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
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3. Half–Reactions
The overall reaction is split into two half–reactions, one involving oxidation and one involving reduction.
8H+ + MnO4– + 5Fe Mn2+ + 5Fe3+ + 4H2O
Reduction: 8H+ + MnO4– + 5e– Mn2+ + 4H2O
Oxidation: 5Fe Fe3+ + 5e–
Identify the species being oxidized and reduced and also the oxidizing and reducing agents in the equation above.
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4. For each half–reaction: Balance all the elements except H and O.
The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
Write separate equations for the oxidation and reduction half–reactions.
For each half–reaction:
Balance all the elements except H and O.
Balance O using H2O.
Balance H using H+.
Balance the charge using electrons.
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5. Add the half–reactions, and cancel identical species.
The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
If necessary, multiply one or both balanced half–reactions by an integer to equalize the number of electrons transferred in the two half–reactions.
Add the half–reactions, and cancel identical species.
Check that the elements and charges are balanced.
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6. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
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7. Interactive Example 18.1 - Balancing Oxidation–Reduction Reactions (Acidic)
Potassium dichromate (K2Cr2O7) is a bright orange compound that can be reduced to a blue-violet solution of Cr3+ ions
Under certain conditions, K2Cr2O7 reacts with ethanol (C2H5OH) as follows:
Balance this equation using the half-reaction method

8. Interactive Example 18.1 - Solution
The reduction half-reaction is:
Chromium is reduced from an oxidation state of +6 in Cr2O72– to one of the +3 in Cr3+
The oxidation half-reaction is:
Carbon is oxidized from an oxidation state of –2 in C2H5OH to +4 in CO2 8

9. Interactive Example 18.1 - Solution (Continued 1)
Balancing all elements except hydrogen and oxygen in the first half-reaction
Balancing oxygen using H2O
Balancing hydrogen using H+ 9

10. Interactive Example 18.1 - Solution (Continued 2)
Balancing the charge using electrons
Next, we turn to the oxidation half-reaction
Balancing carbon 10

11. Interactive Example 18.1 - Solution (Continued 3)
Balancing oxygen using H2O
Balancing hydrogen using H+
We then balance the charge by adding 12e– to the right side 11

12. Interactive Example 18.1 - Solution (Continued 4)
In the reduction half-reaction, there are 6 electrons on the left-hand side, and there are 12 electrons on the right-hand side of the oxidation half-reaction
Thus, we multiply the reduction half-reaction by 2 to give: 12

13. Interactive Example 18.1 - Solution (Continued 5)
Adding the half-reactions and canceling identical species 13

14. Interactive Example 18.1 - Solution (Continued 6)
Check that elements and charges are balanced 14

15. Br–(aq) + MnO4–(aq) Br2(l)+ Mn2+(aq)
EXERCISE!
Balance the following oxidation–reduction reaction that occurs in acidic solution.
Br–(aq) + MnO4–(aq) Br2(l)+ Mn2+(aq)
10Br–(aq) + 16H+(aq) + 2MnO4–(aq) 5Br2(l)+ 2Mn2+(aq) + 8H2O(l)
10Br-(aq) + 16H+(aq) + 2MnO4-(aq)  5Br2(l)+ 2Mn2+(aq) + 8H2O(l)
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16. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
Use the half–reaction method as specified for acidic solutions to obtain the final balanced equation as if H+ ions were present.
To both sides of the equation obtained above, add a number of OH– ions that is equal to the number of H+ ions. (We want to eliminate H+ by forming H2O.)
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17. Check that elements and charges are balanced.
The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
Form H2O on the side containing both H+ and OH– ions, and eliminate the number of H2O molecules that appear on both sides of the equation.
Check that elements and charges are balanced.
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18. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
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19. Balance this equation using the half-reaction method
Interactive Example Balancing Oxidation–Reduction Reactions (Basic)
Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores
An aqueous solution containing cyanide ion is often used to extract the silver using the following reaction that occurs in basic solution:
Balance this equation using the half-reaction method

20. Interactive Example 18.2 - Solution
Balance the equation as if H+ ions were present
Balance the oxidation half-reaction
Balance carbon and nitrogen
Balance the charge

21. Interactive Example 18.2 - Solution (Continued 1)
Balance the reduction half-reaction:
Balance oxygen
Balance hydrogen
Balance the charge

22. Interactive Example 18.2 - Solution (Continued 2)
Multiply the balanced oxidation half-reaction by 4
Add the half-reactions, and cancel identical species

23. Interactive Example 18.2 - Solution (Continued 3)
Add OH– ions to both sides of the balanced equation to eliminate the H+ ions
We need to add 4OH– to each side
4H2O(l)

24. Interactive Example 18.2 - Solution (Continued 4)
Eliminate as many H2O molecules as possible
Check that elements and charges are balanced

25. Figure 18.2 - Galvanic Cells
Galvanic cells can contain a salt bridge
Galvanic cells can contain a porous-disk connection 25

26. Galvanic Cell - Components
A salt bridge or a porous disk is used to permit ions to flow without extensive mixing of the solutions
Salt bridge: Contains a strong electrolyte in a U-tube
Porous disk: Contains tiny passages that allow hindered flow of ions 26

27. Galvanic Cell - Components (Continued)
Anode: Electrode compartment where oxidation occurs
Cathode: Electrode compartment where reduction occurs 27

28. Cell Potential (Ecell)
Pull or driving force on electrons
Termed as the electromotive force (emf) of the cell
Volt (V): Unit of electrical potential
Defined as 1 joule of work per coulomb of charge transferred 28

29. Measuring Cell Potential
Use a voltmeter
Voltmeter: Draws current via a known resistance
Maximum cell potential can be ascertained by measuring it under zero current
Potentiometer: Variable voltage device, which is powered by an external source, inserted in opposition to the cell potential 29

30. AP Learning Objectives, Margin Notes and References
LO 3.12 The student can make qualitative or quantitative predictions about galvanic or electrolytic reactions based on half-cell reactions and potentials and/or Faraday’s laws.
LO 3.13 The student can analyze data regarding galvanic or electrolytic cells to identify properties of the underlying redox reactions.
LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

31. Galvanic Cell
All half-reactions are given as reduction processes in standard tables.
Table 18.1
1 M, 1atm, 25°C
When a half-reaction is reversed, the sign of E° is reversed.
When a half-reaction is multiplied by an integer, E° remains the same.
A galvanic cell runs spontaneously in the direction that gives a positive value for E°cell.
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32. Table 18.1 - Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions32

33. Table Standard Reduction Potentials at 25°C (298 K) for Many Common Half-Reactions (Continued)

34. Example: Fe3+(aq) + Cu(s) → Cu2+(aq) + Fe2+(aq)
Half-Reactions:
Fe3+ + e– → Fe2+ E° = 0.77 V
Cu2+ + 2e– → Cu E° = 0.34 V
To balance the cell reaction and calculate the cell potential, we must reverse reaction 2.
Cu → Cu2+ + 2e– – E° = – 0.34 V
Each Cu atom produces two electrons but each Fe3+ ion accepts only one electron, therefore reaction 1 must be multiplied by 2.
2Fe3+ + 2e– → 2Fe E° = 0.77 V
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35. Overall Balanced Cell Reaction
2Fe3+ + 2e– → 2Fe E° = 0.77 V (cathode)
Cu → Cu2+ + 2e– – E° = – 0.34 V (anode)
Balanced Cell Reaction:
Cu + 2Fe3+ → Cu2+ + 2Fe2+
Cell Potential:
E°cell = E°(cathode) – E°(anode)
E°cell = 0.77 V – 0.34 V = 0.43 V
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36. Line Notation Used to describe electrochemical cells.
Anode components are listed on the left.
Cathode components are listed on the right.
Separated by double vertical lines which indicated salt bridge or porous disk.
The concentration of aqueous solutions should be specified in the notation when known.
Example: Mg(s)|Mg2+(aq)||Al3+(aq)|Al(s)
Mg → Mg2+ + 2e– (anode)
Al3+ + 3e– → Al (cathode)
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37. Interactive Example 18.3 - Galvanic Cells
Consider a galvanic cell based on the following reaction:
The half-reactions are:
Give the balanced cell reaction, and calculate E°for the cell

38. Interactive Example 18.3 - Solution
The half-reaction involving magnesium must be reversed and since this is the oxidation process, it is the anode
Also, since the two half-reactions involve different numbers of electrons, they must be multiplied by integers

39. Interactive Example 18.3 - Solution (Continued)

40. Description of a Galvanic Cell
The cell potential (always positive for a galvanic cell where E°cell = E°(cathode) – E°(anode)) and the balanced cell reaction.
The direction of electron flow, obtained by inspecting the half–reactions and using the direction that gives a positive E°cell.
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41. Description of a Galvanic Cell
Designation of the anode and cathode.
The nature of each electrode and the ions present in each compartment. A chemically inert conductor is required if none of the substances participating in the half–reaction is a conducting solid.
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42. Example 17-4 - Description of a Galvanic Cell
Describe completely the galvanic cell based on the following half-reactions under standard conditions:
Ag+ + e– → Ag E° = 0.80 V (1)
Fe3+ + e– → Fe2+ E° = 0.77 V (2)

43. Since a positive value is required, reaction (2) must run in reverse
Example Solution
Since a positive value is required, reaction (2) must run in reverse
Ag+ + e– → Ag E°(cathode) = 0.80 V
Fe2+ → Fe3+ + e– –E° (anode) = – 0.77 V
Ag+ (aq) + Fe2+(aq) → Fe3+(aq) + Ag(s) = 0.03 V

44. Example 17-4 - Solution (Continued 1)
Ag+ receives electrons, and Fe2+ loses electrons in the cell reaction
The electrons will flow from the compartment containing Fe2+ to the compartment containing Ag+
Oxidation occurs in the compartment containing Fe2+ (electrons flow from Fe2+ to Ag+)
Hence this compartment functions as the anode
Reduction occurs in the compartment containing Ag+, so this compartment functions as the cathode

45. Example 17-4 - Solution (Continued 2)
The electrode in the Ag/Ag+ compartment is silver metal, and an inert conductor, such as platinum, must be used in the Fe2+/Fe3+ compartment
Appropriate counterions are assumed to be present
Line notation
Pt(s)| Fe2+(aq), Fe3+(aq)|| Ag+(aq)|Ag(s)

46. Sketch a cell using the following solutions and electrodes. Include:
CONCEPT CHECK!
Sketch a cell using the following solutions and electrodes. Include:
The potential of the cell
The direction of electron flow
Labels on the anode and the cathode
Zn electrode in 1.0 M Zn2+(aq) and Cu electrode in 1.0 M Cu2+(aq)
Zinc is the anode, copper is the cathode (electrons flow from zinc to copper). The cell potential is 1.10 V.
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47. CONCEPT CHECK!
Consider the cell from part b. What would happen to the potential if you increase the [Cu2+]? Explain. The cell potential should increase.
Since the copper(II) ion is the reactant in the overall equation of the cell, the cell potential should increase (LeChâtelier's principle applies here).
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48. Work and Cell Potential
Cell potential (E) and work (w) have opposite signs
In any real, spontaneous process some energy is always wasted
Actual work realized is always less than the calculated maximum 48

49. Work and Cell Potential (Continued)
Actual work done is given by the following equation:
E - Actual potential difference at which current flows
q - Quantity of charge in coulombs transferred
Faraday (F): Charge on 1 mole of electrons
F = 96,485 C/mol e– 49

50. Cell Potential and Free Energy
Change in free energy equals the maximum useful work that can be obtained from a process
wmax = ΔG
For a galvanic cell
wmax = – qEmax = ΔG
Since q = nF
Therefore, ΔG = – qEmax = – nFEmax 50

51. Maximum Cell Potential
Directly related to the free energy difference between the reactants and the products in the cell
ΔG° = –nFE°
Galvanic cell will run in the direction that provides a positive Ecell value
Positive Ecell value implies negative ΔG value, which is the condition for spontaneity 51

52. Interactive Example 18.5 - Calculating ΔG°for a Cell Reaction
Using the data in Table 18.1, calculate ΔG°for the following reaction:
Is this reaction spontaneous?

53. Interactive Example 18.5 - Solution
The half-reactions are:
We can calculate ΔG° from the equation
ΔG° = –nFE°

54. Interactive Example 18.5 - Solution (Continued 1)
Since two electrons are transferred per atom in the reaction, 2 moles of electrons are required per mole of reactants and products
Thus, n = 2 mol e–, F = 96,485 C/mol e–, and E° = 0.78 V = 0.78 J/C

55. Interactive Example 18.5 - Solution (Continued 2)
The process is spontaneous, as indicated by both the negative sign of ΔG° and the positive sign of
This reaction is used industrially to deposit copper metal from solutions resulting from the dissolving of copper ores

56. AP Learning Objectives, Margin Notes and References
LO 3.12 The student can make qualitative or quantitative predictions about galvanic or electrolytic reactions based on half-cell reactions and potentials and/or Faraday’s laws.
LO 3.13 The student can analyze data regarding galvanic or electrolytic cells to identify properties of the underlying redox reactions.
LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

57. Interactive Example 18.7 - The Effects of Concentration on E
For the cell reaction
Predict whether Ecell is larger or smaller than for the following case:
[Al3+] = 2.0 M, [Mn2+] = 1.0 M
[Al3+] = 1.0 M, [Mn2+] = 3.0 M 57

58. Interactive Example 18.7 - Solution
A product concentration has been raised above 1.0 M
This will oppose the cell reaction and will cause Ecell to be less than (Ecell < 0.48 V)
A reactant concentration has been increased above 1.0 M
Ecell will be greater than (Ecell > 0.48 V) 58

59. Concentration Cells
Cells in which both compartments have the same components at different concentrations
Cell potential is a factor of difference in concentration
Voltages are small 59

60. Example 18.8 - Concentration Cells
Determine the direction of electron flow, and designate the anode and cathode for the cell represented in the figure on the right: 60

61. Example Solution
The concentrations of Fe2+ ion in the two compartments can (eventually) be equalized by transferring electrons from the left compartment to the right
This will cause Fe2+ to be formed in the left compartment, and iron metal will be deposited (by reducing Fe2+ ions to Fe) on the right electrode 61

62. Example 18.8 - Solution (Continued)
Since electron flow is from left to right, oxidation occurs in the left compartment (the anode) and reduction occurs in the right (the cathode) 62

63. Given in a form that is valid at 25°C
Nernst Equation
Explains the relationship between cell potential and concentrations of cell components
Given in a form that is valid at 25°C
Resulting potential calculated from this equation is the maximum potential before any current flow 63

64. Nernst Equation and Equilibrium
A cell will spontaneously discharge until it reaches equilibrium
Q = K (equilibrium constant)
Ecell = 0 64

65. Nernst Equation and Equilibrium (Continued)
Dead battery
Battery in which the cell reaction has reached equilibrium
At equilibrium:
Both cell compartments have the same free energy
ΔG = 0
Cell can no longer work 65

66. CONCEPT CHECK!
Explain the difference between E and E°. When is E equal to zero? When the cell is in equilibrium ("dead" battery). When is E° equal to zero? E  is equal to zero for a concentration cell.
ε is the cell potential at any condition, and ε is the cell potential under standard conditions (1.0 M or 1 atm, and 25C).
ε equals zero when the cell is in equilibrium ("dead" battery). ε is equal to zero for a concentration cell.
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67. Determining Ion Concentration
Cell potentials can help determine concentration of an ion
pH meter - Device used to measure concentration using observed potential
Composed of:
Standard electrode of known potential
Glass electrode: Changes potentials based on H+ ion concentration in a solution
Potentiometer

68. Figure 18.12 - A Glass Electrode
Electrical potential
Depends on the difference in [H+] between the reference solution and the solution into which the electrode is dipped
Varies with the pH of the solution being tested

69. Group of galvanic cells that are connected in series
Battery
Group of galvanic cells that are connected in series
Potentials of each individual cell add up to give the total battery potential
Essential source of portable power in today’s society 69

70. Nickel–cadmium battery Lithium-ion battery
Types of Battery
Lead storage battery
Dry cell battery
Silver cell
Mercury cell
Nickel–cadmium battery
Lithium-ion battery 70

71. Nickel–Cadmium and Lithium-Ion Batteries
Nickel–cadmium battery
Lithium-ion batteries
Li+ ions migrate from the cathode to the anode and enter the interior as the battery is charged
Charge-balancing electrons travel to the anode through the external circuit in the charger
The opposite process occurs on discharge

72. Galvanic cells for which reactants are continuously supplied
Fuel Cells
Galvanic cells for which reactants are continuously supplied
The image depicts a hydrogen–oxygen fuel cell
Anode reaction
2H2 + 4OH– → 4H2O + 4e–
Cathode reaction
4e– + O2 + 2H2O → 4OH–

73. AP Learning Objectives, Margin Notes and References
LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

74. Involves oxidation of the metal.
Process of returning metals to their natural state – the ores from which they were originally obtained.
Involves oxidation of the metal.
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75. The Electrochemical Corrosion of Iron
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76. Corrosion Prevention
Application of a coating (like paint or metal plating)
Galvanizing - Process in which steel is coated with zinc to prevent corrosion
Alloying
Cathodic Protection
Protects steel in buried fuel tanks and pipelines.
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77. Cathodic Protection
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78. Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
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79. Figure 18.19 - A Galvanic Cell and an Electrolytic Cell

80. 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
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81. Interactive Example 18.11 - Electroplating
How long must a current of 5.00 A be applied to a solution of Ag+ to produce 10.5 g silver metal?

82. Interactive Example 18.11 - Solution
In this case, we must use the steps given earlier in reverse

83. Interactive Example 18.11 - Solution (Continued 1)
Each Ag+ ion requires one electron to become a silver atom
Thus 9.73 ×10 –2 mole of electrons is required, and we can calculate the quantity of charge carried by these electrons

84. Interactive Example 18.11 - Solution (Continued 2)
The 5.00 A (5.00 C/s) of current must produce 9.39 × 103 C of charge
Thus,

85. Interactive Example 18.12 - Relative Oxidizing Abilities
An acidic solution contains the ions Ce4+, VO2+, and Fe3+
Using the E°values listed in Table 18.1, give the order of oxidizing ability of these species, and predict which one will be reduced at the cathode of an electrolytic cell at the lowest voltage

86. Interactive Example 18.12 - Solution
The half-reactions and E°values are as follows:
The order of oxidizing ability is therefore:
Ce4+ > VO2+ > Fe3+
The Ce4+ ion will be reduced at the lowest voltage in an electrolytic cell

87. CONCEPT CHECK!
An unknown metal (M) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate g of the metal from a solution containing M(NO3)3. What is the metal? gold (Au)
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:
g / 3.65 × 10–4 mol M = 197 g/mol
Note: Use the red box animation to assist in explaining how to solve the problem.
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88. CONCEPT CHECK!
Consider a solution containing 0.10 M of each of the following: Pb2+, Cu2+, Sn2+, Ni2+, and Zn2+. Predict the order in which the metals plate out as the voltage is turned up from zero. Cu2+, Pb2+, Sn2+, Ni2+, Zn2+ Do the metals form on the cathode or the anode? Explain.
Use the reduction potentials from Table Once that voltage has been surpassed, the metal will plate out. They plate out on the cathode (since they are reduced).
Order: Cu2+, Pb2+, Sn2+, Ni2+, Zn2+
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