1. Prentice Hall © 2003Chapter 20 For the SHE, we assign 2H + (aq, 1M) + 2e -  H 2 (g, 1 atm) E  red = 0.

4. Prentice Hall © 2003Chapter 20 For Zn: E  cell = E  red (cathode) - E  red (anode) 0.76 V = 0 V - E  red (anode). Therefore, E  red (anode) = -0.76 V. Standard reduction potentials must be written as reduction reactions: Zn 2+ (aq) + 2e -  Zn(s), E  red = -0.76 V.

5. Prentice Hall © 2003Chapter 20 Changing the stoichiometric coefficient does not affect E  red. Therefore, 2Zn 2+ (aq) + 4e -  2Zn(s), E  red = -0.76 V.

6. Prentice Hall © 2003Chapter 20 Reactions with E  red < 0 are spontaneous oxidations relative to the SHE. The larger the difference between E  red values, the larger E  cell.

7. Prentice Hall © 2003Chapter 20 Oxidizing and Reducing Agents The more positive E  red the stronger the oxidizing agent on the left. The more negative E  red the stronger the reducing agent on the right.

8. Prentice Hall © 2003Chapter 20

9. Prentice Hall © 2003Chapter 20 More generally, for any electrochemical process

10. Prentice Hall © 2003Chapter 20 Example: For the following cell, what is the cell reaction and E o cell ? Al 3+ (aq) + 3e - → Al(s); E o Al = -1.66 V Fe 2+ (aq) + 2e - → Fe(s); E o Fe = -0.41 V Al(s)|Al 3+ (aq)||Fe 2+ (aq)|Fe(s)

11. Prentice Hall © 2003Chapter 20 EMF and Free-Energy Change We can show that  G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell. We define Since n and F are positive, if  G > 0 then E < 0. Spontaneity of Redox Reactions

12. Prentice Hall © 2003Chapter 20 The Nernst Equation

13. Prentice Hall © 2003Chapter 20 At a temperature of 298 K: This gives:

14. Prentice Hall © 2003Chapter 20 Cell EMF and Chemical Equilibrium A system is at equilibrium when  G = 0. E = 0 V and Q = K eq :

15. Prentice Hall © 2003Chapter 20 Al 3+ (aq) + 3e - → Al(s); E o Al = -1.66 V Fe 2+ (aq) + 2e - → Fe(s); E o Fe = -0.41 V Al(s)|Al 3+ (aq)||Fe 2+ (aq)|Fe(s) In the problem presented before, calculate ΔG o and K eq

16. Prentice Hall © 2003Chapter 20 Example: For the reaction: Zn(s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu(s) Calculate the E cell given that E o cell = +1.10 V, [Cu 2+ ] = 5.0 M and [Zn 2+ ] = 0.050 M

17. Prentice Hall © 2003Chapter 20 Exercise: Calculate the emf generated by the cell that employs the following cell reaction: 2Al(s) + 3I 2 (s) → 2Al 3+ (aq) + 6I - (aq) Take E o cell to be +2.20 V; [Al 3+ ] = 4.0 x 10 -3 M and [I - ] = 0.010M Ans: +2.36 V

18. Prentice Hall © 2003Chapter 20 Battery- self-contained electrochemical power source with one or more voltaic cell. When the cells are connected in series, greater emfs can be achieved. Primary cell – nonrechargeable Secondary cell-rechargeable Batteries

19. Prentice Hall © 2003Chapter 20 2 of 2 PrevPrev Next How Hydrogen Fuels Cells Work Lead-acid battery: Pb and PbO 2 are electrodes and are immersed in sulphuric acid. Electrodes are separated by glass fibres or wood. Alkaline battery: anode is Zn in contact with a concentrated solution of KOH. The cathode is a mixture of MnO2(s) and graphite separated from the anode by a porous fabric

20. Prentice Hall © 2003Chapter 20 2 of 2 PrevPrev Next How Hydrogen Fuels Cells Work Nickel-cadmium, Ni-metal hydride and Li-ion batteries

21. Prentice Hall © 2003Chapter 20 Fuel cells 2 of 2 PrevPrev Next How Hydrogen Fuels Cells Work Fuel cells differ from batteries in that they are not self-contained systems. They use conventional fuels e.g. H 2 and CH 4 to produce electricity.

22. Prentice Hall © 2003Chapter 20 2 of 2 PrevPrev Next How Hydrogen Fuels Cells Work The most common fuel cell is the one involving the reaction of H 2 (g) and O 2 (g) to produce H 2 O(l). This is based on that if H 2 O can be split by electricity, then combining H 2 and O 2 should produce water and electricity.

23. Prentice Hall © 2003Chapter 20 2 of 2 PrevPrev Next How Hydrogen Fuels Cells Work

24. Prentice Hall © 2003Chapter 20 Corrosion of Iron Since E  red (Fe 2+ ) < E  red (O 2 ) iron can be oxidized by oxygen. Cathode: O 2 (g) + 4H + (aq) + 4e -  2H 2 O(l). Anode: Fe(s)  Fe 2+ (aq) + 2e -. Dissolved oxygen in water usually causes the oxidation of iron. Fe 2+ initially formed can be further oxidized to Fe 3+ which forms rust, Fe 2 O 3.xH 2 O(s). Corrosion

26. Prentice Hall © 2003Chapter 20 2 of 2 PrevPrev Next How Hydrogen Fuels Cells Work Galvanised iron, i.e. Fe coated with Zn uses the principle of electrochemistry to protect the iron from corrosion even after the surface coat is broken. The standard red. Pot. for Fe and Zn are: Fe 2+ (aq) + 2e - → Fe(s) E o red = -0.44 V Zn 2+ (aq) + 2e - → Zn(s) E o red = -0.76 V

28. Prentice Hall © 2003Chapter 20 2 of 2 PrevPrev Next How Hydrogen Fuels Cells Work Protecting a metal from corrosion by making it the cathode in an electrochemical cell is called cathodic protection. The metal that is oxidised while protecting the cathode is called the sacrificial anode.

29. Prentice Hall © 2003Chapter 20 Electrolysis of Aqueous Solutions Nonspontaneous reactions require an external current in order to force the reaction to proceed. Electrolysis reactions are nonspontaneous. In voltaic and electrolytic cells: –reduction occurs at the cathode, and –oxidation occurs at the anode. –However, in electrolytic cells, electrons are forced to flow from the anode to cathode.

30. Prentice Hall © 2003Chapter 20 –In electrolytic cells the anode is positive and the cathode is negative. (In galvanic cells the anode is negative and the cathode is positive.)

32. Prentice Hall © 2003Chapter 20 Example, decomposition of molten NaCl. Cathode: 2Na + (l) + 2e -  2Na(l) Anode: 2Cl - (l)  Cl 2 (g) + 2e -. Industrially, electrolysis is used to produce metals like Al.

33. Prentice Hall © 2003Chapter 20 Electroplating Active electrodes: electrodes that take part in electrolysis. Example: electrolytic plating.

34. Prentice Hall © 2003Chapter 20

35. Prentice Hall © 2003Chapter 20 Electroplating Consider an active Ni electrode and another metallic electrode placed in an aqueous solution of NiSO 4 : Anode: Ni(s)  Ni 2+ (aq) + 2e - Cathode: Ni 2+ (aq) + 2e -  Ni(s). Ni plates on the inert electrode. Electroplating is important in protecting objects from corrosion.

36. Prentice Hall © 2003Chapter 20 Quantitative Aspects of Electrolysis We want to know how much material we obtain with electrolysis. Consider the reduction of Cu 2+ to Cu. –Cu 2+ (aq) + 2e -  Cu(s). –2 mol of electrons will plate 1 mol of Cu. –The charge of 1 mol of electrons is 96,500 C (1 F). –Since Q = It, the amount of Cu can be calculated from the current (I) and time (t) taken to plate.

37. Prentice Hall © 2003Chapter 20 Example: When an aqueous solution of CuSO 4 is electrolysed, Cu metal is deposited: Cu 2+ (aq) + 2e - → Cu(s) If a constant current was passed for 5.00 h and 404 mg of Cu metal was deposited, what was the current? Ans: 6.81 x 10 -2 A

38. Prentice Hall © 2003Chapter 20 Example: The half reaction for formation of Mg metal upon electrolysis of molten MgCl 2 is: Mg 2+ (aq) + 2e - → Mg(s) a) Calculate the mass of Mg formed upon passage of a current of 60.0A for a period of 4.00 x 10 3 s b) How many seconds would be required to produce 50.0g of Mg from MgCl 2 if the current is 100.0A? Ans: a) 30.2g, b) 3.97 x 10 3 s

39. Prentice Hall © 2003Chapter 20 Electrical Work ∆G is a measure of maximum useful work, w max, that can be extracted from the process: ∆ G = w max = -nFE For an electrolytic cell, an external source of energy is used to bring about a non-spontaneous electrochemical process. In this case, w = nFE ext

40. Prentice Hall © 2003Chapter 20 Electrical work can be expressed in energy units of watts x time: 1 W = 1 J/s Usually, the kWh is used Example: Calculate the no. of kWh of electricity required to produce 1.00 kg of Mg from electrolysis of molten MgCl 2 if the applied emf is 5.00 V Ans: 11.0 kWh