1. OBJECTIVES Eiectrochemistry is the study of production of electricity from the energy released during a spontaneous reaction and use of electrical energy to Bring about non spontaneous chemical reaction. How Commercially generate electricity It provides many ideas for protection of metal surface from corrosion

2. ELECTROCHEMISTRY CLASS-12 BY P.K. CHOURASIA K,V MANDLA MP

3. Chapter 18Slide 3 Redox Reactions02

4. Chapter 18Slide 4 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

5. Chapter 18Slide 5 Redox Reactions01

6. 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. Chapter 18Slide 7 Oxidation Half-Reaction: Zn(s)  Zn 2+ (aq) + 2 e –. The Zn loses two electrons to form Zn 2+. Redox Reactions02

8. 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. 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) Electrons flow from anode to cathode. Anode is placed on left by convention.

10. Chapter 18Slide 10 Electrochemical Cells Overall: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

11. Chapter 18Slide 11 Electrochemical Cells03 The standard potential of any galvanic cell is the subtraction of the standard half-cell potentials for the oxidation and reduction half-cells. E° cell = E° reduction - E° oxidation Standard half-cell potentials are always quoted as a reduction process. The sign must be changed for the oxidation process.

12. Chapter 18Slide 12 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.

13. Chapter 18Slide 13 Electrochemical Cells05

14. Chapter 18Slide 14 Electrochemical Cells06

15. Chapter 18Slide 15 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)

16. Chapter 18Slide 16 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?

17. Chapter 18Slide 17 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

18. Chapter 18Slide 18 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)

19. Chapter 18Slide 19 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)

20. Chapter 18Slide 20 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

21. Chapter 18Slide 21 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

22. Chapter 18Slide 22 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 –

23. Chapter 18Slide 23 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.

24. Chapter 18Slide 24 The Nernst Equation07 The overall potential is given by: Which rearranges to give an equation for the determination of pH:

25. Chapter 18Slide 25 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

26. Chapter 18Slide 26 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.

27. Chapter 18Slide 27 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?

28. Chapter 18Slide 28 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.

29. Chapter 18Slide 29 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

30. Chapter 18Slide 30 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.

31. Chapter 18Slide 31 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.

32. Chapter 18Slide 32 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.

33. Chapter 18Slide 33 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

34. Chapter 18Slide 34 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.

35. Chapter 18Slide 35 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

36. Chapter 18Slide 36 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.

37. Chapter 18Slide 37 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.

38. Chapter 18Slide 38 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.

39. Chapter 18Slide 39 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)

40. Chapter 18Slide 40 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).

41. Chapter 18Slide 41 Corrosion04 Cathodic Protection with Zinc Layer

42. Chapter 18Slide 42 Corrosion05 Cathodic Protection with Magnesium Anode

43. Chapter 18Slide 43 Electrolysis01

44. Chapter 18Slide 44 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

45. Chapter 18Slide 45 Overall: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Redox Reactions04

46. Chapter 18Slide 46 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)

47. Chapter 18Slide 47 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.

48. Chapter 18Slide 48 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.

49. Chapter 18Slide 49 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.

50. Chapter 18Slide 50 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:

51. Chapter 18Slide 51 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?

52. Chapter 18Slide 52 Electrolysis Applications01 Manufacture of Sodium (Downs Cell):

53. Chapter 18Slide 53 Electrolysis Applications02 Manufacture of Cl 2 and NaOH (Chlor–Alkali):

54. Chapter 18Slide 54 Electrolysis Applications03 Manufacture of Aluminum (Hall–Heroult):

55. Chapter 18Slide 55 Electrolysis Applications04 Electrorefining and Electroplating:

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