1. Dr. Paul Charlesworth Michigan Technological University Dr. Paul Charlesworth Michigan Technological University C h a p t e rC h a p t e r C h a p t e rC h a p t e r 18 Electrochemistry Chemistry 4th Edition McMurry/Fay Chemistry 4th Edition McMurry/Fay

2. Prentice Hall ©2004 Chapter 18Slide 2 Redox Reactions01

3. Prentice Hall ©2004 Chapter 18Slide 3 Redox reaction are those involving the oxidation and reduction of species. OIL – Oxidation Is Loss of electrons. RIG – Reduction Is Gain of electrons. Oxidation and reduction must occur together. They cannot exist alone. Redox Reactions01

4. Prentice Hall ©2004 Chapter 18Slide 4 Redox Reactions02

5. Prentice Hall ©2004 Chapter 18Slide 5 Oxidation Half-Reaction: Zn(s)  Zn 2+ (aq) + 2 e –. The Zn loses two electrons to form Zn 2+. Redox Reactions02

6. Prentice Hall ©2004 Chapter 18Slide 6 Redox Reactions03 Reduction Half-Reaction: Cu 2+ (aq) + 2 e –  Cu(s) The Cu 2+ gains two electrons to form copper.

7. Prentice Hall ©2004 Chapter 18Slide 7 Overall: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Redox Reactions04

8. Prentice Hall ©2004 Chapter 18Slide 8 Electrochemical Cells01 Electrodes: are usually metal strips/wires connected by an electrically conducting wire. Salt Bridge: is a U-shaped tube that contains a gel permeated with a solution of an inert electrolyte. Anode: is the electrode where oxidation takes place. Cathode: is the electrode where reduction takes place.

9. Prentice Hall ©2004 Chapter 18Slide 9 Electrochemical Cells02 Convention for expressing the cell: Anode Half-Cell || Cathode Half-Cell Electrode | Anode Soln || Cathode Soln | Electrode Zn(s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu(s) Pt(s) | H 2 (1 atm) | H + (1 M) || Fe 3+ (aq), Fe 2+ (aq) | Pt(s) Electrons flow from anode to cathode. Anode is placed on left by convention.

10. Prentice Hall ©2004 Chapter 18Slide 10 Electrochemical Cells03 The standard potential of any galvanic cell is the sum of the standard half-cell potentials for the oxidation and reduction half-cells. E° cell = E° oxidation + E° reduction Standard half-cell potentials are always quoted as a reduction process. The sign must be changed for the oxidation process.

11. Prentice Hall ©2004 Chapter 18Slide 11 Electrochemical Cells04 The standard half-cell potentials are determined from the difference between two electrodes. The reference point is called the standard hydrogen electrode (S.H.E.) and consists of a platinum electrode in contact with H 2 gas (1 atm) and aqueous H + ions (1 M). The standard hydrogen electrode is assigned an arbitrary value of exactly 0.00 V.

12. Prentice Hall ©2004 Chapter 18Slide 12 Electrochemical Cells05

13. Prentice Hall ©2004 Chapter 18Slide 13 Electrochemical Cells06

14. Prentice Hall ©2004 Chapter 18Slide 14 Electrochemical Cells07 When selecting two half-cell reactions the more negative value will form the oxidation half-cell. Consider the reaction between zinc and silver: Ag + (aq) + e –  Ag(s)E° = 0.80 V Zn 2+ (aq) + 2 e –  Zn(s)E° = – 0.76 V Therefore, zinc forms the oxidation half-cell: Zn(s)  Zn 2+ (aq) + 2 e – E° = – (–0.76 V)

15. Prentice Hall ©2004 Chapter 18Slide 15 Electrochemical Cells08 What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 2 solution? What is the standard emf of an electrochemical cell made of a Mg electrode in a 1.0 M Mg(NO 3 ) 2 solution and a Ag electrode in a 1.0 M AgNO 3 solution?

16. Prentice Hall ©2004 Chapter 18Slide 16 Spontaneity of a Reaction01 The value of E˚ cell is related to the thermodynamic quantities of ∆G˚ and K. The value of E˚ cell is related to ∆G˚ by: ∆G˚ = –nFE˚ cell The value of K is related to ∆G˚ by: ∆G˚ = –RT ln K

17. Prentice Hall ©2004 Chapter 18Slide 17 Spontaneity of a Reaction02 These are combined: –nFE˚ cell = –RT ln K and rearranged at 25°C to give: E˚ cell = (0.0257  n) ln K or E˚ cell = (0.0592  n) log K (Units = volts, n = moles, F = 96,500 coulombs)

18. Prentice Hall ©2004 Chapter 18Slide 18 Spontaneity of a Reaction03 Calculate the standard free energy change (∆G˚) and the equilibrium constant (K) for the following reactions at 25°C: 1. Sn(s) + 2 Cu 2+ (aq) æ Sn 2+ (aq) + 2 Cu + (aq) 2. Fe 2+ (aq) + 2 Ag(s) æ Fe(s) + 2 Ag + (aq) 3. 4 Fe 2+ (aq) + O 2 (g) + 4 H + (aq)  4 Fe 3+ (aq) + 2 H 2 O (l)

19. Prentice Hall ©2004 Chapter 18Slide 19 Cell potentials can be modified by temperature and composition changes according to the equation: ∆G = ∆G° + RT lnQ  –nFE = –nFE° + RT lnQ Giving: The Nernst Equation01

20. Prentice Hall ©2004 Chapter 18Slide 20 Consider the reaction of metallic zinc with hydrochloric acid. Calculate the cell potential at 25°C when [H + ] = 1.0 M, [Zn 2+ ] = 0.0010 M, and P H 2 = 0.10 atm. Consider the reaction of metallic copper with iron(III) to give copper(II) and iron(II). What is the potential of a cell when [Fe 3+ ] = 0.0001 M, [Cu 2+ ] = 0.25 M, and [Fe 2+ ] = 0.20 M? The Nernst Equation03

21. Prentice Hall ©2004 Chapter 18Slide 21 The Nernst Equation04 A particularly important use of the Nernst equation is in the electrochemical determination of pH. Pt | H 2 (1 atm) | H + (? M) || Reference Cathode E cell = E H 2  H + + E ref The Nernst equation can be applied to the half-reaction: H 2 (g)  2 H + (aq) + 2 e –

22. Prentice Hall ©2004 Chapter 18Slide 22 The Nernst Equation05 For the half-reaction: H 2 (g)  2 H + (aq) + 2 e – E° = 0 V for this reaction (standard hydrogen electrode). P H 2 is 1 atm.

23. Prentice Hall ©2004 Chapter 18Slide 23 The Nernst Equation07 The overall potential is given by: Which rearranges to give an equation for the determination of pH:

24. Prentice Hall ©2004 Chapter 18Slide 24 The Nernst Equation08 A higher cell potential indicates a higher pH, therefore we can measure pH by measuring E cell. A glass electrode (Ag/AgCl wire in dilute HCl) with a calomel reference is the most common arrangement. Glass: Ag(s) + Cl – (aq)  AgCl(s) + e – E° = –0.22 V Calomel: Hg 2 Cl 2 (s) + 2 e –  2 Hg(l) + 2 Cl – (aq) E° = 0.28 V

25. Prentice Hall ©2004 Chapter 18Slide 25 The Nernst Equation09 The glass pH probe is constructed as follows: Ag(s) | AgCl(s) | HCl(aq) | glass | H + (aq) || reference The difference in [H + ] from one side of the glass membrane to the other causes a potential to develop, which adds to the measured E cell.

26. Prentice Hall ©2004 Chapter 18Slide 26 The Nernst Equation10 The following cell has a potential of 0.55 V at 25°C: Pt(s) | H 2 (1 atm) | H + (? M) || Cl – (1 M) | Hg 2 Cl 2 (s) | Hg(l) What is the pH of the solution at the anode? The following cell has a potential of 0.28 V at 25°C: Pt(s) | H 2 (1 atm) | H + (? M) || Pb 2+ (1 M) | Pb(s) What is the pH of the solution at the anode?

27. Prentice Hall ©2004 Chapter 18Slide 27 Batteries01 Batteries are the most important practical application of galvanic cells. Single-cell batteries consist of one galvanic cell. Multicell batteries consist of several galvanic cells linked in series to obtain the desired voltage.

28. Prentice Hall ©2004 Chapter 18Slide 28 Batteries02 Lead Storage Battery: A typical 12 volt battery consists of six individual cells connected in series. Anode: Lead grid packed with spongy lead. Pb(s) + HSO 4 – (aq)  PbSO 4 (s) + H + (aq) + 2 e – Cathode: Lead grid packed with lead oxide. PbO 2 (s) + 3 H + (aq) + HSO 4 – (aq) + 2 e –  PbSO 4 (s) + 2 H 2 O(l) Electrolyte: 38% by mass sulfuric acid. Cell Potential: 1.924 V

29. Prentice Hall ©2004 Chapter 18Slide 29 Batteries03 Zinc Dry-Cell: Also called a Leclanché cell, uses a viscous paste rather than a liquid solution. Anode: Zinc metal can on outside of cell. Zn(s)  Zn 2+ (aq) + 2 e – Cathode: MnO 2 and carbon black paste on graphite. 2 MnO 2 (s) + 2 NH 4 + (aq) + 2 e –  Mn 2 O 3 (s) + 2 NH 3 (aq) + 2 H 2 O(l) Electrolyte: NH 4 Cl and ZnCl 2 paste. Cell Potential: 1.5 V but deteriorates to 0.8 V with use.

30. Prentice Hall ©2004 Chapter 18Slide 30 Batteries04 Alkaline Dry-Cell: Modified Leclanché cell which replaces NH 4 Cl with NaOH or KOH. Anode: Zinc metal can on outside of cell. Zn(s) + 2 OH – (aq)  ZnO(s) + H 2 O(l) + 2 e – Cathode: MnO 2 and carbon black paste on graphite. 2 MnO 2 (s) + H 2 O(l) + 2 e –  Mn 2 O 3 (s) + 2 OH – (aq) Electrolyte: NaOH or KOH, and Zn(OH) 2 paste. Cell Potential: 1.5 V but longer lasting, higher power, and more stable current and voltage.

31. Prentice Hall ©2004 Chapter 18Slide 31 Batteries05 Mercury Dry-Cell: Modified Leclanché cell which replaces MnO 2 with HgO and uses a steel cathode. Anode: Zinc metal can on outside of cell. Zn(s) + 2 OH – (aq)  ZnO(s) + H 2 O(l) + 2 e – Cathode: HgO in contact with steel. 2 HgO (s) + H 2 O(l) + 2 e –  Hg(l) + 2 OH – (aq) Electrolyte: KOH, and Zn(OH) 2 paste. Cell Potential: 1.3 V with small size, longer lasting, and stable current and voltage.

32. Prentice Hall ©2004 Chapter 18Slide 32 Batteries06 Nickel–Cadmium Battery: Modified Leclanché cell which is rechargeable. Anode: Cadmium metal. Cd(s) + 2 OH – (aq)  Cd(OH) 2 (s) + 2 e – Cathode: Nickel(III) compound on nickel metal. NiO(OH) (s) + H 2 O(l) + e –  Ni(OH) 2 (s) + OH – (aq) Electrolyte: Nickel oxyhydroxide, NiO(OH). Cell Potential: 1.30 V

33. Prentice Hall ©2004 Chapter 18Slide 33 Batteries07 Nickel–Metal–Hydride (NiMH): Replaces toxic Cd anode with a hydrogen atom impregnated ZrNi 2 metal alloy. During oxidation at the anode, hydrogen atoms are released as H 2 O. Recharging reverses this reaction.

34. Prentice Hall ©2004 Chapter 18Slide 34 Batteries08 Lithium Ion (Li–ion): The newest rechargeable battery is based on the migration of Li + ions. Anode: Li metal, or Li atom impregnated graphite. Li(s)  Li + + e – Cathode: Metal oxide or sulfide that can accept Li +. MnO 2 (s) + Li + (aq) + e –  LiMnO 2 (s) Electrolyte: Lithium-containing salt such as LiClO 4, in organic solvent. Solid state polymers can also be used. Cell Potential: 3.0 V

35. Prentice Hall ©2004 Chapter 18Slide 35 Batteries09 Fuel Cell: Uses externally fed CH 4 or H 2, which react to form water. Most common is H 2. Anode: Porous carbon containing metallic catalysts. 2 H 2 (s) + 4 OH – (aq)  4 H 2 O(l) + 4 e – Cathode: Porous carbon containing metallic catalysts. O 2 (s) + 2 H 2 O(l) + 4 e –  4 OH – (aq) Electrolyte: Hot aqueous KOH solution. Cell Potential: 1.23 V, but is only 40% of cell capacity.

36. Prentice Hall ©2004 Chapter 18Slide 36 Batteries10 Fuel cells are not batteries because they are not self- contained. Fuel cells typically have about 40% conversion to electricity; the remainder is lost as heat. Excess heat can be used to drive turbine generators.

37. Prentice Hall ©2004 Chapter 18Slide 37 Corrosion01 Corrosion is the oxidative deterioration of metal. 25% of steel produced in USA goes to replace steel structures and products destroyed by corrosion. Rusting of iron requires the presence of BOTH oxygen and water. Rusting results from tiny galvanic cells formed by water droplets.

38. Prentice Hall ©2004 Chapter 18Slide 38 Corrosion02 Oxidation:Fe(s)  Fe 2+ (aq) + 2 e – Reduction:O 2 (g) + 4 H + (aq) + 4 e –  2 H 2 O(l) Overall: 2 Fe(s) + O 2 (g) + 4 H + (aq)  2 Fe 2+ (aq) + 2 H 2 O(l)

39. Prentice Hall ©2004 Chapter 18Slide 39 Corrosion03 Galvanizing: is the coating of iron with zinc. Zinc is more easily oxidized than iron, which protects and reverses oxidation of the iron. Cathodic Protection: is the protection of a metal from corrosion by connecting it to a metal (a sacrificial anode) that is more easily oxidized. All that is required is an electrical connection to the sacrificial anode (usually magnesium or zinc).

40. Prentice Hall ©2004 Chapter 18Slide 40 Corrosion04 Cathodic Protection with Zinc Layer

41. Prentice Hall ©2004 Chapter 18Slide 41 Corrosion05 Cathodic Protection with Magnesium Anode

42. Prentice Hall ©2004 Chapter 18Slide 42 Electrolysis01

43. Prentice Hall ©2004 Chapter 18Slide 43 Electrolysis: is the process in which electrical energy is used to drive a nonspontaneous chemical reaction. An electrolytic cell is an apparatus for carrying out electrolysis. Processes in an electrolytic cell are the reverse of those in a galvanic cell. Electrolysis01

44. Prentice Hall ©2004 Chapter 18Slide 44 Electrolysis05 Electrolysis of Water: Requires an electrolyte species, that is less easily oxidized and reduced than water, to carry the current. Anode: Water is oxidized to oxygen gas. 2 H 2 O(l)  O 2 (g) + 4 H + (aq) + 4 e – Cathode: Water is reduced to hydrogen gas. 4 H 2 O(l) + 4 e –  2 H 2 (g) + 4 OH – (aq)

45. Prentice Hall ©2004 Chapter 18Slide 45 Electrolysis02 Electrolysis of Molten Sodium Chloride:Electrolysis of Aqueous Sodium Chloride: uses different processes to molten sodium chloride. Based on cell potentials, water (–0.83 V) would be preferentially reduced over sodium ions (–2.71 V). Based on cell potentials, water (+1.23 V) should be preferentially oxidized over chloride ions (+1.36 V). The observed product at the anode is Cl 2, not O 2, because of a phenomenon called overvoltage.

46. Prentice Hall ©2004 Chapter 18Slide 46 Electrolysis04 Overvoltage: Additional voltage needed to maintain the rate of electron transfer at the electrode–solution interface. Overvoltage is required when a half-reaction has a significant activation energy, and so a slow rate. Overvoltage for formation of O 2 or H 2 is much greater than for formation of Cl 2.

47. Prentice Hall ©2004 Chapter 18Slide 47 Electrolysis07 Quantitative Electrolysis: The amount of substance produced at an electrode by electrolysis depends on the quantity of charge passed through the cell. Reduction of 1 mol of sodium ions requires 1 mol of electrons to pass through the system. The charge on 1 mol of electrons is 96,500 coulombs.

48. Prentice Hall ©2004 Chapter 18Slide 48 Electrolysis08 To determine the moles of electrons passed, we measure the current and time that the current flows: Charge (C) = Current (A) x Time (s) Because the charge on 1 mol of e – is 96,500 C, the number of moles of e – passed through the cell is:

49. Prentice Hall ©2004 Chapter 18Slide 49 Electrolysis09 A constant current of 30.0 A is passed through an aqueous solution of NaCl for a time of 1.00 hour. How many grams of Na and how many liters of Cl 2 gas at STP are produced? How many amperes must be passed through a Downs cell to produce sodium metal at a rate of 30.0 kg/hour?

50. Prentice Hall ©2004 Chapter 18Slide 50 Electrolysis Applications01 Manufacture of Sodium (Downs Cell):

51. Prentice Hall ©2004 Chapter 18Slide 51 Electrolysis Applications02 Manufacture of Cl 2 and NaOH (Chlor–Alkali):

52. Prentice Hall ©2004 Chapter 18Slide 52 Electrolysis Applications03 Manufacture of Aluminum (Hall–Heroult):

53. Prentice Hall ©2004 Chapter 18Slide 53 Electrolysis Applications04 Electrorefining and Electroplating: