1. Chapter 18 Electrochemistry

2. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 2 Review of Terms  Electrochemistry – the study of the interchange of chemical and electrical energy  Oxidation – reduction (redox) reaction – involves a transfer of electrons from the reducing agent to the oxidizing agent  Oxidation – loss of electrons  Reduction – gain of electrons  Reducing agent – electron donor  Oxidizing agent – electron acceptor

3. 3 2Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg 2+ + 4e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 002+2-

4. 4 Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1.Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 = 0 2.In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3.The oxidation number of oxygen is usually –2. In H 2 O 2 and O 2 2- it is –1.

5. 5 4.The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 5.Group IA metals are +1, IIA metals are +2 and fluorine is always –1.

6. 6 Galvanic Cells spontaneous redox reaction anode oxidation cathode reduction

7. Section 18.2 Galvanic Cells Galvanic Cell  Oxidation occurs at the anode.  Reduction occurs at the cathode.  Salt bridge or porous disk – devices that allow ions to flow without extensive mixing of the solutions.  Salt bridge – contains a strong electrolyte held in a Jello–like matrix.  Porous disk – contains tiny passages that allow hindered flow of ions. Copyright © Cengage Learning. All rights reserved 7

8. Section 18.2 Galvanic Cells Cell Potential  A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment.  The “pull”, or driving force, on the electrons is called the cell potential ( ), or the electromotive force (emf) of the cell.  Unit of electrical potential is the volt (V). Copyright © Cengage Learning. All rights reserved 8

9. 9 Galvanic Cells The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Cell Diagram Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+ ] = 1 M and [Zn 2+ ] = 1 M Zn (s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (s) anode cathode salt bridge phase boundary

10. Section 18.2 Galvanic Cells Voltaic Cell: Cathode Reaction Copyright © Cengage Learning. All rights reserved 10 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE

11. Section 18.2 Galvanic Cells Voltaic Cell: Anode Reaction Copyright © Cengage Learning. All rights reserved 11 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE

12. 12 Standard Reduction Potentials Standard reduction potential (E 0 ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. E 0 = 0 V Standard hydrogen electrode (SHE) 2e - + 2H + (1 M) H 2 (1 atm) Reduction Reaction

13. 13 E 0 is for the reaction as written The more positive E 0 the greater the tendency for the substance to be reduced  A galvanic cell runs spontaneously in the direction that gives a positive value for E° cell. The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0

14. 14 A galvanic cell consists 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. Calculate the standard emf of this cell at 25°C. Problem-1 E° cell = E° cathode - E° anode

15. Example 15 Strategy At first it may not be clear how to assign the electrodes in the galvanic cell. From Table 18.1 (in slide 13), or Appendix A5.5, we write the standard reduction potentials of Ag and Mg and determine which is the anode and which is the cathode. Solution The standard reduction potentials are Ag + (1.0 M) + e - Ag(s) E° = 0.80 V Mg 2+ (1.0 M) + 2e - Mg(s) E° = -2.37 V

16. Example 16 we see that Ag + will oxidize Mg: Anode (oxidation):Mg(s) Mg 2+ (1.0 M) + 2e - Cathode (reduction): 2Ag + (1.0 M) + 2e - 2Ag(s) _________________________________________________ Overall: Mg(s) + 2Ag + (1.0 M) Mg 2+ (1.0 M) + 2Ag(s)

17. Example 17 Note that in order to balance the overall equation we multiplied the reduction of Ag + by 2. We can do so because, as an intensive property, E° is not affected by this procedure. We find the emf of the cell by using the following Equation and Table 18.1: E° cell = E° cathode - E° anode = E° Ag + /Ag - E° Mg 2+ /Mg = 0.80 V - (-2.37 V) = 3.17 V Check The positive value of E° shows that the forward reaction is favored.

18. Section 18.3 Standard Reduction Potentials Problem 2: 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  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 Copyright © Cengage Learning. All rights reserved 18

19. Section 18.3 Standard Reduction Potentials Overall Balanced Cell Reaction Balanced Cell Reaction: Cu + 2Fe 3+ → Cu 2+ + 2Fe 2+  Cell Potential: E° cell = E° (cathode) – E° (anode) E° cell = 0.77 V – 0.34 V = 0.43 V Copyright © Cengage Learning. All rights reserved 19

20. 20 Problem 3 a)Ag electrode in 1.0 M Ag + (aq) and Cu electrode in 1.0 M Cu 2+ (aq) b)Zn electrode in 1.0 M Zn 2+ (aq) and Cu electrode in 1.0 M Cu 2+ (aq)

21. 21 Answer (a)0.46 V (b)1.10 V

22. 22 The faraday is a dimensionless unit of electric charge quantity, equal to approximately 6.02 x 10 23 electric charge carriers. This is equivalent to one mole, also known as Avogadro's constant.moleAvogadro's constant In the International System of Units ( SI ), the coulomb (C) is the preferred unit of electric charge quantity. It is equivalent to one ampere- second (1 A · s) and represents approximately 6.24 x 10 18 electric charge carriers. To convert from coulombs to faradays, multiply by 1.04 x 10 -5. To convert from faradays to coulombs, multiply by 9.65 x 10 4.SI coulombampere- second

23. 23 Spontaneity of Redox Reactions  G = -nFE cell  G 0 = -nFE cell 0 n = number of moles of electrons in reaction F = 96,500 J V mol = 96,500 C/mol  G 0 = -RT ln K = -nFE cell 0 E cell 0 = RT nF ln K (8.314 J/K mol)(298 K) n (96,500 J/V mol) ln K = = 0.0257 V n ln K E cell 0 = 0.0592 V n log K E cell 0

24. 24 Spontaneity of Redox Reactions  G 0 = -RT ln K = -nFE cell 0

25. Example 25 Problem 4 Calculate the equilibrium constant for the following reaction at 25°C: Sn(s) + 2Cu 2+ (aq) Sn 2+ (aq) + 2Cu + (aq)

26. Example 26 18.4 Solution The half-cell reactions are Anode (oxidation): Sn(s) Sn 2+ (aq) + 2e - Cathode (reduction): 2Cu 2+ (aq) + 2e - 2Cu + (aq) E° cell = E° cathode - E° anode = E° Cu 2+ /Cu + - E° Sn 2+ /Sn = 0.15 V - (-0.14 V) = 0.29 V

27. Example 27 18.4 Equation (18.5) can be written In the overall reaction we find n = 2. Therefore,

28. Example 28 18.5 Calculate the standard free-energy change for the following reaction at 25°C: 2Au(s) + 3Ca 2+ (1.0 M) 2Au 3+ (1.0 M) + 3Ca(s)

29. Example 29 18.5 Calculate the standard free-energy change for the following reaction at 25°C: 2Au(s) + 3Ca 2+ (1.0 M) 2Au 3+ (1.0 M) + 3Ca(s)

30. Example 30 18.5 Strategy The relationship between the standard free energy change and the standard emf of the cell is given by Equation (18.3): ΔG° = -nFE° cell. Thus, if we can determine E° cell, we can calculate ΔG°. We can determine the E° cell of a hypothetical galvanic cell made up of two couples (Au 3+ /Au and Ca 2+ /Ca) from the standard reduction potentials in Table 18.1.

31. Example 31 18.5 Solution The half-cell reactions are Anode (oxidation): 2Au(s) 2Au 3+ (1.0 M) + 6e - Cathode (reduction): 3Ca 2+ (1.0 M) + 6e - 3Ca(s) E° cell = E° cathode - E° anode = E° Ca 2+ /Ca - E° Au 3+ /Au = -2.87 V - 1.50 V = -4.37 V

32. Example 32 18.5 Now we use Equation (18.3): ΔG° = -nFE° The overall reaction shows that n = 6, so ΔG° = -(6) (96,500 J/V · mol) (-4.37 V) = 2.53 x 10 6 J/mol = 2.53 x 10 3 kJ/mol Check The large positive value of ΔG° tells us that the reaction favors the reactants at equilibrium. The result is consistent with the fact that E° for the galvanic cell is negative.

33. 33 The Effect of Concentration on Cell Emf Nernst Equation  G =  G 0 + RT ln Q  G = -nFE  G 0 = -nFE 0 -nFE = -nFE 0 + RT ln Q E = E 0 - ln Q RT nF Nernst equation At 298 K - 0.0257 V n ln Q E 0 E = - 0.0592 V n log Q E 0 E =

34. Section 18.5 Dependence of Cell Potential on Concentration Nernst Equation  The relationship between cell potential and concentrations of cell components (p.854-855, Zumdahl)  At 25°C: or (Q= K at equilibrium, and E cell = 0, no chemical driving force to push electrons through the wire) Copyright © Cengage Learning. All rights reserved 34

35. Section 18.5 Dependence of Cell Potential on Concentration A Concentration Cell Copyright © Cengage Learning. All rights reserved 35

36. 36 Concentration Cells Galvanic cell from two half-cells composed of the same material but differing in ion concentrations.

37. Section 18.5 Dependence of Cell Potential on Concentration A concentration cell is constructed using two nickel electrodes with Ni 2+ concentrations of 1.0 M and 1.00 × 10 -4 M in the two half-cells. (When the electrodes are of the same material, cathode goes with higher concentration) Calculate the potential of this cell at 25°C. 0.118 V Copyright © Cengage Learning. All rights reserved 37 EXERCISE!

38. Section 18.5 Dependence of Cell Potential on Concentration The cell potential equals 0.118 V. ε = ε° - (0.0591/n)logQ. For a concentration cell, ε° = 0. ε = 0 - (0.0591/2)log(1.0×10 -4 / 1.0) = 0.118 V

39. Example 39 18.6 Predict whether the following reaction would proceed spontaneously as written at 298 K: Co(s) + Fe 2+ (aq) Co 2+ (aq) + Fe(s) given that [Co 2+ ] = 0.15 M and [Fe 2+ ] = 0.68 M.

40. Example 40 18.6 Strategy Because the reaction is not run under standard-state conditions (concentrations are not 1 M), we need Nernst’s equation to calculate the emf (E) of a hypothetical galvanic cell and determine the spontaneity of the reaction. The standard emf (E°) can be calculated using the standard reduction potentials in Table 18.1. Remember that solids do not appear in the reaction quotient (Q) term in the Nernst equation. Note that 2 moles of electrons are transferred per mole of reaction, that is, n = 2.

41. Example 41 18.6 Solution The half-cell reactions are Anode (oxidation): Co(s) Co 2+ (aq) + 2e - Cathode (reduction): Fe 2+ (aq) + 2e - Fe(s) E° cell = E° cathode - E° anode = E° Fe 2+ /Fe - E° Co 2+ /Co = -0.44 V – (-0.28 V) = -0.16 V

42. Example 42 From the Nernst Equation we write Because E is negative, the reaction is not spontaneous in the direction written. 18.6

43. Example 43 18.7 Consider a galvanic cell. In a certain experiment, the emf (E) of the cell is found to be 0.54 V at 25°C. Suppose that [Zn 2+ ] = 1.0 M and P H 2 = 1.0 atm. Calculate the molar concentration of H +.

44. Example 44 18.7 Strategy The equation that relates standard emf and nonstandard emf is the Nernst equation. The overall cell reaction is Zn(s) + 2H + (? M) Zn 2+ (1.0 M) + H 2 (1.0 atm) Given the emf of the cell (E), we apply the Nernst equation to solve for [H + ]. Note that 2 moles of electrons are transferred per mole of reaction; that is, n = 2.

45. Example 45 18.7 Solution As we saw earlier, the standard emf (E°) for the cell is 0.76 V. From Nernst Equation, we write

46. Example 46 18.7 Check The fact that the nonstandard-state emf (E) is given in the problem means that not all the reacting species are in their standard-state concentrations. Thus, because both Zn 2+ ions and H 2 gas are in their standard states, [H + ] is not 1 M.

47. Example You make a galvanic cell at 25°C containing: – A nickel electrode in 1.0 M Ni 2+ (aq) – A silver electrode in 1.0 M Ag + (aq) Sketch this cell, labeling the anode and cathode, showing the direction of the electron flow, and calculate the cell potential. 1.03 V Copyright © Cengage Learning. All rights reserved 47 CONCEPT CHECK!

48. Example 48 Nickel is the anode, silver is the cathode (electrons flow from nickel to silver). To find the cell potential, use the Nernst equation: ε = ε° - (0.0591/n)logQ. The overall equation is 2Ag + + Ni → 2Ag + Ni 2+. Therefore, Q = [Ni 2+ ] / [Ag + ] 2 = (1.0 M) / (1.0 M) 2. ε = 1.03 V – (0.0591/2)log(1) = 1.03 V.

49. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 49 Half–Reactions  The overall reaction is split into two half–reactions, one involving oxidation and one involving reduction. 8H + + MnO 4 – + 5Fe 2+ Mn 2+ + 5Fe 3+ + 4H 2 O Reduction: 8H + + MnO 4 – + 5e – Mn 2+ + 4H 2 O Oxidation: 5Fe 2+ 5Fe 3+ + 5e – The Half–Reaction Method for Balancing Equations for Oxidation– Reduction Reactions Occurring in Acidic Solution

50. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 50 1.Write separate equations for the oxidation and reduction half–reactions. 2.For each half–reaction: A.Balance all the elements except H and O. B.Balance O using H 2 O. C.Balance H using H +. D.Balance the charge using electrons. The Half–Reaction Method for Balancing Equations for Oxidation– Reduction Reactions Occurring in Acidic Solution

51. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 51 3.If necessary, multiply one or both balanced half – reactions by an integer to equalize the number of electrons transferred in the two half – reactions. 4.Add the half – reactions, and cancel identical species. 5.Check that the elements and charges are balanced. The Half–Reaction Method for Balancing Equations for Oxidation– Reduction Reactions Occurring in Acidic Solution

52. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 52 The Half–Reaction Method for Balancing Equations for Oxidation– Reduction Reactions Occurring in Acidic Solution

53. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 53 Cr 2 O 7 2- (aq) + SO 3 2- (aq) Cr 3+ (aq) + SO 4 2- (aq)  How can we balance this equation?  First Steps:  Separate into half-reactions.  Balance elements except H and O.

54. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 54  Cr 2 O 7 2- (aq) 2Cr 3+ (aq)  SO 3 2- (aq) SO 4 2- (aq)  How many electrons are involved in each half reaction? Method of Half Reactions

55. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 55  6e - + Cr 2 O 7 2- (aq) 2Cr 3+ (aq)  SO 3 2- (aq) + SO 4 2- (aq) + 2e -  How can we balance the oxygen atoms? Method of Half Reactions (continued)

56. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 56  6e - + Cr 2 O 7 2- (aq) Cr 3+ (aq) + 7H 2 O  H 2 O +SO 3 2- (aq) + SO 4 2- (aq) + 2e -  How can we balance the hydrogen atoms? Method of Half Reactions (continued)

57. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 57  This reaction occurs in an acidic solution.  14H + + 6e - + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O  H 2 O +SO 3 2- SO 4 2- + 2e - + 2H +  How can we balance the electrons? Method of Half Reactions (continued)

58. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 58  14H + + 6e - + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O  3[H 2 O +SO 3 2- SO 4 2- + 2e - + 2H + ]  Final Balanced Equation: Cr 2 O 7 2- + 3SO 3 2- + 8H + 2Cr 3+ + 3SO 4 2- + 4H 2 O Method of Half Reactions (continued)

59. Section 18.1 Balancing Oxidation-Reduction Equations Copyright © Cengage Learning. All rights reserved 59 Balance the following oxidation – reduction reaction that occurs in acidic solution. Br – (aq) + MnO 4 – (aq) Br 2 (l)+ Mn 2+ (aq) 10Br – (aq) + 16H + (aq) + 2MnO 4 – (aq) 5Br 2 (l)+ 2Mn 2+ (aq) + 8H 2 O(l) EXERCISE!

60. Section 18.1 Balancing Oxidation-Reduction Equations In Basic solution  Zumdahl (Example 18.2, p.838-839)

61. 61 Balancing Redox Equations 1.Write the unbalanced equation for the reaction in ionic form. The oxidation of Fe 2+ to Fe 3+ by Cr 2 O 7 2- in acid solution? Fe 2+ + Cr 2 O 7 2- Fe 3+ + Cr 3+ 2.Separate the equation into two half-reactions. Oxidation: Cr 2 O 7 2- Cr 3+ +6+3 Reduction: Fe 2+ Fe 3+ +2+3 3.Balance the atoms other than O and H in each half-reaction. Cr 2 O 7 2- 2Cr 3+

62. 62 Balancing Redox Equations 4.For reactions in acid, add H 2 O to balance O atoms and H + to balance H atoms. Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 5.Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe 2+ Fe 3+ + 1e - 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 6.If necessary, equalize the number of electrons in the two half- reactions by multiplying the half-reactions by appropriate coefficients. 6Fe 2+ 6Fe 3+ + 6e - 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O

63. 63 Balancing Redox Equations 7.Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 6Fe 2+ 6Fe 3+ + 6e - Oxidation: Reduction: 14H + + Cr 2 O 7 2- + 6Fe 2+ 6Fe 3+ + 2Cr 3+ + 7H 2 O 8.Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6 x 2 = 24 = 6 x 3 + 2 x 3 9.For reactions in basic solutions, add OH - to both sides of the equation for every H + that appears in the final equation.

64. Example 64 18.1 Write a balanced ionic equation to represent the oxidation of iodide ion (I - ) by permanganate ion ( ) in basic solution to yield molecular iodine (I 2 ) and manganese(IV) oxide (MnO 2 ).

65. Example 65 18.1 Strategy We follow the preceding procedure for balancing redox equations. Note that the reaction takes place in a basic medium. Solution Step 1: The unbalanced equation is + I - MnO 2 + I 2

66. Example 66 Step 2: The two half-reactions are Oxidation: I - I 2 Reduction: MnO 2 18.1 0 +7+4 Step 3: We balance each half-reaction for number and type of atoms and charges. Oxidation half-reaction: We first balance the I atoms: 2I - I 2

67. Example 67 18.1 To balance charges, we add two electrons to the right-hand side of the equation: 2I - I 2 + 2e - Reduction half-reaction: To balance the O atoms, we add two H 2 O molecules on the right: MnO 2 + 2H 2 O To balance the H atoms, we add four H + ions on the left: + 4H + MnO 2 + 2H 2 O

68. Example 68 18.1 There are three net positive charges on the left, so we add three electrons to the same side to balance the charges: + 4H + + 3e - MnO 2 + 2H 2 O Step 4: We now add the oxidation and reduction half reactions to give the overall reaction. In order to equalize the number of electrons, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 as follows: 3(2I - I 2 + 2e - ) 2( + 4H + + 3e - MnO 2 + 2H 2 O) ________________________________________ 6I - +2 + 8H + + 6e - 3I 2 + 2MnO 2 + 4H 2 O + 6e -

69. Example 69 18.1 The electrons on both sides cancel, and we are left with the balanced net ionic equation: 6I - + 2 + 8H + 3I 2 + 2MnO 2 + 4H 2 O This is the balanced equation in an acidic medium. However, because the reaction is carried out in a basic medium, for every H + ion we need to add equal number of OH - ions to both sides of the equation: 6I - + 2 + 8H + + 8OH - 3I 2 + 2MnO 2 + 4H 2 O + 8OH -

70. Example 70 18.1 Finally, combining the H + and OH - ions to form water, we obtain 6I - + 2 + 4H 2 O 3I 2 + 2MnO 2 + 8OH - Step 5: A final check shows that the equation is balanced in terms of both atoms and charges.

71. Section 18.1 Balancing Oxidation-Reduction Equations

72. Section 18.8 Electrolysis Stoichiometry of Electrolysis  How much chemical change occurs with the flow of a given current for a specified time? current and time  quantity of charge  moles of electrons  moles of analyte  grams of analyte Copyright © Cengage Learning. All rights reserved 72

73. Section 18.8 Electrolysis Stoichiometry of Electrolysis  current and time  quantity of charge Coulombs of charge = amps (C/s) × seconds (s)  quantity of charge  moles of electrons Copyright © Cengage Learning. All rights reserved 73

74. Section 18.8 Electrolysis An unknown metal (M) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate 0.0719 g of the metal from a solution containing M(NO 3 ) 3. What is the metal? gold (Au) Copyright © Cengage Learning. All rights reserved 74 CONCEPT CHECK!

75. Section 18.8 Electrolysis The metal has a molar mass of 197 g/mol. The metal is gold (Au). To find the molar mass, divide the number of grams of the metal by the number of moles of metal. To find the moles of metal: 52.8 sec × (2.00 C/sec) × (1 mol e – /96485 C) × (1 mol M/3 mole e – ) = 3.65 × 10 –4 mol M To find the molar mass: 0.0719 g / 3.65 × 10 –4 mol M = 197 g/mol

76. Section 18.8 Electrolysis Consider a solution containing 0.10 M of each of the following: Pb 2+, Cu 2+, Sn 2+, Ni 2+, and Zn 2+. Predict the order in which the metals plate out as the voltage is turned up from zero. Cu 2+, Pb 2+, Sn 2+, Ni 2+, Zn 2+ Do the metals form on the cathode or the anode? Explain. Use the reduction potentials from Table 18.1. Once that voltage has been surpassed, the metal will plate out. They plate out on the cathode (since they are reduced). Order: Cu 2+, Pb 2+, Sn 2+, Ni 2+, Zn 2+ Copyright © Cengage Learning. All rights reserved 76 CONCEPT CHECK!

77. Section 18.9 Commercial Electrolytic Processes Electroplating a Spoon Copyright © Cengage Learning. All rights reserved 77

78. Section 18.6 Batteries Schematic of the Hydrogen-Oxygen Fuel Cell

79. 79 Electrolysis of Water

80. Section 18.7 Corrosion The Electrochemical Corrosion of Iron (p.862-863) Copyright © Cengage Learning. All rights reserved 80

81. Section 18.6 Batteries One of the Six Cells in a 12–V Lead Storage Battery Copyright © Cengage Learning. All rights reserved 81

82. Section 18.6 Batteries A Common Dry Cell Battery