1. Chapter 20 Electrochemistry and Oxidation-Reduction

2. Overview 4 Electrochemistry deals with chemical changes produced by an electric current and with the production of electricity by chemical reactions.

3. 4 Energy changes are measured electrically whenever possible 4 This allows for very accurate measurements.   G and equilibrium constants can be measured electrochemically. Overview

4. 4 When a strip of magnesium is placed in HCl, a reaction takes place which we have represented as a single replacement reaction: Mg (s) + 2HCl (aq) MgCl 2(aq) + H 2(g) Galvanic Cells and Cell Potential

5. 4 Looking at it as an Oxidation- Reduction reaction it becomes: Mg (s) + 2H 3 O +1 (aq) + 2Cl -1 (aq) H 2(g) + Mg +2 (aq) + 2Cl -1 (aq) + 2H 2 O ( ) Galvanic Cells and Cell Potential

6. 4 This reaction has two halves –the magnesium oxidizing (gaining charge by losing electrons) Mg (s) Mg +2 (aq) +2e -1 –the reduction of hydrogen (losing charge by gaining electrons) 2H 3 O +1 (aq) + 2e -1 H 2(g) + 2H 2 O ( ) Galvanic Cells and Cell Potential

7. 4 The two half-reactions can be run in two separate containers with only a wire (e -1 pathway) and a salt bridge (ion pathway) to connect them. Galvanic Cells and Cell Potential

8. Mg MgCl 2 HCl Salt Bridge e -1 + ions - ions Bubbles of H 2 OxidationReduction

9. 4 This is an example of a galvanic cell (sometimes called a voltage cell). 4 A battery is also such an example. –It converts chemical energy into electrical energy. Galvanic Cells and Cell Potential

10. –The anode is where oxidation takes place. –The cathode is where reduction takes place. –Remember, Anode and oxidation both start with a vowel. Galvanic Cells and Cell Potential

11. Cell Potentials 4 Cell Potential (E cell ) is the “driving force that pushes electrons through the external circuit. 4 E cell is measured in Volts (V) 4 This potential can be measured with a Voltmeter

12. 4 A galvanic cell has a positive voltage when the cell reaction is spontaneous. 4 The cell has a negative voltage when the cell reaction is not spontaneous. Cell Potentials

13. 4 Each half-cell also has a voltage potential. 4 We can predict the potential of a cell by using the half-cell potentials provided in the appendix. Cell Potentials

14. 4 The cell potential is the sum of the half-cell potential of the anode, E ox, and the half-cell potential of the cathode, E red. E cell = E ox + E red Cell Potentials

15. 4 For the reaction: Zn (s) + Cu +2 (aq) Zn +2 (aq) + Cu (s) we can predict the overall potential of the cell by finding the half-cell potentials of each half-cell and adding the two. Cell Potentials

17. Standard Electrode Potentials 4 The term electrode is used to represent a complete half-cell. –Zinc in a solution of zinc sulfate is such an example. –Platinum wire (inert conductor)in a solution of bromine and the bromide ion is another.

18. 4 There is no direct method of determining the potential of such an electrode. 4 To determine the potential, the hydrogen electrode is assigned a potential of Zero, and the other electrodes are compared to this. Standard Electrode Potentials

19. 4 The standard hydrogen electrode is a gas electrode. 4 Such an electrode is set up by bubbling the gas around an inert conductor. Standard Electrode Potentials

20. 4 The net ionic equation for the electrode is: 2H 3 O +1 + 2e -1 H 2 + H 2 O where the hydrogen is being reduced. Standard Electrode Potentials

21. 4 By international standards, the values of electrode potentials are given for the reduction process. 4 Standard reduction potentials, E red, are measured with respect to hydrogen at 25 o C, 1 atm, and 1M solution of ions. Standard Electrode Potentials

22. 4 When a potential is measured for an oxidation process, it can be easily converted. 4 The reduction potential and the oxidation potential of the same electrode have the same absolute value but of opposite sign. Standard Electrode Potentials

23. 4 When a table of standard reduction potentials is constructed (appendix H) from most negative to most positive, the activity series correlates with many chemical properties of the elements. 4 These correlations are: Standard Electrode Potentials

24. –The metals with large negative reduction potentials at the top of the series are good reducing agents in the free state. They are the metals most easily oxidized to their ions by the removal of electrons. Standard Electrode Potentials

25. –The elements with large positive reduction potentials at the bottom of the series are good oxidizing agents when in the oxidized form - that is, when the metals are in the form of ions and the nonmetals are in the elemental state. Standard Electrode Potentials

26. –The reduced form of any element reduces the oxidized form of any element below it. For example, metallic zinc reduces copper(II) ions according to the equation: Zn + Cu +2 Zn +2 + Cu Standard Electrode Potentials

27. Calculation of Cell Potentials 4 The values in Table 20.1 and Appendix H can be sued to determine standard state cell potentials. 4 Two points are important to remember when using these values:

28. –E o values are for reduction half- reactions, and the sign of a reduction potential must be reversed when it is used as a potential for an oxidation half-reaction. –Changing the stoichiometric coefficients of a half-cell equation does not change the value of E o Calculation of Cell Potentials

29. Ex) Write the cell reaction and determine the standard state potential for the cell diagrammed below: Calculation of Cell Potentials

30. Co Solution of Co +2 Solution of Fe +3 and Fe +2 e -1 Anode Cathode Calculation of Cell Potentials

31. 4 The two half-reactions are: Calculation of Cell Potentials

32. 4 For simplicity, a line notation can be used to represent the electrodes in a galvanic cell. 4 For the previous reaction we can represent the cell as follows: Calculation of Cell Potentials e -1

33. Ex) Determine the standard cell potential, and write equations for the half-reactions and the cell reaction for the cell described by the following line notation: Calculation of Cell Potentials e -1

34. 4 The species to the left of the double line are involved with the anode half-reaction: Fe (s) Fe +2 (aq) + 2e -1 Calculation of Cell Potentials

35. 4 The species to the right of the double line are involved in the cathode half-reaction: MnO 4 -1 (aq) + 8H 3 O +1 (aq) + 5e -1 Mn +2 (aq) + 12H 2 O ( ) Calculation of Cell Potentials

36. 4 The total equation would then be: 5Fe (s) + 2MnO 4 -1 (aq) + 16H 3 O +1 (aq) + 10e -1 5Fe +2 (aq) + 2Mn +2 (aq) + 24H 2 O ( ) Calculation of Cell Potentials

37. 4 The standard cell potential would then be: Calculation of Cell Potentials

38. Cell Potential, Electrical Work, and Free Energy 4 Galvanic Cells are sources of energy that can do work. 4 The amount of work depends on the cell potential. 4 Since the work is done on the surroundings, it will have a negative value (-).

39. 4 We can show the relationship between the work available and the cell potential by the following equation: Cell Potential, Electrical Work, and Free Energy

40. –Where n = number of moles of electrons –and F = a constant called a faraday (96.485 kJ/Vmole.  The maximum amount of work available, w max, is  G. Cell Potential, Electrical Work, and Free Energy

41. 4 The equation can then be modified to be: 4 At standard state conditions we have: Cell Potential, Electrical Work, and Free Energy

42. Ex) Calculate the standard free energy change at 25 o C for the reaction: Cd (s) + Pb +2 (aq) Cd +2 (aq) + Pb (s) Cell Potential, Electrical Work, and Free Energy

44. 4 Now we can use the formula:  The negative value for  G means that the reaction is spontaneous. Cell Potential, Electrical Work, and Free Energy

45. The Effect of Concentration on Cell Potentials 4 We know that most reactions do not occur at standard state. 4 We can derive a relationship between the cell potential at nonstandard states.

46. 4 The equation is called the Nernst Equation. The Effect of Concentration on Cell Potentials

47. 4 Where: –R = the gas constant (8.314 J/K) –T = the Kelvin temperature –F = the faraday constant (96.485kJ/Vmol) –n = the number of moles of electrons –Q = the reaction quotient. The Effect of Concentration on Cell Potentials

48. 4 For reactions at 25 o C, we can rearrange the equation to be: –the constant, 0.05916 V, contains the value of RT/F and has been converted to the logarithmic scale. The Effect of Concentration on Cell Potentials

49. Ex) Calculate the potential at 25 o C for the cell The cell reaction and cell potential at standard states is: Cd (s) + Pb +2 (aq) Cd +2 (aq) + Pb (s) The Effect of Concentration on Cell Potentials

50. 4 Since the cell reaction is at 25 o C, we can use the simplified equation: –in this case, n = 2 The Effect of Concentration on Cell Potentials

52. 4 The cell potential decreases from 0.277V at standard state to 0.179V at nonstandard concentrations. The Effect of Concentration on Cell Potentials

53. 4 The magnitude of the cell potential is measure of spontaneity of a reaction. –A decrease in potential indicates a decrease in spontaneity. –Also, an increase in Q indicates a decrease in spontaneity. The Effect of Concentration on Cell Potentials

54. Cell Potential and the Equilibrium Constant  Since  G is related to the equilibrium constant and the cell potential, we can see that the cell potential of an Redox reaction is related to the equilibrium constant, K.

55. 4 The generalized formula for this is: 4 At 25 o C, we can simplify this: Cell Potential and the Equilibrium Constant

56. Ex) Determine the equilibrium constant at 25oC for the reaction: Br 2( ) + 2Cl -1 (aq) 2Br -1 (aq) + Cl 2(g) Cell Potential and the Equilibrium Constant

57. 4 The standard cell potential can be determined as follows: Cell Potential and the Equilibrium Constant

58. 4 We see that two electrons are involved in the reaction so n=2 Cell Potential and the Equilibrium Constant

60. Balancing Oxidation- Reduction Reactions 4 To balance Redox Reactions, we can categorize them into three groups: –Molecular Redox Reactions –Acidic Solutions –Basic Solutions

61. 4 A Molecular Redox Reaction is when two ions in a reaction have a change in their oxidation reactions. 4 One will increase (oxidize) while the other will decrease (reduce) in Oxidation. Balancing Oxidation- Reduction Reactions

62. –I will use the following example and show step-by-step how to balance the equation: Sn + HNO 3 SnO 2 + NO 2 + H 2 O Balancing Oxidation- Reduction Reactions

63. Step 1: Assign oxidation numbers to each element to identify the elements being oxidized and those being reduced: Sn + HNO 3 SnO 2 + NO 2 + H 2 O Balancing Oxidation- Reduction Reactions 0 +1 +5 -2+4-2+4-2+1-2

64. Step 2: Now write two equations using only the elements that change in ON. Then add electrons to bring the equation into electrical balance: Sn 0 Sn +4 + 4 e -1 oxidation N +5 + 1 e -1 N +4 reduction Balancing Oxidation- Reduction Reactions

65. Step 3: Multiply the two equations by the smallest whole numbers to cancel out the electrons: 1(Sn 0 Sn +4 + 4 e -1 )oxidation (N +5 + 1 e -1 N +4 )4reduction Balancing Oxidation- Reduction Reactions

66. Step 4: Transfer the coefficient in front of each substance in the half- reactions to the corresponding substance in the original equation: Sn + 4HNO 3 SnO 2 + 4NO 2 + H 2 O Balancing Oxidation- Reduction Reactions

67. Step 5: Balance the remaining elements which were not oxidized or reduced in the usual manner: Sn + 4HNO 3 SnO 2 + 4NO 2 + 2H 2 O (balanced Balancing Oxidation- Reduction Reactions

68. 4 An Ionic Redox Reaction involves ions. 4 Ionic Redox Reactions are usually found in acidic or basic solutions, so the addition of H 2 O, H 3 O +1 and/or OH -1 is done to balance the overall oxidation numbers. Balancing Oxidation- Reduction Reactions

69. –Starting with an example of an ionic Redox reaction in an acidic solution, I will use the following example and show you the steps as before: MnO 4 -1 + S -2 Mn +2 + S 0 (acidic solution) Balancing Oxidation- Reduction Reactions

70. Step 1: Assign oxidation numbers to each element to identify the elements being oxidized and those being reduced: MnO 4 -1 + S -2 Mn +2 + S 0 Balancing Oxidation- Reduction Reactions +7-2 +2 0

71. Step 2: Now write two equations using only the elements that change in ON. Then add electrons to bring the equation into electrical balance: Mn +7 + 5e -1 Mn +2 (reduction) S -2 S 0 + 2e -1 (oxidation) Balancing Oxidation- Reduction Reactions

72. Step 3: Multiply the two equations by the smallest whole numbers to cancel out the electrons: 2(Mn +7 + 5e -1 Mn +2 )(reduction) (S -2 S 0 + 2e -1 )5(oxidation) Balancing Oxidation- Reduction Reactions

73. Step 4: Transfer the coefficient in front of each substance in the half- reactions to the corresponding substance in the original equation: 2MnO 4 -1 + 5S -2 2Mn +2 + 5S 0 Balancing Oxidation- Reduction Reactions

74. Step 5: Since the reaction takes place in an acidic environment, we now add hydronium ions to the more negative side to balance the charges: 16 H 3 O +1 + 2MnO 4 -1 + 5S -2 2Mn +2 + 5S 0 Balancing Oxidation- Reduction Reactions

75. Step 6: As an aqueous solution, water is also available as part of the reaction. Balance the Hydrogens and Oxygens by adding water to either side of the equation: 16 H 3 O +1 + 2MnO 4 -1 + 5S -2 2Mn +2 + 5S 0 + 24 H 2 O Balancing Oxidation- Reduction Reactions

76. –Steps 1 through 4 and step 6 are the same for a basic solution. In step 5, instead of adding a hydronium ion, we add an hydroxide ion. As an example: CrO 4 -2 + Fe(OH) 2 Cr(OH) 3 + Fe(OH) 3 Balancing Oxidation- Reduction Reactions

77. –After step 4 we get: CrO 4 -2 + 3Fe(OH) 2 Cr(OH) 3 + 3Fe(OH) 3 Balancing Oxidation- Reduction Reactions

78. Step 5: Since the reaction takes place in a basic environment, we now add hydroxide ions to the more positive side to balance the charges: CrO 4 -2 + 3Fe(OH) 2 Cr(OH) 3 + 3Fe(OH) 3 + 2OH -1 Balancing Oxidation- Reduction Reactions

79. Step 6: As an aqueous solution, water is also available as part of the reaction. Balance the Hydrogens and Oxygens by adding water to either side of the equation: 4H 2 O + CrO 4 -2 + 3Fe(OH) 2 Cr(OH) 3 + 3Fe(OH) 3 + 2OH -1 Balancing Oxidation- Reduction Reactions