1. CHEM 180/181Chapter 20 Chapter 20 Electrochemistry

2. CHEM 180/181Chapter 20 Topics in Electrochemistry Oxidation-Reduction (Redox) Numbers Balancing Redox Reactions Voltaic Cells Cell EMF Spontaneity of Redox Reactions Batteries Corrosion Electrolysis

3. CHEM 180/181Chapter 20 Oxidation and Reduction Oxidation (loss of e - ) Na Na + + e - Reduction (gain of e - ) Cl + e - Cl - Oxidation-reduction (redox) reactions occur when electrons are transferred from an atom that is oxidized to an atom that is reduced. Electron transfer can produce electrical energy spontaneously, but sometimes electrical energy is used to make them occur (nonspontaneous).

4. CHEM 180/181Chapter 20

5. CHEM 180/181Chapter 20 Oxidation-Reduction (Redox) Reactions BOTH reduction and oxidation must occur. A substance that gives up electrons is oxidized and is called a reducing agent or reductant (causes another substance to be reduced). A substance that accepts electrons is reduced and is therefore called an oxidizing agent or oxidant (causes another substance to be oxidized).

6. CHEM 180/181Chapter 20 Determine if it’s a redox reaction by keeping track of the oxidation states of all elements involved. Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) (0) (+1) (+2) (0) Quick hint: If a reaction includes an ELEMENT that turns into an ion, it’s a redox reaction. Oxidation-reduction (Redox) Reactions

7. CHEM 180/181Chapter 20 1.Atoms in elemental form, oxidation number is zero. (Cl 2, H 2, P 4, Ne are all zero) 2.Monoatomic ion, the oxidation number is the charge on the ion. (Na + : +1; Al 3+ : +3; Cl - : -1) 3.O is usually -2. But in peroxides (like H 2 O 2 and Na 2 O 2 ) it has an oxidation number of -1. 4.H is +1 when bonded to nonmetals and -1 when bonded to metals. (+1 in H 2 O, NH 3 and CH 4 ; -1 in NaH, CaH 2 and AlH 3 ) 5.The oxidation number of F is -1. 6.The sum of the oxidation numbers for the molecule is the charge on the molecule (zero for a neutral molecule). Oxidation Number Guidelines

8. CHEM 180/181Chapter 20 Example Determine the oxidation state of all elements in ammonium thiosulfate (NH 4 ) 2 (S 2 O 3 ). (NH 4 ) 2 (S 2 O 3 ) NH 4 + S 2 O 3 2- -3 +1-2 -6 total +2 NH 4 + S 2 O 3 2- +4 sum = overall charge on ion -3 +4

9. CHEM 180/181Chapter 20 Determining Oxidation States What is the oxidation state of Mn in MnO 4 - ? Answer: +7

10. CHEM 180/181Chapter 20 Balancing oxidation-reduction equations Balancing chemical equations follows law of conservation of mass. AND, gains and losses of electrons must also be balanced.

11. CHEM 180/181Chapter 20 Half-Reactions Separate oxidation and reduction processes in equation, Sn 2+ (aq) + 2Fe 3+ (aq) Sn 4+ (aq) + 2Fe 2+ (aq) Oxidation: Sn 2+ (aq) Sn 4+ (aq) + 2e - Reduction: 2Fe 3+ (aq) + 2e - 2Fe 2+ (aq) Overall, the number of electrons lost in the oxidation half reaction must equal the number gained in the reduction half reaction.

12. CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Acidic Consider : MnO 4 - (aq) + C 2 O 4 2- (aq) Mn 2+ (aq) + CO 2 (g) Unbalanced half-reactions: MnO 4 - (aq) Mn 2+ (aq) C 2 O 4 2- (aq) CO 2 (g) First, balance everything EXCEPT hydrogen and oxygen. Deal with half-reactions SEPARATELY. (acidic)

13. CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Acidic To balance O: Add 4H 2 O to products to balance oxygen in reactants. MnO 4 - (aq) Mn 2+ (aq) + 4H 2 O(l) To balance H: Add 8H + to reactant side to balance the 8H in water. 8H + (aq) + MnO 4 - (aq) Mn 2+ (aq) + 4H 2 O(l)

14. CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Acidic Balance charge: Add up charges on both sides. Add 5 electrons to reactant side. 5e - + 8H + (aq) + MnO 4 - (aq) Mn 2+ (aq) + 4H 2 O(l) Mass balance of C in oxalate half-reaction. C 2 O 4 2- (aq) 2CO 2 (g) Balance charge by adding two electrons to the products. C 2 O 4 2- (aq) 2CO 2 (g) + 2e - Last step: Cancel electrons and add reactions together.

15. CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Acidic 5e - + 8H + (aq) + MnO 4 - (aq) Mn 2+ (aq) + 4H 2 O(l) C 2 O 4 2- (aq) 2CO 2 (g) + 2e - Top reaction times 2. Bottom reaction times 5. ALL PARTS!! 10e - + 16H + (aq) + 2MnO 4 - (aq) 2Mn 2+ (aq) + 8H 2 O(l) 5C 2 O 4 2- (aq) 10CO 2 (g) + 10e -. 16H + (aq) + 2MnO 4 - (aq) + 5C 2 O 4 2- (aq) 2Mn 2+ (aq) + 8H 2 O(l) + 10CO 2 (g) **Electrons have cancelled out**

16. CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Acidic Summary 1. Divide equation into two incomplete half-reactions. 2. Balance each half-reaction (a) balance elements other than H and O. (b) balance O atoms by adding H 2 O. (c) balance H atoms by adding H+ (basic conditions will require further work at this step). (d) balance charge by adding e - to the side with greater overall positive charge.

17. CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Summary 3.Multiply each half-reaction by an integer so that the number of electrons lost in one half-reaction equals the number gained in the other half-reaction. 4. Add the two half-reactions and cancel out all species appearing on both sides of the equation. 5. Check equation to make sure there are same number of atoms of each kind and the same total charge on both sides. Errors can be caught!!

18. CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Basic Balancing process is started using H + and H 2 O, then adjusting with OH - to uphold reaction conditions (H + does not exist in basic solutions). Balance the following reaction: H 2 O 2 (aq) + ClO 2 (aq) ClO 2 - (aq) + O 2 (g) (basic)

19. CHEM 180/181Chapter 20 Basic Redox Reactions Split into two half-reactions. H 2 O 2 (aq) O 2 (g) ClO 2 (aq) ClO 2 - (aq) Balance elements, then oxygen by adding H 2 O. Then, add H + to balance H, just like an acidic redox reaction. H 2 O 2 (aq) O 2 (g) + 2H + ClO 2 (aq) ClO 2 - (aq) Balancing Equations by the Method of Half-Reactions: Basic

20. CHEM 180/181Chapter 20 Basic Redox Reactions Add OH - to both sides, enough to neutralize all H + (basic reactions cannot support H + ). 2OH - + H 2 O 2 (aq) O 2 (g) + 2H + + 2OH - ClO 2 (aq) ClO 2 - (aq) Combine H + and OH - to form H 2 O. 2OH - + H 2 O 2 (aq) O 2 (g) + 2H 2 O Balancing Equations by the Method of Half-Reactions: Basic

21. CHEM 180/181Chapter 20 Basic Redox Reactions Balance charge by adding e -. 2OH - + H 2 O 2 (aq) O 2 (g) + 2H 2 O + 2e - 1e - + ClO 2 (aq) ClO 2 - (aq) Multiply each reaction so both have same e -. Then add them together, cancelling where possible. 2OH - + H 2 O 2 (aq) + 2ClO 2 (aq) O 2 (g) + 2H 2 O + 2ClO 2 - (aq) Double-check your answer! Balancing Equations by the Method of Half-Reactions: Basic

22. CHEM 180/181Chapter 20 Balance this redox equation Cu(s) + NO 3 - (aq) Cu 2+ (aq) + NO 2 (g) (acidic) Ans: Cu(s) + 2NO 3 - (aq) + 4H + (aq) Cu 2+ (aq) + 2NO 2 (aq) + 2H 2 O(l)

23. CHEM 180/181Chapter 20 Balance this redox equation NO 2 - (aq) + Al(s) NH 3 (aq) + Al(OH) 4 - (aq) (basic) 2Al(s) + NO 2 - (aq) +OH - (aq) + 5H 2 O(aq) 2Al(OH) 4 - (aq) + NH 3 (aq)

24. CHEM 180/181Chapter 20 Voltaic (aka galvanic) cells: electrochemical reactions in which electron transfer occurs via an external circuit. The energy released in a voltaic cell reaction can be used to perform electrical work. Reactions are spontaneous. Voltaic Cells

25. CHEM 180/181Chapter 20 For example Anode: Zn(s)  Zn 2+ (aq) + 2e - (oxidation half-reaction) Cathode: Cu 2+ (aq) + 2e -  Cu(s) (reduction half reaction) Salt bridge: cations move from anode to cathode, anions move from cathode to anode. External circuit (wire): electrons move from anode to cathode (between two solid metal electrodes). Electron transfer can naturally occur in OTHER forms besides a wire (direct electron transport between solution and metal). Voltaic Cells: Components

26. CHEM 180/181Chapter 20 Anions and cations move through a porous barrier or salt bridge. Cations move into the cathodic compartment to neutralize the excess negatively charged ions. Anions move into the anodic compartment to neutralize the excess Zn 2+ ions formed by oxidation. Anode: Zn(s)  Zn 2+ (aq) + 2e - (oxidation half-reaction) Voltaic Cells: Ion Flow

27. CHEM 180/181Chapter 20 Simplified Voltaic Cell In any voltaic cell, the electrons flow from the anode through the external circuit to the cathode. Anode thus labeled with negative sign (-) and cathode with positive sign (+). Anions always migrate toward the anode and cations towards the cathode.

28. CHEM 180/181Chapter 20 As oxidation occurs, Zn(s) is converted to Zn 2+ and 2e -. The electrons flow from anode (where they’re produced) towards the cathode where they are consumed in the reduction reaction. Voltaic Cells: Electron Flow We expect the Zn electrode to lose mass and the Cu electrode to gain mass.

29. CHEM 180/181Chapter 20 If a strip of Zn is placed in a solution of CuSO 4, Cu is deposited on the Zn and the Zn dissolves, forming Zn 2+. Zn is spontaneously oxidized to Zn 2+ by Cu 2+ (Cu is the oxidizing agent). The Cu 2+ is spontaneously reduced to Cu 0 by Zn (Zn is the reducing agent). The entire process is spontaneous. Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) One-Pot Voltaic Cells

30. CHEM 180/181Chapter 20 The Atomic Level A Cu 2+ (aq) ion comes into contact with a Zn(s) atom on the surface of the electrode. Two electrons are directly transferred from the Zn(s), forming Zn 2+ (aq), to the Cu 2+ (aq) forming Cu(s).

31. CHEM 180/181Chapter 20 Electromotive force (emf): the force required to push electrons through the external circuit. Potential difference: difference per electrical charge between two electrodes. Measured in volts. One volt is the potential required to impart one joule of energy to a charge of one coulomb: Cell Electromotive Force (EMF)

32. CHEM 180/181Chapter 20 Cell potential (E cell ): the emf of a cell. Aka cell voltage and is positive for spontaneous cell reactions. E cell strictly depends on reactions that occur at the cathode and anode, the concentration of reactants and products, and the temperature and is described by the equation: E cell = E red (cathode) - E red (anode) Cell EMF For 1M solutions at 25˚C and 1 atm (standard conditions), the standard emf (standard cell potential), E cell, is written as E ˚ cell.

33. CHEM 180/181Chapter 20 Emf of a voltaic cell depends on the particular cathode and anode half-cell reactions involved. Standard reduction potential measured for each half-cell and then used to determine E˚ cell (emf). Convenient and extensive tabulation of electrochemical data. Standard Reduction Potentials

34. CHEM 180/181Chapter 20 Standard Reduction Potentials Potential associated with each half-cell reaction is chosen to be the potential for reduction to occur at that electrode. Hence, Standard Reduction Potentials. Standard reduction potentials, E˚ red, are measured relative to the standard hydrogen electrode (SHE).

35. CHEM 180/181Chapter 20 SHE is assumed as the cathode (for consistency). Pt electrode in a tube containing 1M H + solution. H 2 (g) is bubbled through the tube and equilibrium is established. Standard Hydrogen Electrode (SHE)

36. CHEM 180/181Chapter 20 For the SHE: 2H + (aq, 1M) + 2e -  H 2 (g, 1 atm) E˚ red = 0.00 V Standard reduction potentials can then be calculated using the SHE (E˚= 0.00 V) as E˚ red (cathode). Each calculated E˚ is rewritten as a reduction and tabulated. Determining Standard Reduction Potentials

37. CHEM 180/181Chapter 20 Finding the Standard Reduction Potential E˚ cell is measured, 0.00 V (SHE) is used for E˚(cathode), and the reduction potential for Zn is found.

38. CHEM 180/181Chapter 20 We measure E˚ cell relative to the SHE: E˚ cell = E˚ red (cathode) - E˚ red (anode) 0.76V = 0.00 V - E˚ red (anode). Therefore, E˚ red (anode) = -0.76V And we find that -0.76V can be assigned to reduction of zinc. Zn 2+ (aq) + 2e -  Zn(s) E˚ red = -0.76 V Zn Standard Reduction Potential

39. CHEM 180/181Chapter 20 Since E˚ red = -0.76 V (negative!) we conclude that the reduction of Zn 2+ in the presence of the SHE is not spontaneous. However, the oxidation of Zn with the SHE is spontaneous. Reactions with E˚ red > 0 are spontaneous reductions relative to the SHE. E˚ red < 0 are spontaneous oxidations. Changing the stoichiometric coefficient does not affect E˚ red. 2Zn 2+ (aq) + 4e -  2Zn(s) E˚ red = -0.76 V Zn Standard Reduction Potential

40. CHEM 180/181Chapter 20 Full list in Appendix E

41. CHEM 180/181Chapter 20 The larger the difference between E˚ red values, the larger E˚ cell. Spontaneous voltaic cell: E˚ red (cathode) is more positive than E˚ red (anode), resulting in a positive E˚cell. Recall EMF Trends

42. CHEM 180/181Chapter 20 Zn(s) + Cu 2+ (aq, 1M) Zn 2+ (aq, 1M) + Cu(s) E˚ cell = 1.10V Given: Zn 2+ + 2e - Zn(s) E˚ red = -0.76V Calculate the E˚ red for the reduction of Cu 2+ to Cu. Cu 2+ + 2e - Cu(s) E˚ cell = E˚ red (cathode) – E˚ red (anode) 1.10V = E˚ red (cathode) – (-0.76V) E˚ red (cathode) = 1.10V – 0.76V = 0.34V Q: What about E˚ red for the oxidation of Cu(s) – reverse reaction? Calculating E˚ red from E˚ cell

43. CHEM 180/181Chapter 20 Oxidizing and Reducing Agents We can use E˚ red values to understand aqueous reaction chemistry. e.g. Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Since Cu 2+ is responsible for the oxidation of Zn(s), Cu 2+ is called the oxidizing agent. Since Zn(s) is responsible for the reduction of Cu 2+, Zn(s) is called the reducing agent. E˚ red for Cu 2+ (0.34V) indicates that, compared to E˚ red for Zn (-0.76V), it will be reduced.

44. CHEM 180/181Chapter 20 Example Problem: Cell emf Calculate the standard emf for the following: Ni(s) + 2Ce 4+ (aq) Ni 2+ (aq) + 2Ce 3+ (aq) Ask yourself…What are the half-reactions? Which one is oxidized? Reduced? Ni 2+ (aq) + 2e - Ni(s) E˚ red = -0.28V Ce 4+ (aq) + 1e - Ce 3+ (aq) E˚ red = 1.61V E˚ cell = E˚ red (cathode) - E˚ red (anode) E˚ cell = 1.61V-(-0.28) = 1.89V is the oxidizing agent.

45. CHEM 180/181Chapter 20 Which reaction will undergo reduction? What reaction occurs at the anode? Sn 4+ (aq) + 2e - Sn 2+ (aq) E˚ red = 0.154V MnO 4 - (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O(l) E˚ red = 1.51V MnO 4 - (aq) reaction has more positive E˚ red, so it will be reduced more favorably. Sn 2+ will be oxidized at the anode. E˚ cell = 1.51-0.154 = 1.36V Example Problem: Cell emf

46. CHEM 180/181Chapter 20 The more positive E˚ red, the greater the tendency for the reactant in the half-reaction to be reduced. Therefore, the stronger the oxidizing agent. The more negative E˚ red, the greater the tendency for the product in the half-reaction to be oxidized. This means the product is a reducing agent. Oxidizing and Reducing Agents: EMF

47. CHEM 180/181Chapter 20 A species at higher left of the table of standard reduction potentials will spontaneously oxidize a species that is lower right in the table. F 2 will oxidize H 2 or Li. Ni 2+ will oxidize Al(s).

48. CHEM 180/181Chapter 20 Example Problem: Redox Agents Which is the strongest reducing agent? Oxidizing agent? Ce 4+, Br 2, H 2 O 2, or Zn? E˚ red = 1.61V for Ce 4+ (aq) 1.065V for Br 2 (l) 1.776 for H 2 O 2 (aq) -0.763V for Zn(s) Most easily reduced = most positive = H 2 O 2. So, H 2 O 2 is the strongest oxidizing agent. Most easily oxidized = most negative = Zn. So, Zn is the strongest reducing agent.

49. CHEM 180/181Chapter 20 In a spontaneous voltaic (galvanic) cell, E˚ red (cathode) is more positive than E˚ red (anode). A positive E˚ cell indicates a spontaneous voltaic cell process. A negative E˚ cell indicates a non-spontaneous process. It all relates back to Gibbs Free Energy. Spontaneity of Redox Reactions

50. CHEM 180/181Chapter 20 Cell emf and free-energy change are related by  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.  G units are J (assumed per mole). 1F = 96,500 C/mol = 96,500 J/Vmol If standard conditions:  G˚ = -nFE˚ cell If E cell > 0 then  G < 0, both of which indicate spontaneous processes. Spontaneity of Redox Reactions

51. CHEM 180/181Chapter 20 Example Problem: Cell spontaneity Calculate ΔG˚ for the following reaction. Is it spontaneous? Cl 2 (g) + 2I - (aq) 2Cl - (aq) + I 2 (s) E˚ cell = 0.823V ΔG˚ = -nFE˚ cell ΔG˚ = -(2 mol)(96,500 J/Vmol)(0.823 V) = -156 kJ Since ΔG˚ is negative (and E˚ cell is positive), it is spontaneous.

52. CHEM 180/181Chapter 20 Defn: Self-contained electrochemical power source with one (or more) complete voltaic cell. When multiple cells or multiple batteries are connected in series, greater emfs can be achieved. Cathode labeled with a plus sign; the anode with a minus sign. Batteries

53. CHEM 180/181Chapter 20 Spontaneous redox reactions, as in voltaic cells, serve as the basis for battery operation. Specific reactions at anode and cathode determine the voltage of the battery, and the usable life of the battery depends on the quantity of these substances. Primary cells cannot be recharged. Secondary cells are rechargeable using an external power source after its emf has dropped below a usable level. How Batteries Work

54. CHEM 180/181Chapter 20 A 12V car battery consists of 6 cathode/anode pairs, connected in series, each producing 2V. Cathode: PbO 2 (s) 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(s) in sulfuric acid Pb(s) + SO 4 2- (aq)  PbSO 4 (s) + 2e - The overall electrochemical reaction is PbO 2 (s) + Pb(s) + 2SO 4 2- (aq) + 4H + (aq)  2PbSO 4 (s) + 2H 2 O(l) Lead-Acid Batteries

55. CHEM 180/181Chapter 20 Lead-Acid Batteries E  cell = E  red (cathode) - E  red (anode) = (+1.685 V) - (-0.356 V) = +2.041 V. Wood or fiberglass spacers prevent electrode contact. H 2 SO 4 consumed during discharge, (voltage may vary with use). Recharging reverses forward reaction: PbO 2 (s) + Pb(s) + 2SO 4 2- (aq) + 4H + (aq)  2PbSO 4 (s) + 2H 2 O(l)

56. CHEM 180/181Chapter 20 Most common nonrechargeable (primary) battery (100 billion produced annually). Anode: Powdered Zn in gel in contact with cKOH solution Zn(s) + 2OH - (aq)  Zn(OH) 2 (s) + 2e - Cathode: MnO 2 and C paste 2MnO 2 (s) + 2H 2 O(l) + 2e -  2MnO(OH)(s) + 2OH - (aq) Alkaline Batteries

57. CHEM 180/181Chapter 20 Lightweight batteries for use in cell phones, notebook computers, etc. Nickel-cadmium (NiCad) battery still common. Cadmium oxidized at anode, nickel oxyhydroxide [NiO(OH)(s)] reduced at cathode. Cathode: 2NiO(OH)(s) + 2H 2 O(l) + 2e -  2Ni(OH) 2 (s) + 2OH - (aq) Anode: Cd(s) + 2OH - (aq)  Cd(OH) 2 (s) + 2e - Rechargeable Batteries

58. CHEM 180/181Chapter 20 Solid reaction products adhere to the electrodes, permitting electrode reactions to be reversed during recharging. Drawbacks of NiCad battery: Toxicity of Cd (disposal problems), relatively heavy. Rechargeable Batteries

59. CHEM 180/181Chapter 20 Cathode reaction in nickel metal hydride (NiMH) battery is same as for Ni-Cd battery. 2NiO(OH)(s) + 2H 2 O(l) + 2e -  2Ni(OH) 2 (s) + 2OH - (aq) Anode consists of a metal alloy, e.g. ZrNi 2, capable of absorbing hydrogen atoms. During oxidation, the hydrogen atoms lose electrons, and the resulting H + ions react with OH - to form water (drives cathode reaction to the right). Another type: Li-ion battery - lightweight with very high energy density. NiMH and Li-ion Batteries

60. CHEM 180/181Chapter 20 Direct, efficient production of electricity. NOT batteries as they are not self-contained (some products must be released). On Apollo moon flights, the H 2 -O 2 fuel cell was the primary source of electricity. Cathode: 2H 2 O(l) + O 2 (g) + 4e -  4OH - (aq) Anode: 2H 2 (g) + 4OH - (aq)  4H 2 O(l) + 4e - What’s the overall reaction? Fuel Cells

61. CHEM 180/181Chapter 20 Generally the result of an undesirable redox reaction. Spontaneous! A metal is attacked by some substance in its environment to form an unwanted compound. For many metals oxidation is thermodynamically favored at RT. Corrosion can form an insulating protective coating, preventing further oxidation. (e.g., Al forms a hydrated form of Al 2 O 3 ) ~20% of iron produced annually in USA is used to replace rusted items!! Corrosion

62. CHEM 180/181Chapter 20 Rusting of iron requires both oxygen and water. Other factors that can accelerate rusting of iron: - pH of solution (acidic) -presence of salts (ocean) -contact with metals more difficult to oxidize than iron -stress on the iron Rusting of iron is electrochemical and iron itself conducts electricity. Corrosion of Iron

63. CHEM 180/181Chapter 20 Recall: E˚ red Fe 2+ (-0.44V) < E˚ red O 2 (1.23V), so iron can be oxidized by oxygen. Cathode: O 2 (g) + 4H + (aq) + 4e -  2H 2 O(l) E˚ red =1.23V Anode: Fe(s)  Fe 2+ (aq) + 2e - E˚ red =-0.44V Dissolved oxygen in water usually causes Fe oxidation. Fe 2+ initially formed can be further oxidized to Fe 3+ which forms rust, Fe 2 O 3. xH 2 O(s). Corrosion of Iron

64. CHEM 180/181Chapter 20 Corrosion of Iron

65. CHEM 180/181Chapter 20 Oxidation occurs more readily at the site with the greatest concentration of O 2 (water-air junction). Salts produce the necessary electrolyte required to complete the electrical circuit. Preventing Corrosion of Iron Corrosion can be prevented by coating the iron with paint or another metal. Galvanized iron (commonly used for nails) is coated with a thin layer of zinc. Corrosion of Iron

66. CHEM 180/181Chapter 20 Zinc protects iron since Zn is more easily oxidized (E˚ red Zn is more negative than E˚ red Fe). Anode: Zn 2+ (aq) +2e -  Zn(s) E˚ red = -0.76 V Cathode: Fe 2+ (aq) + 2e -  Fe(s) E˚ red = -0.44 V Zn is oxidized INSTEAD OF IRON, preventing corrosion. Protects even if the surface coat is broken. Corrosion Prevention: Galvanization

67. CHEM 180/181Chapter 20 Preventing Corrosion of Iron Corrosion Prevention: Galvanization

68. CHEM 180/181Chapter 20 To protect underground pipelines. The water pipe is the cathode and a more active metal is used as the anode (cathodic protection). Often, Mg is used as the sacrificial anode: Mg 2+ (aq) +2e -  Mg(s) E˚ red = -2.37 V Fe 2+ (aq) + 2e -  Fe(s) E˚ red = -0.44 V Since E˚ red of Mg is more negative, it will oxidize first. Corrosion Prevention: Sacrificial Anodes

69. CHEM 180/181Chapter 20 Corrosion Prevention: Sacrificial Anodes

70. CHEM 180/181Chapter 20 Defn: Use of electrical energy to drive a non-spontaneous redox reaction. 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. Anode is now designated as POSITIVE(+). Electrolysis

71. CHEM 180/181Chapter 20 Used industrially to produce metals such as Na and Al. Requires high temperatures. Example: Electrolysis of molten NaCl: NaCl(s)  Na + (l) + Cl - (l) Two reactants: Na + and Cl - Cathode: 2Na + (l) + 2e -  2Na(l) E˚ red = -2.71V Anode: 2Cl - (l)  Cl 2 (g) + 2e - E˚ red = - 1.359V This is clearly a non-spontaneous reaction. E˚ cell = -2.71- (1.359) = -4.069V (which is a + ΔG˚ cell ) Electrolysis of Molten Salts

72. CHEM 180/181Chapter 20 Electrolysis of Molten Solutions Electrons still flow from anode to cathode, but are forced.

73. CHEM 180/181Chapter 20 Aqueous solution electrolysis is complicated by water. Ask: Is water oxidized (to form O 2 ) or reduced (to form H 2 ) in reference to the other components? e.g. For an aqueous solution of NaF in an electrolytic cell, possible reactants are Na +, F - and H 2 O. Both Na + and H 2 O can be reduced but not F -. Thus possible reactions at cathode are: Na + (aq) + e - Na(s) E˚ red = -2.71 V 2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) E˚ red = -0.83 V Electrolysis of Aqueous Solutions

74. CHEM 180/181Chapter 20 More positive the E˚ red favors reduction reactions. Reduction of H 2 O (E˚ red = -0.83V) occurs at cathode with H 2 gas produced. At anode, either F - or H 2 O must be oxidized since Na + cannot lose additional electrons. 2F - (aq) F 2 (g) + 2e - E˚ red = +2.87 V 4OH - (aq) O 2 (g) + 2H 2 O(l) + 4e - E˚ red = +0.40 Thus oxidation of H 2 O occurs at anode (more negative E˚ red favors oxidation reactions). Electrolysis of Aqueous Solutions

75. CHEM 180/181Chapter 20 Cathode 4H 2 O(l) + 4e - 2H 2 (g) + 4OH - (aq) E˚ red = -0.83 V Anode 4OH - (aq) O 2 (g) + 2H 2 O(l) + 4e - E˚ red = - 0.40 V 2H 2 O(l) 2H 2 (g) + O 2 (g) E˚ cell = -1.23 V Process is non-spontaneous (as expected). Electrolysis of Aqueous Solutions Please note – sign on anode half-reaction!!

76. CHEM 180/181Chapter 20 Active electrodes - take part in electrolysis. Example: electrolytic plating. Defn: Electrolysis used to deposit a thin layer of one metal on another in order to improve beauty or resistance to corrosion. e.g. electroplating nickel on a piece of steel. Electroplating

77. CHEM 180/181Chapter 20 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) With voltage supplied, Ni plates on the steel electrode. Electroplating is important in protecting objects from corrosion (chrome). Electroplating E˚cell = 0V, so outside source of voltage needed!

78. CHEM 180/181Chapter 20 How much material can we obtain with electrolysis? Consider the reduction of Cu 2+ to Cu(s). –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 (=1F). –Using Q = It The amount of Cu can be calculated from the current used (I) and time (t) allowed to plate. Quantitative Aspects of Electrolysis

79. CHEM 180/181Chapter 20 Example Q = It Q = (10.0 A)(3600 s) = 3.60x10 4 C Moles e - = CF = (3.60x10 4 C)(1 mol e - / 96,500 C) = 0.373 mol e - Al 3+ + 3e - Al Moles Al = (0.373 mol e - )(1 mol Al / 3 mol e - ) = 0.124 mol Al Grams Al = (0.124 mol Al)(27.0 g / mol) = 3.36 g Al Ans. 3.36g Calculate the number of grams of aluminum produced in 1.00h by the electrolysis of molten AlCl 3 if the electrical current is 10.0 A.

80. CHEM 180/181Chapter 20 Free-energy is a measure of the maximum amount of useful work that can be obtained from a system. We know If work is negative, then work is performed by the system and E is positive (which means it’s spontaneous) Electrical Work

81. CHEM 180/181Chapter 20 The emf can be thought of as a measure of the driving force for a redox reaction to proceed. In order to drive non-spontaneous, electrolytic reactions, the external, applied emf must be greater than E cell. If E cell = -1.35 V, then external emf must be at least +1.36 V. Note: work can be expressed in Watts (W) 1 W = 1 J/s. Electrical Work

82. CHEM 180/181Chapter 20 Batteries & Corrosion: 20.66, 20.68, 20.71, 20.72 Recommended Problems DO as many problems as possible!! Not just these… Redox reactions: 20.3, 20.7, 20.9 Voltaic Cells: 20.13, 20.21, 20.23 Electrolysis: 20.75, 20.79

83. CHEM 180/181Chapter 20 End of Chapter 20 Electrochemistry