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

2. Oxidation-Reduction Reactions
An oxidation occurs when an atom or ion loses electrons.
A reduction occurs when an atom or ion gains electrons.

3. Oxidation-Reduction Reactions
One cannot occur without the other.

4. Oxidation-Reduction Reactions
Oxidation Is Loss of e- Reduction Gain of e-
2Ca + O2  2CaO
Ca begins neutral and ends as Ca2+. Loss of e-
O2 begins neutral and ends as O2-. Gain of e-.
Ca is oxidized and O2 is reduced.

5. Oxidation Numbers
To determine if an oxidation-reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.

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

7. Oxidation and Reduction
A species is oxidized when it loses electrons.
Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

8. Oxidation and Reduction
A species is reduced when it gains electrons.
Here, each of the H+ gains an electron and they combine to form H2.
OIL RIG or LEO the lion goes GER

9. 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.

10. Assigning Oxidation Numbers
Rules on page 139.
Elements in their elemental form have an oxidation number of 0.
The oxidation number of a monatomic ion is the same as its charge.

11. Assigning Oxidation Numbers
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.

12. Assigning Oxidation Numbers
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.

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

14. For the reaction given below, what substance is oxidized and what is reduced?
3 NO Cr2O H Cr NO H2O
N of NO2- is reduced, Cr of Cr2O72- is oxidized
N of NO2- is oxidized, Cr of Cr2O72- is reduced
O of NO2- is oxidized, Cr of Cr2O72- is reduced
Cr3+ is reduced, N of NO2- is oxidized
N of NO3- is oxidized, Cr3+ is reduced

15. For the reaction given below, what substance is oxidized and what is reduced?
3 NO Cr2O H Cr NO H2O
N of NO2- is reduced, Cr of Cr2O72- is oxidized
N of NO2- is oxidized, Cr of Cr2O72- is reduced
O of NO2- is oxidized, Cr of Cr2O72- is reduced
Cr3+ is reduced, N of NO2- is oxidized
N of NO3- is oxidized, Cr3+ is reduced

16. Activity Series More easily oxidized metals at top.
Look at the top of the series. Where are these elements on the per. table? More stable as elements or compounds?

17. Activity Series Metals can be oxidized by elements below it
Cu + 2Ag+  2Ag + Cu2+ is a possible reaction but 2Ag + Cu2+  Cu + 2Ag+ is not. Why?

18. Give it some thought…
Which of the following metals is the easiest to oxidize? Which is the most difficult to oxidize?
Gold
Lithium
Iron
Sodium
Aluminum

19. Give it some thought…
Which of the following metals will be oxidized by Pb(NO3)2? Zn Cu Fe

20. Give it some thought…
Does a reaction occur when an aqueous solution of NiCl2 is added to a test tube containing strips of metallic zinc?

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

22. 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.

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

24. Half-Reaction Method Balance each half-reaction.
Balance elements other than H and O.
Balance O by adding H2O.
Balance H by adding H+.
Balance charge by adding electrons.
Multiply the half-reactions by integers so that the electrons gained and lost are the same.

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

26. MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)
Half-Reaction Method
Consider the reaction between MnO4− and C2O42− :
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)

27. Half-Reaction Method MnO4− + C2O42-  Mn2+ + CO2
First, we assign oxidation numbers.
MnO4− + C2O42-  Mn2+ + CO2 +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.

28. Oxidation Half-Reaction
C2O42−  CO2
To balance the carbon, we add a coefficient of 2:
C2O42−  2 CO2

29. Oxidation Half-Reaction
C2O42−  2 CO2
The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.
C2O42−  2 CO2 + 2 e−

30. Reduction Half-Reaction
MnO4−  Mn2+
The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side.
MnO4−  Mn H2O

31. Reduction Half-Reaction
MnO4−  Mn H2O
To balance the hydrogen, we add 8 H+ to the left side.
8 H+ + MnO4−  Mn H2O

32. Reduction Half-Reaction
8 H+ + MnO4−  Mn H2O
To balance the charge, we add 5 e− to the left side.
5 e− + 8 H+ + MnO4−  Mn H2O

33. Combining the Half-Reactions
Now we evaluate the two half-reactions together:
C2O42−  2 CO2 + 2 e−
5 e− + 8 H+ + MnO4−  Mn H2O
To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.

34. Combining the Half-Reactions
5 C2O42−  10 CO e−
10 e− + 16 H+ + 2 MnO4−  2 Mn H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn H2O + 10 CO2 +10 e−

35. Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the electrons. Subtracting them, we are left with:
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn H2O + 10 CO2

36. Time to think…
Complete and balance this equation by the method of half-reactions:

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

38. 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.

39. Voltaic Cells

40. Voltaic Cells A typical cell looks like this.
The oxidation occurs at the anode.
The reduction occurs at the cathode.
AN OX
RED CAT

41. Voltaic Cells
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.

42. 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.

43. 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.

44. 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.

45. Time to think… Why do anions in a salt bridge move toward the anode?
Oxidation at anode, electrons being removed, anions migrate toward anode to maintain charge balance.

46. Time to think… The two half-reactions in a voltaic cell are
Indicate which reaction occurs at the anode and which at the cathode.
Which electrode is consumed in the cell reaction?
Which electrode is positive?

47. 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.

48. 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 Ecell.

49. Cell Potential
Cell potential is measured in volts (V).
1 V = 1 J C

50. Standard Reduction Potentials
Reduction potentials for many electrodes have been measured and tabulated.

51. 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−  H2 (g, 1 atm)

52. Finding E°cell
Select the half-reaction with the more positive voltage as the reduction.
Reverse the half-reaction with the less positive voltage. Don’t forget to change the sign when you flip the oxidation reaction.
Calculate the cell voltage by adding the reduction half reaction to the oxidation half reaction.

53. Step 1: Write out the two half reactions and look up their voltage.
What is the reaction that occurs when an Ag half cell is coupled with a Cu half cell? What is the reaction that occurs when an Ag half cell is coupled with a Cu half cell?
Step 1:
Write out the two half reactions and look up their voltage.
Ag+/Ag
Ag+ + 1e-  Ag(s) E°red(volts) =
Cu2+/Cu
Cu2+ + 2e-  Cu(s) E°red(volts) =
Select the reaction that has a larger voltage as reduction (your cathode)

54. What is the reaction that occurs when an Ag half cell is coupled with a Cu half cell? What is the reaction that occurs when an Ag half cell is coupled with a Cu half cell?
Step 2:
Reverse the reaction that has a lower voltage and make it your oxidation reaction.
Ag+/Ag
Ag+ + 1e-  Ag(s) E°red(volts) =
Cu /Cu2+
Cu(s)  Cu2+ + 2e- E°ox(volts) =
Step 3:
Add the reduction voltage to the oxidation voltage.
E°cell = E°red + E°ox
= = E°cell

55. Writing the complete reaction.
What is the reaction that occurs when an Ag half cell is coupled with a Cu half cell? What is the reaction that occurs when an Ag half cell is coupled with a Cu half cell?
Writing the complete reaction.
Ag+/Ag
2Ag+ + 2e-  E°red(volts) =
(NOTE: Unlike in thermochemistry, changing the coefficients does not change the voltage.)
Cu /Cu2+
Cu(s)  Cu2+ + 2e- E°ox(volts) =
OVERALL
2Ag+ + Cu(s)  Cu2+ + 2Ag(s) E°cell =

56. Analyzing the complete reaction.
What is the reaction that occurs when an Ag half cell is coupled with a Cu half cell? What is the reaction that occurs when an Ag half cell is coupled with a Cu half cell?
Analyzing the complete reaction.
OVERALL
2Ag+ + Cu(s)  Cu2+ + 2Ag(s) E°cell =
DON’T FORGET:
RED CAT (reduction at cathode)
AN OX (anode is oxidation)
Gets Reduced.
It is an oxidizing agent.
Gets Oxidized.
It is a reducing agent.

57. Standard Reduction Potentials
The strongest oxidizers have the most positive reduction potentials.
The strongest reducers have the most negative reduction potentials.

58. Time to think…

59. Time to think…

60. Time to think…
Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction
ANS: 0.79 V

61. Time to think…
Using data in Table 20.1, calculate the standard emf for a cell that employs the following overall cell reaction:
ANS: V

62. Oxidizing and Reducing Agents
The greater the difference between the two, the greater the voltage of the cell.

63. Time to think…
Using standard reduction potentials (Table 20.1), determine whether the following reaction are spontaneous under standard conditions.

64. Time to think…
Using the standard reduction potentials listed in Appendix E, determine which of the following reactions are spontaneous under standard conditions:
Reactions (b) and (c) are spontaneous.

65. Free Energy G for a redox reaction can be found by using the equation
G = −nFE
where n is the number of moles of electrons transferred, and F is a constant, the Faraday.
1 F = 96,485 C/mol = 96,485 J/V-mol

66. Free Energy Under standard conditions, G = −nFE
This is why your E°cell voltage MUST be positive for the reaction to be spontaneous.

68. Time to think…
Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change, G°, and the equilibrium constant, K, at room temperature (T = 298 K) for the reaction:
Go = x 105 J/mol
K = 9 x 1029

69. Time to think…

70. Nernst Equation Remember that G = G + RT ln Q This means
−nFE = −nFE + RT ln Q

71. Nernst Equation
Dividing both sides by −nF, we get the Nernst equation:
E = E − RT nF
ln Q
or, using base-10 logarithms,
E = E −
2.303 RT nF
log Q

72. Nernst Equation At room temperature (298 K), 2.303 RT F = 0.0592 V
Thus the equation becomes
E = E −
0.0592 n
log Q

73. Concentration Cells Ecell 
Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.
For such a cell,
would be 0, but Q would not.
Ecell 
Therefore, as long as the concentrations are different, E will not be 0.

74. Applications of Oxidation-Reduction Reactions

75. Batteries

76. Alkaline Batteries

77. Hydrogen Fuel Cells

78. Corrosion and…

79. …Corrosion Prevention

80. ELECTROLYSIS AND ELECTROLYTIC CELLS [NON spontaneous cells] Purpose: To produce purified forms of elements

81. Electrolytic cells Nonspontaneous Usually one cell
Two of the same electrodes (graphite is a good choice)
Must have a power source
Use energy to create a chemical change

82. Voltaic cells are spontaneous
Separated into two cells
Is a battery
Anode is negative
Cathode is positive
AnOx
Red Cat
Fat Cat
Electrolytic cells are forced to work by using a power source
Single container
NEEDS a battery
Anode is positive
Cathode is negative
AnOx
Red Cat
Fat Cat
EPA (electrolytic positive anode)

84. If there is no water present and
you have a pure molten ionic
compound, then…
The cation will be reduced (gain
electrons/go down in charge).
The anion will be oxidized (lose
electrons/go up in charge).

85. Cell with molten NaCl

86. Diagram an electrolytic cell of a molten solution of KI
Diagram an electrolytic cell of a molten solution of KI. Be sure to include anode, cathode, and overall cell reactions. Clearly state any observations and label electron flow.

87. Memorize these equations:
To predict products in aqueous solutions:
No IA or IIA metal will ever be reduced
No polyatomic ion will ever oxidize
Instead– use water
Memorize these equations:
Oxidation 2H2O  O H e-
Reduction 2H2O e-  H OH-

88. Cell with aqueous NaCl

89. If water is present and you
have an aqueous solution of the
ionic compound, then…
You’ll need to figure out if the ions
are reacting, or the water is reacting.
You can always look at a reduction
potential table to figure it out.

90. Draw a diagram of an electrolytic cell containing an aqueous solution of KI. Be sure to label anode, cathode, electron flow and show all half-reactions and the overall reaction. Clearly state any observations at each electrode.

91. Quantitative Aspects of Electrolysis

92. Faraday’s Law
The amount of substance being oxidized or reduced at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell.
Use dimensional analysis
Conversions to memorize:
1 volt = 1 Joule/Coulomb
1 Amp = 1 Coulomb/second (current)
Faraday = 96,500 Coulombs/mole of e-
mole of e- come from balanced redox equation

93. Calculate the number of grams of aluminum produced in 1
Calculate the number of grams of aluminum produced in 1.00 hr by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.
3.36 g

94. The half-reaction for the formation of magnesium metal upon electrolysis of molten MgCl2 is Mg2+ + 2e-  Mg. Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 x 103 s.
How many seconds would be required to produce 50.0g of Mg from MgCl2 if the current is A?
a) g Mg b) x 103 sec

95. A steady current of 1.00 ampere is passed through an electrolytic cell containing a 1 molar solution of AgNO3 and having a silver anode and a platinum cathode until 1.54 grams of silver is deposited.
How long does the current flow to obtain this deposit?
What weight of chromium would be deposited in a second cell containing 1-molar chromium III nitrate and having a chromium anode and a platinum cathode by the same current in the same time as was used in the silver cell?
If both electrodes were platinum in this second cell, what volume of O2 gas measured at STP would be released at the anode while the chromium is being deposited at the cathode? The current and the time are the same as in (a).