1. Chapter 18 Electrochemistry Lesson 1

2. Electrochemistry 18.1Balancing Oxidation–Reduction Reactions 18.2 Galvanic Cells 18.3 Standard Reduction Potentials 18.4Cell Potential, Electrical Work, and Free Energy 18.5Dependence of Cell Potential on Concentration 18.6Batteries 18.7Corrosion 18.8Electrolysis 18.9Commercial Electrolytic Processes

3. Electrochemical Reactions In electrochemical reactions, electrons are transferred from one species to another.

4. Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

5. Oxidation and Reduction A species is oxidized when it loses electrons. –Here, zinc loses two electrons to go from neutral zinc metal to the Zn 2+ ion.

6. Oxidation and Reduction A species is reduced when it gains electrons. –Here, each of the H + gains an electron and they combine to form H 2.

7. Oxidation and Reduction What is reduced is the oxidizing agent. –H + oxidizes Zn by taking electrons from it. What is oxidized is the reducing agent. –Zn reduces H + by giving it electrons.

8. Assigning Oxidation Numbers 1.Elements in their elemental form have an oxidation number of 0. 2.The oxidation number of a monatomic ion is the same as its charge.

9. Assigning Oxidation Numbers 3.Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. –Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1. –Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.

10. Assigning Oxidation Numbers 3.Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. –Fluorine always has an oxidation number of −1. –The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.

11. Assigning Oxidation Numbers 4.The sum of the oxidation numbers in a neutral compound is 0. 5.The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

12. Balancing Oxidation-Reduction Equations Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.

13. Balancing Oxidation-Reduction Equations This involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.

14. Half-Reaction Method 1.Assign oxidation numbers to determine what is oxidized and what is reduced. 2.Write the oxidation and reduction half- reactions.

15. Half-Reaction Method 3.Balance each half-reaction. a.Balance elements other than H and O. b.Balance O by adding H 2 O. c.Balance H by adding H +. d.Balance charge by adding electrons. 4.Multiply the half-reactions by integers so that the electrons gained and lost are the same.

16. Half-Reaction Method 5.Add the half-reactions, subtracting things that appear on both sides. 6.Make sure the equation is balanced according to mass. 7.Make sure the equation is balanced according to charge.

17. Half-Reaction Method Consider the reaction between MnO 4 − and C 2 O 4 2− : MnO 4 − (aq) + C 2 O 4 2− (aq)  Mn 2+ (aq) + CO 2 (aq)

18. Half-Reaction Method First, we assign oxidation numbers. MnO 4 − + C 2 O 4 2-  Mn 2+ + CO 2 +7+3+4+2 Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized.

19. Oxidation Half-Reaction C 2 O 4 2−  CO 2 To balance the carbon, we add a coefficient of 2: C 2 O 4 2−  2 CO 2

20. Oxidation Half-Reaction C 2 O 4 2−  2 CO 2 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side. C 2 O 4 2−  2 CO 2 + 2 e −

21. Reduction Half-Reaction MnO 4 −  Mn 2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side. MnO 4 −  Mn 2+ + 4 H 2 O

22. Reduction Half-Reaction MnO 4 −  Mn 2+ + 4 H 2 O To balance the hydrogen, we add 8 H + to the left side. 8 H + + MnO 4 −  Mn 2+ + 4 H 2 O

23. Reduction Half-Reaction 8 H + + MnO 4 −  Mn 2+ + 4 H 2 O To balance the charge, we add 5 e − to the left side. 5 e − + 8 H + + MnO 4 −  Mn 2+ + 4 H 2 O

24. Combining the Half-Reactions Now we evaluate the two half-reactions together: C 2 O 4 2−  2 CO 2 + 2 e − 5 e − + 8 H + + MnO 4 −  Mn 2+ + 4 H 2 O To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.

25. Combining the Half-Reactions 5 C 2 O 4 2−  10 CO 2 + 10 e − 10 e − + 16 H + + 2 MnO 4 −  2 Mn 2+ + 8 H 2 O When we add these together, we get: 10 e − + 16 H + + 2 MnO 4 − + 5 C 2 O 4 2−  2 Mn 2+ + 8 H 2 O + 10 CO 2 +10 e −

26. Combining the Half-Reactions 10 e − + 16 H + + 2 MnO 4 − + 5 C 2 O 4 2−  2 Mn 2+ + 8 H 2 O + 10 CO 2 +10 e − The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16 H + + 2 MnO 4 − + 5 C 2 O 4 2−  2 Mn 2+ + 8 H 2 O + 10 CO 2

27. The Half–Reaction Method for Balancing Redox Equations in Basic Solution 1.Use the half–reaction method as specified for acidic solutions to obtain the final balanced equation as if H + ions were present. 2.To both sides of the equation, add a number of OH – ions that is equal to the number of H + ions present. (You want to eliminate H + by turning is into H 2 O) 3.Form H 2 O on the side containing both H + and OH – ions, and eliminate the number of H 2 O molecules that appear on both sides of the equation. 4.Check that elements and charges are balanced.

28. Balancing in Basic Solution 16 H + + 2 MnO 4 − + 5 C 2 O 4 2−  2 Mn 2+ + 8 H 2 O + 10 CO 2 How would our last answer look in basic solution?

29. Galvanic Cell A device in which chemical energy is converted to electrical energy. It uses a spontaneous redox reaction to produce a current that can be used to generate energy or to do work.

30. A Galvanic Cell

31. In Galvanic Cell: Oxidation occurs at the anode. Reduction occurs at the cathode. Salt bridge or porous disk allows ions to flow without extensive mixing of the solutions.  Salt bridge – contains a strong electrolyte held in a gel–like matrix.  Porous disk – contains tiny passages that allow hindered flow of ions.

32. Galvanic Cell (Voltaic Cell)

33. Voltaic Cells In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

34. Voltaic Cells We can use that energy to do work if we make the electrons flow through an external device. We call such a setup a voltaic cell (Galvanic Cells).

35. Voltaic Cells A typical cell looks like this. The oxidation occurs at the anode. The reduction occurs at the cathode.

36. Voltaic Cells

37. As one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

38. Voltaic Cells Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. –Cations move toward the cathode. –Anions move toward the anode.

39. Voltaic Cells In the cell, then, electrons leave the anode and flow through the wire to the cathode. As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

40. Voltaic Cells As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

41. Electromotive Force (emf) Water only spontaneously flows one way in a waterfall. Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

42. Electromotive Force (emf) The potential difference between the anode and cathode in a cell is called the electromotive force (emf). It is also called the cell potential, and is designated E cell.

43. Cell Potential Cell potential is measured in volts (V). 1 V = 1 JCJC

44. Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated (Table 18.1 on p. 845).

45. Standard Hydrogen Electrode Their values are referenced to a standard hydrogen electrode (SHE). By definition, the reduction potential for hydrogen is 0 V: 2 H + (aq, 1M) + 2 e −  H 2 (g, 1 atm)

46. Standard Cell Potentials The cell potential at standard conditions can be found through this equation: E cell  = E red (cathode) − E red (anode)  Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

47. Cell Potentials For the oxidation in this cell, For the reduction, E red = −0.76 V  E red = +0.34 V 

48. Cell Potentials E cell  = E red  (cathode) − E red  (anode) = +0.34 V − (−0.76 V) = +1.10 V

49. Oxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials.

50. Standard Reduction Potentials Values of E ° in standard table of half-cell potentials are for reduction half-cell reactions  Table 18.1  1 M, 1 atm, 25°C When a half-reaction is reversed, the algebraic sign of E ° changes. When a half-reaction is multiplied by an integer, the value of E ° remains the same. A galvanic cell runs spontaneously in the direction that gives a positive value for E ° cell

51. Example: Fe 3+ (aq) + Cu (s)  Cu 2+ (aq) + Fe 2+ (aq) Half-Reactions:  Fe 3+ + e –  Fe 2+ E ° = 0.77 V  Cu 2+ + 2e –  Cu E ° = 0.34 V To balance the cell reaction and calculate the cell potential, we must reverse reaction 2.  Cu  Cu 2+ + 2e – – E ° = – 0.34 V Each Cu atom produces two electrons but each Fe 3+ ion accepts only one electron, therefore reaction 1 must be multiplied by 2.  2Fe 3+ + 2e –  2Fe 2+ E ° = 0.77 V

52. Standard Cell Potential 2Fe 3+ + 2e –  2Fe 2+ ; E ° = 0.77 V (cathode) Cu  Cu 2+ + 2e – ; – E ° = – 0.34 V (anode) Balanced Cell Reaction: Cu + 2Fe 3+  Cu 2+ + 2Fe 2+ Cell Potential: E E ° cell = E °(cathode) – E °(anode) E ° cell = 0.77 V – 0.34 V = 0.43 V

53. Cell Notations for Galvanic Cells Used to describe electrochemical cells. Anode components are listed on the left. Cathode components are listed on the right. Separated by double vertical lines. The concentration of aqueous solutions should be specified in the notation when known. Example: Mg (s) | Mg 2+ (aq) || Al 3+ (aq) | Al (s)  Mg  Mg 2+ + 2e – (anode)  Al 3+ + 3e –  Al(cathode)

54. Description of a Galvanic Cell The cell potential is always positive for a galvanic cell, where E ° cell = E ° (cathode half-cell) – E ° (anode half-cell) (as given in the table of half-cell potentials) Anode is the negative terminal and cathode is the positive terminal. Electron flows from the anode (-) to cathode (+). Current flows from cathode(+) to anode(-). Positive ions flows from anode to cathode half-cells and negative ions flows from cathode to anode though the “salt bridge”.

55. Cell Potential, Free Energy, and Electrical Work Maximum cell potential and free energy Directly related to the free energy difference between the reactants and the products in the cell reaction.  ΔG° = –nF E °  F = 96,485 C/mol e – Work: In any real, spontaneous process some energy is always lost (wasted) – the actual work obtained is always less than the calculated (maximum) value.

56. There are two kinds electrochemical cells. Electrochemical cells containing nonspontaneous chemical reactions are called electrolytic cells. Electrochemical cells containing spontaneous chemical reactions are called voltaic or galvanic cells. 1.Galvanic cell (voltaic cell)— energy released during a spontaneous reaction (  G < 0) generates electricity 2.Electrolytic cell—consumes electrical energy from an external source to cause a nonspontaneous redox reaction to occur (  G > 0) 3.The cathode is negative in electrolytic cells and positive in voltaic cells. 4.The anode is positive in electrolytic cells and negative in voltaic cells. Electrochemistry