration and Ecell • Example. Calculate the cell potential for the following: Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) Where [Cu2+] = 0.30 M and [Fe2+] = 0.10 M

45. Concentration and Ecell (cont.) • First, need to identify the 1/2 cells Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) E°1/2 = 0.34 V Cu2+(aq) + 2e- Cu(s) E°1/2 = -0.44 V Fe2+(aq) + 2e- Fe(s) Fe(s) Fe 2+(aq) + 2e- E°1/2 = +0.44 V Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) E°cell = +0.78 V

46. Concentration and Ecell (cont.) • Now, calculate Ecell Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) E°cell = +0.78 V Ecell = E°cell - (0.05916/n)log(Q) Ecell = 0.78 V - (0.05916 /2)log(0.33) Ecell = 0.78 V - (-0.014 V) = 0.794 V

47. Concentration and Ecell (cont.) • If [Cu2+] = 0.30 M, what [Fe2+] is needed so that Ecell = 0.76 V? Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) E°cell = +0.78 V Ecell = E°cell - (0.05916/n)log(Q) 0.76 V = 0.78 V - (0.05916/2)log(Q) 0.02 V = (0.05916/2)log(Q) 0.676 = log(Q) 4.7 = Q

48. Concentration and Ecell (cont.) Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) 4.7 = Q [Fe2+] = 1.4 M

49. Comparing Q and Ecell From the last two problems: E°cell = +0.78V Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) Ecell = E°cell - (0.05916/n)log(Q)

50. Example 20.9 Page 657 E°cell = +0.179V