1. Oxidation/Reduction Chapter 20

2. Two Types of Chemical Rxns 1.Exchange of Ions – no change in charge/oxidation numbers – Acid/Base Rxns NaOH + HCl

3. Two Types of Chemical Rxns – Precipitation Rxns Pb(NO 3 ) 2 (aq) + KI(aq) – Dissolving Rxns CaCl 2 (s) 

4. Two Types of Chemical Rxns 2.Exchange of Electrons – changes in oxidation numbers/charges Fe(s) + CuSO 4 (aq)  FeSO 4 (aq) + Cu(s) Remove spectator ions Fe + Cu 2+  Fe 2+ + Cu Protons Electrons

5. Review of Oxidation Numbers Oxidation numbers – the charge on an ion or an assigned charge on an atom. Al Cl 2 P 4 Mg 2+ Cl -

6. Review of Oxidation Numbers Calculate the oxidation numbers for: HClO Cr 3+ S 8 Fe 2 (SO 4 ) 3 Mn 2 O 3 SO 3 2- KMnO 4 NO 3 - HSO 4 -

7. Oxidation 1.Classical Definition –addition of oxygen 2.Modern Definition – an increase in oxidation number Fe + O 2  Fe 2 O 3 CO + O 2  CO 2 CH 3 CH 2 OH  CH 3 CHO  CH 3 COOH

8. Oxidation Fe + O 2  (limited oxygen) Fe + O 2  (excess oxygen)

9. Oxidation C + O 2  (limited oxygen) C + O 2  (excess oxygen)

10. Reduction 1.Classical –addition of hydrogen 2.Modern –decrease (reduction) in oxidation number N 2 + 3H 2  2NH 3 (Haber process) R-C=C-R+ H 2  | | H H (unsaturated fat)(saturated fat)

11. Oxidizing/Reducing Agents Oxidation and Reduction always occur together. Oxidizing Agents – Get reduced – Gain electrons Reducing Agents – Get oxidized – Lose electrons

12. Oxidizing/Reducing Agents 0+2 Cu+O 2  CuO 0 -2 Got oxidized, reducing agent Got reduced, oxidizing agent

14. Identify the Oxidizing/Reducing Agents in the following (Calculate the ox. numbers also). Cu + S 8  Cu 2 S H 2 + O 2  H 2 O Cu + AgNO 3  Cu(NO 3 ) 2 + Ag H 2 O + Al + MnO 4 -  Al(OH) 4 - + MnO 2

15. Identify the Oxidizing/Reducing Agents in the following (Calculate the ox. numbers also). Al + O 2  Al 2 O 3 Li + N 2  Li 3 N Cu(NO 3 ) 2 + Fe  Fe(NO 3 ) 3 + Cu KMnO 4 + FeSO 4  Fe 2 (SO 4 ) 3 + Mn + K +

16. Balancing Redox Reactions Half-Reaction Method Break eqn into oxidation half and reduction half Easy Examples: – Al + Fe 2+  Fe + Al 3+ – Cu + Zn 2+  Cu 2+ + Zn – Mg + Na +  Mg 2+ + Na

17. Balancing Redox Reactions What’s really happening: Cu 2+ + Zn  Cu + Zn 2+

19. Balancing Redox Reactions Steps for more complicated examples 1.Balance all atoms except H and O 2.Balance charge with electrons 3.Balance O with water 4.Balance H with H + =================================== 5. (Add OH - to make water in basic solutions)

21. Balancing Redox Reactions Example 1: MnO 4 - + C 2 O 4 2-  Mn 2+ + CO 2 1. Separate into half reactions MnO 4 -  Mn 2+ C 2 O 4 2-  CO 2

22. Balancing Redox Reactions MnO 4 -  Mn 2+ +7 +2 5e - + MnO 4 -  Mn 2+ 5e - + MnO 4 -  Mn 2+ + 4H 2 O 8H + + 5e - + MnO 4 -  Mn 2+ + 4H 2 O

23. Balancing Redox Reactions C 2 O 4 2-  CO 2 C 2 O 4 2-  2CO 2 +3 +4 (1 e - per carbon) C 2 O 4 2-  2CO 2 + 2e -

24. Balancing Redox Reactions C 2 O 4 2-  2CO 2 + 2e - (X 5) 8H + + 5e - + MnO 4 -  Mn 2+ + 4H 2 O (X 2) 5C 2 O 4 2-  10CO 2 + 10e - 16H + + 10e - + 2MnO 4 -  2Mn 2+ + 8H 2 O 16H + +5C 2 O 4 2- +2MnO 4 -  2Mn 2+ + 10CO 2 + 8H 2 O

25. Balancing Redox Reactions (Acidic Solutions) Cr 2 O 7 2- + Cl -  Cr 3+ + Cl 2 14 H + + Cr 2 O 7 2- + 6Cl -  2Cr 3+ + 3Cl 2 + 7H 2 O Cu + NO 3 -  Cu 2+ + NO 2 Cu + 4H + + 2NO 3 -  Cu 2+ + 2NO 2 + 2H 2 O Mn 2+ + BiO 3 -  Bi 3+ + MnO 4 - 14H + + 2Mn 2+ + 5BiO 3 -  5Bi 3+ + 2MnO 4 - + 7H 2 O

26. Balancing Redox Reactions (Basic Solutions) Add OH - AT THE VERY END ONLY!!!!! NO 2 - + Al  NH 3 + Al(OH) 4 - 5H 2 O + OH - + NO 2 - + 2Al  NH 3 + 2Al(OH) 4 - Cr(OH) 3 + ClO -  CrO 4 2- + Cl 2 2Cr(OH) 3 + 6ClO -  2CrO 4 2- + 3Cl 2 + 2OH - + 2H 2 O

27. Balance in Both Acidic and Basic Solutions F - + MnO 4 -  MnO 2 + F 2 HNO 2 + H 2 O 2  O 2 + NO H is +1

28. F - + MnO 4 -  MnO 2 + F 2 8H + + 6F - + 2MnO 4 -  2MnO 2 + 3F 2 + 4H 2 O 4H 2 O + 6F - + 2MnO 4 -  2MnO 2 + 3F 2 + 8OH - HNO 2 + H 2 O 2  O 2 + NO 2HNO 2 + H 2 O 2  O 2 + 2NO + 2H 2 O

29. Voltaic (Galvanic) Cells Voltaic(Galvanic) Cells – redox reactions that produce a voltage – Spontaneous reactions (  G<0) – Voltage of the cell (E o cell ) is positive – Batteries Electrolytic cells – redox reactions that must have a current run through them. –  G>0 and E o cell is negative. – Often used to plate metals

30. Voltaic (Galvanic) Cells History Galvani (died 1798)– uses static electricity to move the muscles of dead frogs Volta (1800) – Created the first battery

31. Voltaic (Galvanic) Cells Voltaic cell 1.Anode – Oxidation site 2.Cathode – Reduction site (RC cola) 3.Salt bridge – completes the circuit

32. Voltaic (Galvanic) Cells Cell Notation Zn | Zn 2+ (aq) ||Cu 2+ (aq) | Cu AnodeZn  Zn 2+ + 2e - CathodeCu 2+ + 2e -  Cu CellCu 2+ + Zn  Cu + Zn 2+

33. Voltaic (Galvanic) Cells

34. Hydrogen Electrode 1.Standard Electrode 2.Voltage(potential) = 0 Volts 2H + (aq) + 2e -  H 2 (g)0 volts H 2 (g)  2H + (aq) + 2e - 0 volts 3. Often used in electrodes (like pH)

36. Standard Reduction Potentials Rules – Flipping an equation changes the sign of E – Multiplying an equation does not change the magnitude of E

37. Calculating Cell Potential A cell is composed of copper metal and Cu 2+ (aq) on one side, and zinc metal and Zn 2+ (aq) on the other. Calculate the cell potential. Zn 2+ + 2e -  Zn-0.76 V Cu 2+ + 2e -  Cu+0.34 V flip the zinc equation Zn  Zn 2+ + 2e - +0.76 V Cu 2+ + 2e -  Cu+0.34 V Zn + Cu 2+  Zn 2+ + Cu+1.10 V

38. What is the cell emf of a cell made using Cu and Cu 2+ in one side and Al and Al 3+ in the other? Write the complete cell reaction. ANS: 2.00 V

39. Calculate the standard emf for the following reaction. Hint: break into half-reactions. 2Al(s) + 3I 2 (s)  2Al 3+( aq) + 6I - (aq)

40. A voltaic cell is based on the following half reactions. In + (aq)  In 3+ (aq) + 2e - Br 2 (l) + 2e -  2Br - (aq)+1.06 V If the overall cell voltage is 1.46 V, what is the reduction potential for In 3+ ?

41. Calculate the standard emf for the following reaction. Cr 2 O 7 2- + 14H + + 6I -  2Cr 3+ + 3I 2 + 7H 2 O

42. Two half reactions in a voltaic cell are: Zn 2+ (aq) + 2e -  Zn(s) Li + (aq) + e -  Li(s) a)Calculate the cell emf. b)Which is the anode? Which is the cathode? c)Which electrode is consumed? d)Which electrode is positive? e)Sketch the cell, indicating electron flow.

43. Given the following half-reactions: Pb 2+ + 2e -  Pb Ni 2+ + 2e -  Ni a.Calculate the cell potential ( E o ). b.Label the cathode and anode. c.Identify the oxidizing and reducing agents. d.Which electrode is consumed? e.Which electrode is plated? f.Sketch the cell, indicating the direction of electron flow.

44. Strengths of Oxidizing and Reducing Agents larger the reduction potential, stronger the oxidizing agent – Wants to be reduced, can oxidize something else. lower the reduction potential, stronger the reducing agent – Would rather be oxidized

45. F 2 + 2e -  2F - +2.87 V Stronger Cl 2 + 2e -  2Cl - +1.36 V Oxidizing. Agents.. Li + + e -  Li-3.05V

46. Example 1 Which of the following is the strongest oxidzing agent? Which is the strongest reducing agent? NO 3 - Cr 2 O 7 2- Ag +

47. Which of the following is the strongest reducing agent? Which is the strongest oxidizing agent? I 2 (s)Fe(s)Mn(s)

48. Can copper metal (Cu(s)) act as an oxidizing agent?

49. Spontaneity Voltaic CellsElectrolytic Cells Positive emf Spontaneous Can produce electric current Batteries Negative emf Not spontaneous Must “pump” electricity in Electrolysis

50. Example 1 Are the following cells spontaneous as written? a)Cu + 2H +  Cu 2+ + H 2 b)Cl 2 + 2I -  2Cl - + I 2 c)I 2 + 5Cu 2+ + 6H 2 O  2IO 3 - + 5Cu + 12H + d)Hg 2+ + 2I -  Hg + I 2

51. EMF and  G o  G = -nFE n=number of electrons transferred E =Cell emf F =96,500 J/V-mol(Faraday’s Constant) Positive Voltage gives a negative  G (spont)

52. Calculate the cell potential and free energy change for the following reaction: 4Ag + O 2 + 4H +  4Ag + + 2H 2 O ANS: +0.43 V, -170 kJ/mol

53. Calculate  G and the EMF for the following reaction. Also, calculate the K. 3Ni 2+ + 2Cr(OH) 3 + 10OH -  3Ni + 2CrO 4 2- + 8H 2 O ANS: +87 kJ/mol, -0.15 V, 6 X 10 -16

54. EMF and K  G o = -RTlnK(  G o = -nFE o ) -nFE o = -RTlnK lnK = nFE o (assume 298 K) RT log K = nE o 0.0592

55. Example 1 Calculate  G, cell voltage and the equilibrium constant for the following cell: O 2 + 4H + + 4Fe 2+  4Fe 3+ + 2H 2 O ANS: -177 kJ/mol, 0.459 V, 1 X 10 31

56. Example 2 If the equilibrium constant for a particular reaction is 1.2 X 10 -10, calculate the cell potential. Assume n = 2.

57. Concentration Cells: Nernst Equation  G =  G o + RT lnQ -nFE = -nFE o + RT lnQ E = E o -RT lnQ(assume 298 K) nF E = E o -0.0592 log Q n Can adjust the voltage of any cell by changing concentrations

58. Using the Nernst Eqn Suppose in the following cell, the concentration of Cu 2+ is 5.0 M and the concentration of Zn 2+ is 0.050 M. Calculate the cell voltage. Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)+1.10 V

59. E = E o -0.0592 V log Q n E = 1.10V - 0.0592 V log [Cu][Zn 2+] 2[Cu 2+ ][Zn] E = 1.10V - 0.0592 V log [Zn 2+] 2[Cu 2+ ] E = 1.10V - 0.0592 V log [0.050] 2[5.0] E = +1.16 V

60. Example 1 Calculate the emf at 298 K generated by the following cell (E o = 0.79 V) where: [ Cr 2 O 7 2- ]= 2.0 M, [ H + ]=1.0 M, [ I - ]=1.0 M and [ Cr 3+ ]= 1.0 X 10 -5 M. Cr 2 O 7 2- + 14H + + 6I -  2Cr 3+ + 3I 2 + 7H 2 O ANS: 0.89 V

61. Example 2 Calculate the emf at 298 K generated by the following cell (E o = 2.20 V) where: [ Al 3+ ]= 0.004 M and [ I - ]=0.010 M. 2Al(s) + 3I 2 (s)  2Al 3+( aq) + 6I - (aq) ANS: +2.36 V

62. Example 3 If the voltage of a Zn-H + cell is 0.45 V at 298 K when [Zn 2+ ]=1.0 M and P H2 =1.0 atm, what is the concentration of H + ? Note that atm can be used just like molarity. Zn(s) + 2H + (aq)  Zn 2+ (aq) + H 2 (g) ANS: 5.2 X 10 -6 M

63. Example 4 What pH is required if we want a voltage of 0.542 V and [Zn 2+ ]=0.10 M and P H2 =1.0 atm? Zn(s) + 2H + (aq)  Zn 2+ (aq) + H 2 (g) ANS: 5.84 X 10 -5 M, pH = 4.22

64. Batteries Lead Acid Battery 12 Volt DC Discharges when starting the car, recharges as you drive (generator). Running reaction backward. PbO 2 (s) + Pb(s) +2HSO 4 - (aq) + 2H + (aq)  2PbSO 4 (s) +2H 2 O(l)

65. Alkaline Batteries Basic Zinc can acts as the anode 2MnO 2 (s)+2H 2 O(l)+2e -  2MnO(OH)(s) + 2OH - (aq) Zn(s) + 2OH - (aq)  Zn(OH) 2 (s) + 2e - Rechargable uses Ni-Cd

66. Corrosion Iron rusts in acidic solns (not above pH=9) Water needs to be present Salts accelerate the process O 2 + 4H + + 4e -  2H 2 O Fe  Fe 2+ + 2e - (The Fe 2+ eventually goes to Fe 3+, Fe 2 O 3 )

68. Preventing Corrosion Paint Sometimes oxide layer(Al 2 O 3 ) Galvanizing (coating Fe with Zn) Fe 2+ + 2e -  FeE = -0.44 V Zn 2+ + 2e -  ZnE = -0.76 V Zinc is more easily oxidized (Zn  Zn 2+ + 2e - E = +0.76 V)

70. Cathodic protection (sacrificial anode) Magnesium used in water pipes Magnesium rods used in hot water heaters

71. An iron gutter is nailed using aluminum nails. Will the nail or the iron gutter corrode first? Fe 2+ + 2e -  FeE = -0.44 V Al 3+ + 3e -  AlE = -1.66 V Al will corrode first (Al  Al 3+ + 3e -,E = +1.66 V)

72. Which of the following metals could provide cathodic protection to iron: AlCuNiZn

73. Electrolysis and Electroplating Plating of silver on silverware

74. Electrolytic cells Must run electricity through them Running a voltaic cell backwards Used to produce sodium metal Na + (aq) + e -  Na(s)-2.71 V Cl 2 (g) + 2e -  2Cl - (aq)+1.36 V

75. As a voltaic cell 2Na(s)  2Na + (aq) + 2e - +2.71 V Cl 2 (g) + 2e -  2Cl - (aq)+1.36 V 2Na(s) + Cl 2 (s)  2NaCl(aq)+4.07 V As an electrolytic cell 2Na + (aq) + 2e -  2Na(s)-2.71 V 2Cl - (aq)  Cl 2 (g) + 2e - -1.36 V 2NaCl(aq)  2Na(s) + Cl 2 (s)-4.07 V

76. Quantitative Electrolysis Electric current = Amperes 1 ampere = 1Coloumb I = Q 1 secondt 1 F = 96,500 C/mol – One mole of electrons has a charge of 96,500 C – One electron has a charge of 1.602 X 10 -19 C

77. What mass of aluminum can be produced in 1.00 hour by a current of 10.0 A? Al 3+ + 3e -  Al

78. Moles of e- = (36,000C)(1 mol e - ) = 0.373 mol e - (96,500 C) Al 3+ + 3e -  Al 0.373 mol Al 3+ + 3e -  Al 0.373 mol0.124 mol Al  3.36 g Al

79. Example 2 What mass of magnesium can be produced in 4000 s by a current of 60.0 A? Mg 2+ + 2e -  Mg ANS: 30.2 g Mg

80. What current is required to plate 6.10 grams of gold in 30.0 min? Au 3+ + 3e -  Au

81. How long would it take to plate 50.0 g of magnesium from magnesium chloride if the current is 100.0 A?

82. Given the following: Ag + (aq) + e -  Ag(s)+0.799V Fe 3+ (aq) + e -  Fe 2+ (aq)+0.771 V a.Write the reaction that occurs. b.Calculate the standard cell potential. c.Calculate  G rxn for the reaction from the cell potential. d.Calculate  for the reaction. e.Predict the sign of  S rxn. f.Sketch the cell, labeling anode, cathode, and the direction of electron flow.

83. Do SO 3 and SO 3 2- have the same molecular shape? How about SO 2 ?

84. 16. a) Not redox b) I oxidized (-1 to +5), Cl reduced (+1 to -1) c) S oxidized (+4 to +6), N reduced (+5 to +2) d) Br oxidized (-1 to 0), S reduced (+6 to +4) 20 a. Mo 3+ + 3e -  Mo b. H 2 O + H 2 SO 3  SO 4 2- + 2e - + 4H + c. 4H + + 3e - + NO 3 -  NO + 2H 2 O d. 4H + + 4e - + O 2  2H 2 O e. 4OH - + Mn 2+  MnO 2 + 2e - + 2H 2 O f. 5OH - + Cr(OH) 3  CrO 4 2- + 3e - + 4H 2 O g. 2H 2 O + 4e - + O 2  4OH -

85. 22. a. 3NO 2 - + Cr 2 O 7 2- +8H +  3NO 3 - + 2Cr 3+ + 4H 2 O b.2HNO 3 + 2S +H 2 O  2H 2 SO 3 + N 2 O c.2Cr 2 O 7 2- + 3CH 3 OH + 16H +  4Cr 3+ 3HCO 2 H + 11H 2 O d. 2MnO 4 - + 10Cl - + 16H +  2Mn 2+ + 5Cl 2 + 8H 2 O e.NO 2 - + 2Al + 2H 2 O  NH 4 + + 2AlO 2 - f.H 2 O 2 + 2ClO 2 + 2OH -  O 2 + 2ClO 2 - + 2H 2 O

86. 26. a) Al oxidzed, Ni2+ reduced b) Al  Al3+ + 3e-Ni2+ + 2e-  Ni c) Al anode, Ni cathode d) Al negative, Ni positive e) Electrons flow towards the Ni electrode f) Cations migrate towards Ni electrode

87. 34a) Cd is anode, Pd is cathod b) E red = 0.63 V 36 a) 2.87 V b) 3.21 Vc) -1.211 V d) 0.636V 38 a) 1.35 Vb) 0.29 V

88. 41 a) Mgb) Cac) H 2 d) H 2 C 2 O 4 42 a) Cl 2 b) Cd 2+ c) BrO 3 - d) O 3 44. a) Ce 3+ (weak reductant) b) Ca (strong reductant) c) ClO 3 - (strong oxidant) d) N 2 O 5 ( oxidant) 46a) H 2 O 2 strongest oxidizing agent b) Zn strongest reducing agent 50.a) 3.6 X 10 8 b) 10 41 c) 10 103 52. 0.292 V 54 a) 4 X 10 15 b) 2 X10 65 c) 7.3 X10 49

89. 62a) 2.35 Vb) 2.48 Vc) 2.27 V 64. a) 0.771 Vb) 1.266 V 88. a) 173 gb) 378 min 90.E = 1.10 VW max = -212 kJ/mol Cu W = -1.67 X 10 5 J

90. 1a) 14H + + Cr 2 O 7 2- + 3Fe  2Cr 3+ + 3Fe 2+ + 7H 2 O b)2Br - + F 2  2F - + Br 2 c)4OH - + 2Cr(OH) 3 + ClO 3 -  2CrO 4 2- + Cl - + 5H 2 O 2b) 0.463 Vc) -89.4 kJ/mold) 4.4 X 1015 e) 0.442 V 3)F 2 is str. oxidizing agent, Li, str. reducing agent 4)b) 78 minutes c) 1.19 g d) 0.695 g

91. In a measuring cup: 5 mL of oil 5 mL of ethanol 5 mL of 50% NaOH solution (approximately 30 drops). Place in beaker Heat the mixture, stirring with popsicle stick. Remove from heat. After ~5 minutes, add 10 mL of saturated salt solution. Collect some of the solid and test the pH of your soap. Compare the pH to that of commercial bar soap and liquid detergent solution. See if it lathers.