1. Chapter 20 Electrochemistry Lecture Presentation © 2012 Pearson Education, Inc.

2. Electrochemistry Redox – Bozeman #31 Electrochemistry – Bozeman #34

3. Electrochemistry 20.1 Electrochemical Reactions In electrochemical reactions, electrons are transferred from one species to another. We call these reactions “redox” because one species is “oxidized” (loses electrons) while the other is “reduced”. (gains electrons) © 2012 Pearson Education, Inc.

4. Electrochemistry Now, where’d those electrons go? 2 Fe + O 2  2 FeO 0 0 +2 -2

5. Electrochemistry Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers. © 2012 Pearson Education, Inc.

6. Electrochemistry 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. © 2012 Pearson Education, Inc.

7. Electrochemistry 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, which has an oxidation number of  1. –Hydrogen is  1 when bonded to a metal, and +1 when bonded to a nonmetal. © 2012 Pearson Education, Inc.

8. Electrochemistry 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. © 2012 Pearson Education, Inc.

9. Electrochemistry 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. © 2012 Pearson Education, Inc.

10. Electrochemistry  Examples: For polyatomic ions the sum of the oxidation #’s = charge Ex: SO 4 2- NH 4 + PO 4 3- Ex: O 2 K 2 O H 2 SO 4 (NH 4 ) 2 S 0-2 +1-2+1 -8+2? +6 -2 +1 -2+8 -6 -3 -2 -8 +5 _2_2 -2 -8 +6 +1 +4 -3

11. Electrochemistry Compounds are neutral Oxygen is usually -2 Gainer spot is negative O is +2 Peroxide: O = -1 Elemental state = 0 N = +4, O = -2

12. Electrochemistry Elemental State = 0 Cr = +6 in both

13. Electrochemistry 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. We say “LEO” says “GER” “LEO”  Loss of Electrons is Oxidation

14. Electrochemistry 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. “LEO” says “GER” “GER”  Gain of Electrons is Reduction

15. Electrochemistry 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. © 2012 Pearson Education, Inc.

16. Electrochemistry SnO 2 + Na  SnO + Na 2 O CuCl 2 + Cr  CrCl 3 + Cu Practice: +4 +20+1 Oxidized: Reduced: Oxidizing agent: Reducing agent: Na 0 Sn +4 Na 0 Oxidized: Reduced: Oxidizing agent: Reducing agent: +20 0+3 Cr 0 Cu +2 Cr 0 Sn gains 2 e - Na lose 1e - Cu gains 2 e - Cr loses 3e -

17. Electrochemistry Conserv. of charge means conserv. of electrons Oxidized = “LEO” GIN-LIP LEO-GER

18. Electrochemistry Ni 0  Ni +2 GIN- LIP LEO, LIP Answer is on reactant side GIN-LIP LEO-GER

19. Electrochemistry Cannot happen separately Look for elemental state GIN-LIP LEO-GER

20. Electrochemistry Look for something in its Elemental state No elemental state? No change in Ox. #’s GIN-LIP LEO-GER

21. Electrochemistry Ni 0  Ni +2 GIN - LIP Loss Is Positive Mn +4  Mn +2 GIN- LIP LEO – GER Gain e - Reduction Gain Is Negative GIN-LIP LEO-GER

22. Electrochemistry Oxidizing agents gain electrons GER Ox. Agent is reduced GER….GIN Givers are losers GIN-LIP LEO-GER

23. Electrochemistry Givers are losers Red agents lose e - Are oxidized, LEO LIP Red agents are losers Get stronger left to right on PT Harder to lose e - GIN-LIP LEO-GER

24. Electrochemistry

25. Half Reactions Half reaction: equation showing only the loss (oxidation) Or gain (reduction) of electrons Ex: Na + Cl 2  NaCl Oxidation: 001-1+ Reduction: + 1 e - Na 0  Na + Cl 0  Cl - + 1 e - Notice: Balance of charge AND mass ( 0 ) = (+1) + (-1) ( 0 ) + (-1) = (-1)

26. Electrochemistry SnO 2 + Na  SnO + Na 2 O CuCl 2 + Cr  CrCl 3 + Cu Practice: +4 +20+1 +20 0+3 -2 Oxidation: Reduction: Na 0  Na + + 1 e - Sn +4  Sn +2 + 2 e - Oxidation: Reduction: Cr 0  Cr +3 + 3 e - Cu +2  Cu 0 + 2 e -

27. Electrochemistry Regents: Balancing Redox reactions Ex: Na + Cl 2  NaCl Na 0  Na + + e - Oxidation: Reduction: Cl 0 + e-  Cl-22 2 ( ) + 2 e - 2 Na 0  2 Na + 22 2 Redox reactions can be balanced by balancing the loss of electrons with the gain of electrons We balance the atoms first, then the electrons Insert the new coefficients into the equation and check 2 e - gained by Cl 2 requires 2 e - lost by 2 Na

28. Electrochemistry KNO 3 + C  CO 2 + NO 2 + K 2 O Practice: Write the half reactions (and balance?) +5 0 +4 Oxidation: Reduction: N +5  N +4 + e - C 0  C +4 +4 e - 4 ( ) 4 N +5 + 4 e -  4 N +4 4 42 Gain 1e - Lose 4 e - 4x

29. Electrochemistry HNO 3 + I 2  HIO 3 + NO 2 + H 2 O 0 +5 +4 Oxidation:I 0  I +5 + e - Reduction: N +5  N +4 + 1 e-10 ( ) 10 N +5 + 10 e -  10 N +4 10 4 2 2 2 Try Another One ?

30. Electrochemistry Ionic equations Fe + Cu +2  Cu + Fe +3 Fe + CuCl 2  Cu + FeCl 3 0 0 Oxidation: Fe 0  Fe +3 Reduction: Cu +2  Cu 0 2 ( ) 3 ( ) 32 6 e- lost = 6 e- gained 2e- gained 3e- lost + 3 e- + 2 e - Spectators are missing in NIE’s so you need a different strategy 3 2

31. Electrochemistry 0+10+1 LEO - GER, GIN - LIP Cl 0  Cl - + 1e - Check: GIN-LIP LEO-GER 0 =

32. Electrochemistry 00 Cr +3  Cr 0 + 3e - Zn 0  Zn +2 + 2e - GIN-LIP LEO-GER

33. Electrochemistry 2 e - 3 e - 22 3 3 4Al 0  2Al 2 +3 + 12e - +3

34. Electrochemistry 20.2 Balancing Oxidation-Reduction Equations Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method. Ex: Sn 2+ + Fe 3+  Sn 4+ + Fe 2+ Oxidation: Sn 2+  Sn 4+ + 2e - Reduction: Fe 3+ + 1e -  Fe 2+

35. Electrochemistry © 2012 Pearson Education, Inc. This method 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. 20.2 Balancing Oxidation-Reduction Equations Ex: Sn 2+ + 2Fe 3+  Sn 4+ + 2Fe 2+ Oxidation: Sn 2+  Sn 4+ + 2e - Reduction: 2Fe 3+ + 2e -  2Fe 2+

36. Electrochemistry The Half-Reaction Method © 2012 Pearson Education, Inc. Consider the titration of Na 2 C 2 O 4 with KMnO 4(aq) MnO 4  (aq) + C 2 O 4 2  (aq)  Mn 2+ (aq) + CO 2 (aq)

37. Electrochemistry MnO 4  + C 2 O 4 2   Mn 2+ + CO 2 +7+3+4+2 1.Assign oxidation numbers to determine what is oxidized and what is reduced. Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized.

38. Electrochemistry 2. Write the oxidation and reduction half-reactions. Oxidation: C 2 O 4 2   CO 2 Reduction: MnO 4   Mn 2+ MnO 4  + C 2 O 4 2   Mn 2+ + CO 2 +7+3+4+2

39. Electrochemistry To balance the carbon, we add a coefficient of 2: C 2 O 4 2   2CO 2 © 2012 Pearson Education, Inc. 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.

40. Electrochemistry C 2 O 4 2   2CO 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   2CO 2 + 2e  © 2012 Pearson Education, Inc.

41. Electrochemistry 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+ + 4H 2 O © 2012 Pearson Education, Inc.

42. Electrochemistry MnO 4   Mn 2+ + 4H 2 O To balance the hydrogen, we add 8H + to the left side: 8H + + MnO 4   Mn 2+ + 4H 2 O © 2012 Pearson Education, Inc.

43. Electrochemistry 8H + + MnO 4   Mn 2+ + 4H 2 O To balance the charge, we add 5e  to the left side: 5e  + 8H + + MnO 4   Mn 2+ + 4H 2 O © 2012 Pearson Education, Inc.

44. Electrochemistry C 2 O 4 2   2CO 2 + 2e  5e  + 8H + + MnO 4   Mn 2+ + 4H 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: © 2012 Pearson Education, Inc. 4. Balance electrons lost and gained 5( ) 2( )

45. Electrochemistry 5C 2 O 4 2   10CO 2 + 10e  10e  + 16H + + 2MnO 4   2Mn 2+ + 8H 2 O When we add these together, we get: 10e  + 16H + + 2MnO 4  + 5C 2 O 4 2   2Mn 2+ + 8H 2 O + 10CO 2 +10e  © 2012 Pearson Education, Inc. 5. Add the half-reactions, subtracting things that appear on both sides.

46. Electrochemistry Combining the Half-Reactions 10e  + 16H + + 2MnO 4  + 5C 2 O 4 2   2Mn 2+ + 8H 2 O + 10CO 2 +10e  The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16H + + 2MnO 4  + 5C 2 O 4 2   2Mn 2+ + 8H 2 O + 10CO 2 © 2012 Pearson Education, Inc.

47. Electrochemistry

51. 20.3 Voltaic Cells In spontaneous oxidation- reduction (redox) reactions, electrons are transferred and energy is released. © 2012 Pearson Education, Inc.

52. Electrochemistry Redox in Action: Voltaic cells How is a battery Like a waterfall? Gravity pulls water down Kinetic Energy is generated as electricity Electrostatic forces pull electrons from “weaker” atoms like Zn to stronger ions like Cu 2+ generating electricity

53. Electrochemistry Zn (s) + CuSO 4(aq)  Cu (s) + ZnSO 4(aq) 0 0 +2 2 e - Oxidation: Zn 0  Zn +2 + 2 e - Reduction: Cu +2 + 2 e -  Cu 0 Electrons flow spontaneously from zinc to copper as predicted from the activity series e- The reaction involved is a simple single replacement, but the two participants (Zn and Cu 2+ ) are placed in separate containers.

54. Electrochemistry Voltaic Cells We can use this energy to do work if we make the electrons flow through an external device. We call such a setup a voltaic cell. © 2012 Pearson Education, Inc. You may recognize this “cell” as the typical double A or triple A “battery”. They are technically called “cells”

55. Electrochemistry A typical cell looks like this. The oxidation occurs at the anode. The reduction occurs at the cathode. Remember the phrase “RED CAT, AN OX” Electrons are lost at the “anode” Electrons are gained at the “cathode”

56. Electrochemistry Once even 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. © 2012 Pearson Education, Inc. e-e- + -

57. Electrochemistry Therefore, we use a “salt bridge”, - a U-shaped tube that contains a salt solution, to keep the charges balanced. –Anions (NO 3 - ) move toward the anode. –Cations (Na + ) move toward the cathode. NO 3 - This anion flow balances the electron flow in the wire Excess positive charge starts to build up as Zn 2+ is produced on this side Positive ions are lost on this side as Cu 2+ ion become Cu(s) leaving excess negative ions Note: Na + ions flow also, but I don’t think of this as being as important + -

58. Electrochemistry 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. A porous barrier can substitute for a salt bridge. Solutions don’t mix but are in contact and ions can be pulled across to balance charge The anode gradually dissolves as Zn is oxidized

59. Electrochemistry 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. The cathode gets larger as Cu 2+ is reduced and new Cu(s) collects on its surface A Cell that uses Zn and Cu 2+ is called a Daniel cell, but any pair of chemicals that replace spontaneously can be used – but each pair creates a different “voltage”

60. Electrochemistry

61. -2 -14+12 _2_2 +6 +3 Gaining 2(3 e - ) 0 Losing 6(1 e - )

62. Electrochemistry

63. I - (aq) anode Cr 2 O 7 2- (aq) cathode e-e-

64. Electrochemistry Summary At the anode electrons are lost: At the cathode electrons are gained:Cr 2 O 7 2-  Cr 3+ 6 I -  3I 2 + 6 e - + 7H 2 O+ 14H + + 6 e - Balance with 6 electrons lost Balance O’s with H 2 O …and then balance H 2 O with H + Finally balance with 6 electrons gained I - (aq) anode Cr 2 O 7 2- (aq) cathode e-e-

65. Electrochemistry For the AP (and probably Regents also) it is not necessary to assign charges to the electrodes. Its an academic exercise at best anyway.

66. Electrochemistry Chemical change makes electricity AnOx = oxidation = LEO, loss of electrons Chemical change makes electricity AnOx = loss of e - Take a break with some regents level questions

67. 67 Pb loses e - to Cu

68. 68 Pb loses e - Pb to Ag Pb  Pb 2+ + 2e -

69. Electrochemistry 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.

70. Electrochemistry 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.

71. Electrochemistry Cell Potential Cell potential (Emf) is measured in volts (V). 1 Volt = 1 Joule Coloumb (energy) (electron charge) (force) “The greater the energy separation between the points in the circuit (b/t anode and cathode) the greater the electromotive force exerted by the electrons” per unit of

72. Electrochemistry Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated. e-e- Electrons flow from lower to higher reduction (gain) potential Note: this table is different from the activity series (regents ref tbl J)

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

74. Electrochemistry Standard Cell Potentials The cell potential at standard conditions can be found through this equation: E cell  = E red (cathode)  E red (anode)  © 2012 Pearson Education, Inc.

75. Electrochemistry Cell Potentials For the oxidation in this cell, For the reduction, E red =  0.76 V  E red = +0.34 V  © 2012 Pearson Education, Inc.

76. Electrochemistry Cell Potentials E cell  = E red  (cathode)  E red  (anode) = +0.34 V  (  0.76 V) = +1.10 V

77. Electrochemistry

87. Oxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials. © 2012 Pearson Education, Inc.

88. Electrochemistry Oxidizing and Reducing Agents The greater the difference between the two, the greater the voltage of the cell. © 2012 Pearson Education, Inc.

89. Electrochemistry

92. Weak: Good Reducing (losing) agents Stronger: better Oxidizing (gainer) agents strongest: best Oxidizing (gaining) agents Regents reference table

93. Electrochemistry Reduce = give e - to. Who will give e - to Ni? Best gainer?

94. Electrochemistry The weakest gainer is the best loser. Take e - from Zn, but not Pb?

95. Electrochemistry Will a reaction occur? 20.5 E 0 and ΔG 0 Ex: 2Ag(s) + Ni + (aq)  2Ag + (aq) + Ni (s) Ox: 2Ag(s)  2Ag + + 2e - E 0 Red = + 0.80 V Red: Ni + (aq) + 2e -  Ni (s) E 0 Red = - 0.28 V E 0 cell = E 0 Red Ni + - E 0 Red Ag = (- 0.28 V) - (0.80 V) = -1.08 E 0 cell negative value is not favorable (spontaneous)

96. Electrochemistry Activity series vs. E 0 volt table? Higher replaces lower

97. Electrochemistry Free Energy and E  G for a redox reaction can be found by using the equation  G =  nFE where n is the number of moles of electrons transferred, F is Faraday’s constant: 1 Faraday = 96,485 Coulombs / mole e - 96,485 Joules/Volt-mole -(EFn) =  G -(Joule energy/Coulomb)( Coulombs/mole e - )(# moles e - ) In other words, the greater the Voltage (E) and the greater the moles of electrons transferred, the greater the Free energy change

98. Electrochemistry  G =  nFE Notice: When E is (+) spontaneous (electrons flowing “downhill”)  G will be (-) also spontaneous (loss of free energy makes reaction favorable – ie reaction will happen) (+)(-)

99. Electrochemistry K eq and E 0  G 0 =  nFE 0  G 0 =  RTlnK K = [products] [reactants]  RTlnK =  nFE 0 We can calculate K if we know E 0 Through calculating  G 0 Notice: As K increases, E 0 voltage will increase

100. Electrochemistry

101. I.e. voltage doesn’t depend on stoichiometry

102. Electrochemistry So its technically expressed as -170 kJ/mol Rxn “mol Rxn: “mole reactions’ means molar quantities expressed in the balanced equation

103. Electrochemistry

107. Don’t forget to write the formulas! ΔG 0 =-nFE 0 ΔG 0 = -RTlnK

108. Electrochemistry 20.6 Nonstandard cells: Nernst Equation Remember that  G =  G  + RT ln Q This means −nFE = −nFE  + RT ln Q For calculating ΔG under nonstandard conditions: aka not at equilibrium

109. Electrochemistry Nernst Equation Dividing both sides by −nF, we get the Nernst equation: E = E  − RT nF ln Q

110. Electrochemistry For the Daniel Cell Zn 0 + Cu 2+  Zn 2+ + Cu 0 [Zn 2+ ] product Q = [Cu 2+ ] reactant E = E  − RT nF ln Q As [Cu 2+ ] increases, Q decreases; And E increases E = E  − RT nF ln small E = E 0 - - #

111. Electrochemistry For the Daniel Cell Zn 0 + Cu 2+  Zn 2+ + Cu 0 [Zn 2+ ] product Q = [Cu 2+ ] reactant E = E  − RT nF ln Q As [Cu 2+ ] decreases, Q increases; And E decreases E = E  − RT nF ln large E = E 0 - + #

112. Electrochemistry For the Daniel Cell Zn 0 + Cu 2+  Zn 2+ + Cu 0 [Zn 2+ ] product Q = [Cu 2+ ] reactant As Q gets smaller, further from equilibrium E 0 voltage increases E = E  − RT nF ln Q

113. Electrochemistry 20.9 Electrolysis Reaction made to occur by external electrical current

114. Electrochemistry Electrolysis Negative ions migrate to anode are oxidized

115. Electrochemistry Electrolysis Positive ions migrate to cathode and are reduced

116. Electrochemistry Electrolysis Salt is decomposed into separate elements

117. Electrochemistry Electrolysis Notice: half reactions combine to form overall reaction 2 NaCl(l)  2 Na(s) + Cl 2 (g) 2 moles of electrons required

118. Electrochemistry Amperage = Charge / unit of time 1 amp = 1 coulomb/ second Electric current q t I =

119. Electrochemistry

123. Electrolytic Cell for Electroplating