1. Electrochemistry is the study of chemical reactions that produce electrical effects.

2. General Information Charge – property of matter There are two kinds of charge –electron → - 1.60 x 10 -19 C –proton → +1.60 x 10 -19 C 1.0 Coulomb of charge is equivalent to 6.25 x 10 18 electrons

3. General Information Current –flow of charge –measured in amperes (A) 1 Amp –flow of 1.0 C of charge per second past a given point (C/s)

4. General Information Circuit –a closed pathway for charge to travel –may be a solid or fluid path Conductors in which only electrons can move Solution of positive and negative ions that are free to move

5. Voltaic Cell harnessed chemical reaction which produces an electric current converts chemical potential energy into electrical potential energy.

6. Voltaic Cell

7. Salt bridge (U-tube) –connects the two half cells –allows ions to be exchanged but prevents mixing of the solutions This reaction is spontaneous –oxidation (RA) → Zn (s) ↔ Zn +2 (aq) + 2e - –reduction (OA) → Cu 2+ (aq) + 2e - ↔ Cu (s)

8. Voltaic Cell The two metal rods are called electrodes; this is where the electrons enter and leave the cell. –The electrode where oxidation occurs is called the anode. –The electrode where reduction occurs is called the cathode.

9. Voltaic Cell Anode –electrons leave the cell from the anode –positive ions are produced –Zn (s) → Zn 2+ (aq) + 2e - Cathode –electrons enter the cathode –solid metal is produced –Cu 2+ (aq) + 2e - → Cu (s)

10. Voltaic Cell anode cathode

11. Voltaic Cell At any time, in a given half cell, electrical neutrality is maintained by the movement of ions across the salt bridge. Negative ions (anions) drift toward the anode. Positive ions (cations) drift toward the cathode

12. The Zn/Cu Voltaic Cell

13. Problem Draw a diagram of a copper/aluminum voltaic cell. Label the anode and cathode and show the direction of electron flow. Write the equations for the reactions that occur and predict the cell voltage.

14. Solution Locate the reactions in data book. –Cu 2+ (aq) + 2e - → Cu (s) +0.34 –Al 3+ (aq) + 3e - → Al (s) -1.66 For a voltaic cell reverse the lower reaction. ( Al is lower on table –oxidized) –Cu 2+ (aq) + 2e - → Cu (s) +0.34 –Al (s) → Al 3+ (aq) + 3e - +1.66 +2.00

15. Net ionic reaction Cu 2+ (aq) + 2e - Cu (s) Al (s) Al 3+ (aq) + 3e - 3 X 2 X Remember in all redox equations the electrons are removed 3 Cu 2+ + 6e - 3 Cu (s) 2 Al (s) 2Al 3+ (aq) + 6e - 3 Cu 2+ + 2 Al (s) 3 Cu (s) + 2Al 3+ (aq) ADD

16. copper aluminum Cu 2+ Al 3+ 2.00 V

17. Voltaic Cell the shorthand representation of an electrochemical cell showing the two half- cells connected by a salt bridge or porous barrier, such as: Zn (s) /ZnSO 4(aq) //CuSO 4(aq) /Cu (s) anode cathode Phase boundary Liquid junction

18. Remember Redox reactions can be viewed as a competition for electrons. Therefore the reaction is always between the strongest oxidizing agent and the strongest reducing agent.

19. Problem A strip of metal X is placed in XNO 3 solution. It makes an electrochemical cell with Al 3+ (aq) half cell. It is observed that Al (s) is deposited on the aluminum electrode and the cell voltage is 0.75 V. Calculate the reduction potential of X.

20. Solution Al is reduced: Al 3+ (aq) + 3e - → Al (s) -1.66 X is oxidized: X (s) → X + (aq) + e - ? Solve for ?: -1.66 + ? = +0.75 ? = 2.41 V X is being oxidized, reduction potential is – 2.41 V ( ) x 3 Al 3+ (aq) + 3X (s) → Al (s) + 3X + (aq) 0.75 WS 15-2

21. Standard Electrode Potentials A measure of the relative tendency of substances to gain electrons. The standard was chosen to be the hydrogen half cell. The hydrogen half cell was chosen to be zero. This does not mean H + will not gain electrons.

22. Hydrogen Electrode consists of a platinum electrode covered with a fine powder of platinum around which H 2(g) is bubbled. Its potential is defined as zero volts. Hydrogen Half-Cell H 2(g) → 2 H + (aq) + 2 e - reversible reaction

23. Hydrogen Electrode Non reactive

24. Standard Electrode Potentials Since the hydrogen half cell has a relative value of 0.00 V any half cell connected to it will have its standard potential read off of the voltmeter that connects the two half cells.

25. A Voltaic Cell With Zinc and the Standard Hydrogen Electrode. [Zn 2+ ]= [H + ] = 1.000 M

26. Standard Electrode Potentials Any half cell that gives up electrons to H + will have a negative potential since it has less potential to gain electrons than H +.

27. Standard Electrode Potentials Any half cell that takes electrons from H 2(g) has a positive potential since it has a greater potential to gain electrons than H +.

28. Standard Electrode Potentials Examples: –Cu/Cu 2+ → E o = 0.34 V 0.34 V greater potential than H + /H 2 –Al/Al 3+ → E o = -1.66 1.66 V less potential than H + /H 2

29. If the reduction of mercury (I) in a voltaic cell is desired, the half reaction is: Which of the following reactions could be used as the anode (oxidation)? WS 15-22

30. Multiple cells If the cells are connected to the anode then add the potentials. If the cells are connected at the cathode then subtract the potentials.

32. Electrolytic cells Electrical energy is converted to chemical energy Non-spontaneous reactions Reaction is still between the strongest OA and RA. (OA is not strong enough) Require outside energy source

33. Electrolytic cells The E o value for an electrolytic cell is negative. The E o value is the minimum voltage that must be applied to the cell to force it to react.

34. Electrolytic cells In diagrams –cell with ammeter → voltaic –cell with battery → electrolytic anode (+)cathode (-)

35. Electroplating/Electrolysis Electroplating is the process of

36. Electroplating/Electrolysis 1)Current = rate of flow of charge 2) I = q / t 3)1 e - = 1.6 x 10 -19 C 4)1 mol e - = 6.02 x 10 23 e - x 1.6 x 10 -19 C = 9.65 x 10 4 C current (A)charge (C) time (s) Q = 9.65 x 10 4 C/mol (Faraday’s constant)

37. Electroplating/Electrolysis 5)To calculate number of mol of electrons: n e = q / Q

38. n e = 18000 C / 9.65 x 10 4 C/mol n e = 0.1865 mol of e - 2.50 A of current runs for 2.0 hours through a silver nitrate solution. Find the mass of silver produced. q = It q = (2.50 A) x (7200 s) q = 18 000 C Ag + (aq) + 1e - → Ag (s) 0.1865mol 0.1865 mol m = n x M m = 0.1865 mol x 107.87 g/mol m = 20 g

39. Short cut m = I · t · M m = (2.50 A) (7200 s) (107.87 g/mol) (9.65 x 10 4 C/mol) (1) m = 20 g Q · V I → current (Amps) t → time (s) M → molar mass (g/mol) Q → molar charge (C/mol) V → voltage

40. m = I · t · M 10.0 g = (1.50 A) t (107.87 g/mol) (9.65 x 10 4 C/mol) (1.00) t = 5963.9589 s = 1.66 h Q · V WS 15-33 A person wants to plate an ornament with 10.0 g of silver. The ornament is placed in a salt of silver and a 1.5 A current is placed through the solution. The voltage is 1.00. How long will the plating process take?

41. Copper plating a key - + power e- inert anode cathode Cu 2+ NO 3 -

42. Molten or Aqueous Aqueous NaCl → Na + + Cl - Species present –Na + (aq) –Cl - (aq) –H 2 O (l) Molten NaCl → Na + + Cl - Species present –Na + (l) –Cl - (l)

43. Calculate the net potentials and write the reactions that occur when: –NaI (s) is electrolyzed –A solution of NaI (aq) is electrolyzed

44. NaI (s) is electrolyzed Na + (l) I - (l) Na + (l) + e - → Na (s) -2.71 2I - (l) → I 2(s) + 2e - -0.54 2Na + (l) + 2I - (l) → 2Na (s) + I 2(s) -3.25 OARA

45. A solution of NaI (aq) is electrolyzed H 2 O (l) Na + (aq) I - (aq) 2H 2 O (l) + 2e - → H 2(g) + 2OH - (aq) -0.83 2I - (aq) → I 2(s) + 2e - -0.54 2H 2 O (l) + 2I - (aq) → H 2(g) + 2OH - (aq) + I 2(s) -1.37 OARA

46. Electrolysis of Brine Na + Cl - H 2 O OA RA 2H 2 O (l) + 2Cl - (aq) → H 2(g) + 2OH - (aq) + Cl 2(g)

47. Note The Cl - ion is a stronger reducing agent than H 2 O during electrolysis of brine even though the table indicates that it is not.

48. Electrolysis of molten NaCl

49. Changing Concentration A change in concentration around the cathode or anode will change the value of E o for half cells.

50. Cu (s) → Cu 2+ (aq) + 2e - An increase in concentration of copper ions will cause a shift to the left making less products. Increase in concentration around anode causes a decrease in E o. Cu (s) Cu 2+ Anode

51. Cu 2+ (aq) + 2e - → Cu (s) An increase in concentration of copper ions will cause a shift to the right making more products. Increase in concentration around cathode causes a increase in E o. Cu (s) Cu 2+ Cathode

52. Since E  (Fe 2+ /Fe) < E  (O 2 /H 2 O) iron can be oxidized by oxygen. Cathode –O 2(g) + 4 H + (aq) + 4 e -  2 H 2 O (l) Anode –Fe (s)  Fe 2+ (aq) + 2 e - Fe 2+ initially formed – further oxidized to Fe 3+ which forms rust.

53. Rusting (Corrosion) of Iron

54. Preventing the Corrosion of Iron Corrosion can be prevented by coating the iron with paint or another metal. Galvanized iron - Fe is coated with Zn. Zn protects the iron (Zn - anode and Fe - the cathode) Zn 2+ (aq) +2e-  Zn (s), E  (Zn 2+ /Zn) = -0.76 V Fe 2+ (aq) + 2e-  Fe (s), E  (Fe 2+ /Fe) = -0.44 V

55. Preventing the Corrosion of Iron

56.  To protect underground pipelines, a sacrificial anode is added.  The water pipe - turned into the cathode and an active metal is used as the sacrificial anode.  Mg is used as the sacrificial anode: Mg 2+ (aq) +2e -  Mg (s), E  (Mg 2+ /Mg) = -2.37 V Fe 2+ (aq) + 2e -  Fe (s), E  (Fe 2+ /Fe) = -0.44 V

57. Corrosion Prevention

61. Living Battery The eel generates electric charge in a battery of biological electrochemical cells, each cell providing about 0.15 V and an overall potential difference of ~ 700 V. Note that the eel's head is the cathode(+) and its tail the anode(-). The cells extend over the length of the eel.

62. Types of Batteries Batteries –device that converts chemical energy into electricity Primary Cells –non-reversible electrochemical cell –non-rechargeable cell Secondary Cells –reversible electrochemical cell –rechargeable cell

63. Types of Batteries Primary Cells "dry" cell & alkaline cell 1.5 v/cell mercury cell 1.34 v/cell fuel cell 1.23v/cell Secondary Cells lead-acid (automobile battery) 2 v/cell NiCad 1.25 v/cell

64. Lead-Acid Battery A 12 V car battery - 6 cathode/anode pairs each producing 2 V. Cathode: PbO 2 on a metal grid in sulfuric acid PbO 2(s) + SO 4 2- (aq) + 4H + (aq) + 2e -  PbSO 4(s) + 2H 2 O (l) Anode: Pb Pb (s) + SO 4 2- (aq)  PbSO 4(s) + 2e -

65. A Picture of a Car Battery

66. An Alkaline Battery Anode: Zn cap Zn (s)  Zn 2+ (aq) + 2e - Cathode: MnO 2, NH 4 Cl and carbon paste 2 NH 4 + (aq) + 2 MnO 2(s) + 2e -  Mn 2 O 3(s) + 2NH 3(aq) + 2H 2 O (l) Graphite rod in the center - inert cathode. Alkaline battery, NH 4 Cl is replaced with KOH. Anode: Zn powder mixed in a gel:

67. “Dry” Cell

68. The Alkaline Battery

69. “New” Super Iron Battery BaFeO 4 + 3/2 Zn → 1/2 Fe 2 O 3 + 1/2 ZnO + BaZnO 2 Environmentally friendlier than MnO 2 containing batteries.

70. Fuel Cells Direct production of electricity from fuels occurs in a fuel cell. H 2 - O 2 fuel cell was the primary source of electricity on Apollo moon flights. Cathode: reduction of oxygen 2 H 2 O (l) + O 2(g) + 4 e -  4 OH - (aq) Anode: 2 H 2(g) + 4 OH - (aq)  4 H 2 O (l) + 4 e -