2. LO 3.2 The student can translate an observed chemical change into a balanced chemical equation and justify the choice of equation type (molecular, ionic, or net ionic) in terms of utility for the given circumstances. (Sec 18.1)
LO 3.8 The student is able to identify redox reactions and justify the identification in terms of electron transfer. (Sec 18.1)
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. (Sec , 18.8)
LO 3.13 The student can analyze data regarding galvanic or electrolytic cells to identify properties of the underlying redox reactions. (Sec , 18.8)
LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable. (Sec 18.8)

3. 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. (Sec , 18.7)

4. AP Learning Objectives, Margin Notes and References
LO 3.2 The student can translate an observed chemical change into a balanced chemical equation and justify the choice of equation type (molecular, ionic, or net ionic) in terms of utility for the given circumstances.
LO 3.8 The student is able to identify redox reactions and justify the identification in terms of electron transfer.

5. 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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6. 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–
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7. 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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8. 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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9. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
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10. Cr2O72-(aq) + SO32-(aq) Cr3+(aq) + SO42-(aq)
How can we balance this equation?
First Steps:
Separate into half-reactions.
Balance elements except H and O.
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11. Method of Half Reactions
Cr2O72-(aq) 2Cr3+(aq)  
SO32-(aq) SO42-(aq)
How many electrons are involved in each half reaction?
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12. Method of Half Reactions (continued)
6e- + Cr2O72-(aq) 2Cr3+(aq) 
SO32-(aq) + SO42-(aq) + 2e-
How can we balance the oxygen atoms?
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13. Method of Half Reactions (continued)
6e- + Cr2O72-(aq) Cr3+(aq) + 7H2O 
H2O +SO32-(aq) SO42-(aq) + 2e-
How can we balance the hydrogen atoms?
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14. Method of Half Reactions (continued)
This reaction occurs in an acidic solution.
14H+ + 6e- + Cr2O Cr3+ + 7H2O 
H2O +SO SO e- + 2H+
How can we balance the electrons?
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15. Method of Half Reactions (continued)
14H+ + 6e- + Cr2O Cr3+ + 7H2O
3[H2O +SO SO e- + 2H+]
Final Balanced Equation:
Cr2O SO H Cr3+ + 3SO H2O
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16. 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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17. 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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18. 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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19. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
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20. Galvanic Cell
Device in which chemical energy is changed to electrical energy.
Uses a spontaneous redox reaction to produce a current that can be used to do work.
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21. A Galvanic Cell
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22. 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.
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23. 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).
1 joule of work per coulomb of charge transferred.
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24. Voltaic Cell: Cathode Reaction
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25. Voltaic Cell: Anode Reaction
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26. 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.

27. 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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28. 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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29. 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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30. CONCEPT CHECK!
Order the following from strongest to weakest oxidizing agent and justify. Of those you cannot order, explain why. Fe Na F- Na+ Cl2
Cl2 is a better oxidizing agent than Na+ (Cl2 has the larger reduction potential). The others cannot be ordered because we do not know their reduction potentials (although we can predict that F- will not easily be reduced, we do not have knowledge of quantitative proof from Table 18.1).

31. 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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32. 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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33. 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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34. 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
Ag electrode in 1.0 M Ag+(aq) and Cu electrode in 1.0 M Cu2+(aq)
Copper is the anode, silver is the cathode (electrons flow from copper to silver). The cell potential is 0.46 V.
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35. 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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36. 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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37. 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.

38. Work Work is never the maximum possible if any current is flowing.
In any real, spontaneous process some energy is always wasted – the actual work realized is always less than the calculated maximum.
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39. Maximum Cell Potential
Directly related to the free energy difference between the reactants and the products in the cell.
ΔG° = –nFE°
F = 96,485 C/mol e–
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40. 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.

41. A Concentration Cell
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42. Nernst Equation
The relationship between cell potential and concentrations of cell components
At 25°C: or
(at equilibrium)
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43. 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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44. EXERCISE!
A concentration cell is constructed using two nickel electrodes with Ni2+ concentrations of 1.0 M and × 10-4 M in the two half-cells.
Calculate the potential of this cell at 25°C.
0.118 V
The cell potential equals 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) = V
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45. You make a galvanic cell at 25°C containing:
CONCEPT CHECK!
You make a galvanic cell at 25°C containing:
A nickel electrode in 1.0 M Ni2+(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
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 + Ni2+. Therefore , Q = [Ni2+] / [Ag+]2 = (1.0 M) / (1.0 M)2.
ε = 1.03 V – (0.0591/2)log(1) = 1.03 V.
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46. One of the Six Cells in a 12–V Lead Storage Battery
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47. A Common Dry Cell Battery

48. A Mercury Battery
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49. Schematic of the Hydrogen-Oxygen Fuel Cell

50. 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.

51. 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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52. The Electrochemical Corrosion of Iron
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53. Corrosion Prevention
Application of a coating (like paint or metal plating)
Galvanizing
Alloying
Cathodic Protection
Protects steel in buried fuel tanks and pipelines.
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54. Cathodic Protection
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55. 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 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable.
Additional AP References
LO 5.15 (see Appendix 7.11, “Non-Spontaneous Reactions”)

56. Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
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57. 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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58. Stoichiometry of Electrolysis
current and time  quantity of charge
Coulombs of charge = amps (C/s) × seconds (s)
quantity of charge  moles of electrons
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59. 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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60. 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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61. Production of aluminum Purification of metals Metal plating
Electrolysis of sodium chloride
Production of chlorine and sodium hydroxide
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62. Producing Aluminum by the Hall-Heroult Process
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63. Electroplating a Spoon
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64. The Downs Cell for the Electrolysis of Molten Sodium Chloride
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