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
Primarily concerned with the generation of an electrical current fro a spontaneous chemical reaction and, the opposite process, the use of current to produce a chemical reaction.
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. Synopsis of Assigning Oxidation Numbers (as a Reminder)
Elements = 0
Monatomic ion = charge
F: –1
O: –2 (unless peroxide = –1)
H: +1 (unless a metal hydride = –1)
The sum of the oxidation numbers equals the overall charge (0 in a compound).

7. Oxidation Numbers
To keep track of what loses electrons and what gains them, we assign oxidation numbers.
If the oxidation number increases for an element, that element is oxidized.
If the oxidation number decreases for an element, that element is reduced.

8. Oxidation and Reduction
A species is oxidized when it loses electrons.
Zinc loses two electrons, forming the Zn2+ ion.
A species is reduced when it gains electrons.
H+ gains an electron, forming H2.
An oxidizing agent causes something else to be oxidized (H+); a reducing agent causes something else to be reduced (Zn).

9. The oxidation and reduction are written and balanced separately.
Half-Reactions
The oxidation and reduction are written and balanced separately.
We will use them to balance a redox reaction.
For example, when Sn2+ and Fe3+ react,

10. 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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11. 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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12. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
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13. The Half-Reaction Method
Consider the reaction between MnO4– and C2O42–:
MnO4–(aq) + C2O42–(aq)  Mn2+(aq) + CO2(aq)
Assigning oxidation numbers shows that Mn is reduced (+7  +2) and C is oxidized (+3  +4).

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

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

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

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

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

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

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

21. Combining the Half-Reactions
10e– + 16 H+ + 2 MnO4– + 5 C2O42– 
2 Mn H2O + 10 CO2 +10e–
The only thing that appears on both sides is the electrons. Subtracting them, we are left with
16 H+ + 2 MnO4– + 5 C2O42– 
2 Mn H2O + 10 CO2
(Verify that the equation is balanced by counting atoms and charges on each side of the equation.)

22. 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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23. 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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24. 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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25. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
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26. CN–(aq) + MnO4–(aq) → CNO– (aq) + MnO2(s)
Balancing Redox Equations in Acidic Solution
Complete and balance this equation for a redox reaction that takes place in basic solution:
CN–(aq) + MnO4–(aq) → CNO– (aq) + MnO2(s)
(basic solution)

27. Galvanic Cell/ Voltaic 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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28. A Galvanic Cell
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29. Galvanic Cell Oxidation occurs at the anode.
Reduction occurs at the cathode.
Salt bridge or porous disk – devices that allow ions to flow to neutralize the charge imbalance 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.
anions flow to anode, cations flow to cathode. Choose salt with similar charges that don’t react with electrodes.
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30. Copper ions deposit as solid Cu on the cathode.

31. 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. The emf is the potential difference between the anode and cathode.
Unit of electrical potential is the volt (V).
1 joule of work per coulomb of charge transferred. (1 V = 1 J/C).
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32. Electromotive Force (emf)
Water flows spontaneously one way in a waterfall.
Comparably, electrons flow spontaneously one way in a redox reaction, from high to low potential energy.

33. Describing a Voltaic Cell-draw it
The oxidation–reduction reaction
Cr2O72– (aq) + 14 H+ (aq) + 6 I– (aq) → 2 Cr3+ (aq) + 3 I2(s) + 7 H2O(l)
is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs of the electrodes.

34. The following two half-reactions occur in a voltaic cell:
Ni(s) → Ni2+ (aq) + 2 e– (electrode = Ni)
Cu2+(aq) + 2 e– → Cu(s) (electrode = Cu)
Which one of the following descriptions most accurately describes what is occurring in the half-cell containing the Cu electrode and Cu2+ (aq) solution?
(a) The electrode is losing mass and cations from the salt bridge are flowing into the half-cell.
(b) The electrode is gaining mass and cations from the salt bridge are flowing into the half-cell.
(c) The electrode is losing mass and anions from the salt bridge are flowing into the half-cell.
(d) The electrode is gaining mass and anions from the salt bridge are flowing into the half-cell.

35. Voltaic Cell: Cathode Reaction
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36. Voltaic Cell: Anode Reaction
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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. Galvanic Cell
All half-reactions are given as reduction processes in standard tables. These values are compared to the reduction of hydrogen as a standard.
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
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39. Standard Hydrogen Electrode
Their reference is called the standard hydrogen electrode (SHE).
By definition as the standard, the reduction potential for hydrogen is 0 V:
2 H+(aq, 1M) + 2e–  H2(g, 1 atm)

40. A galvanic cell involving the reactions Zn  Zn2++ 2e− (at the anode) and 2H+ + 2e−  H2 (at the cathode) has a potential of 0.76 V. The standard hydrogen electrode where H2(g) at 1 atm is passed over a platinum electrode in contact with 1 M H+ ions. This electrode process (assuming ideal behavior) is arbitrarily assigned a value of exactly zero volts.
Figure 18-5 p843

42. Oxidizing and Reducing Agents
The more positive the value of E°red, the greater the tendency for reduction under standard conditions.
The strongest oxidizers have the most positive reduction potentials.
The strongest reducers have the most negative reduction potentials.

43. Standard Cell Potentials
The cell potential at standard conditions can be found through this equation:
Ecell
= Ered (cathode) – Ered (anode)
Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

44. Cell Potentials For the anode in this cell, E°red = –0.76 V
For the cathode, E°red = V
So, for the cell, E°cell = E°red (cathode) – E°red (anode) = V – (–0.76 V) = V

45. 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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46. 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°(cathode) – E°(anode)
= 0.77 V – 0.34 V = 0.43 V
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47. 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)
Anode/ what’s being oxidized/ to what it becomes// what’s being reduced/ to what it becomes/ cathode
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48. 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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49. 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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50. 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).

51. A voltaic cell is based on the two standard half-reactions
Determining Half-Reactions at Electrodes and Calculating Cell Potentials
A voltaic cell is based on the two standard half-reactions
Cd2+(aq) + 2 e– → Cd(s)
Sn2+(aq) + 2 e– → Sn(s)
Use data in Appendix A5.5 to determine (a) which half-reaction occurs at the cathode and which occurs at the anode and (b) the standard cell potential.
a) Cathode- Sn Anode- Cd b) Ecell= .26V

52. 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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53. CONCEPT CHECK!
Consider the cell from part a. What would happen to the potential if you increase the [Ag+]? Explain. The cell potential should increase.
Since the silver 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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54. 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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55. Ag+ + Fe2+ → Fe3+ +Ag Ecell=0.03V pt/Fe2+/Fe3+//Ag+/Ag

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. 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.
Work is never maximum because the entropy of the universe must increase for any spontaneous process.
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58. Work
The work done when e- a transferred through a wire depends on the thermodynamic driving force (the push) behind the electrons. This “push” or emf is defined in terms of a potential difference (in volts) between 2 points in the circuit.
(1 V = 1 J/C) Thus 1 joule of work is produced or required (depending on the direction of flow) when 1 coulomb of charge is transferred between 2 points in the circuit that differ by 1 volt.
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59. Free Energy and Redox
Spontaneous redox reactions produce a positive cell potential, or emf.
E° = E°red (reduction) – E°red (oxidation)
Note that this is true for ALL redox reactions, not only for voltaic cells. Dropped cell notation (cathode and anode)
Since Gibbs free energy is the measure of spontaneity, positive emf corresponds to negative ΔG.

60. Free Energy and Redox (Yet a 5th way to calculate G!!)
How do they relate? ΔG = –nFE
n = moles of electrons
F = is the Faraday constant, 96,485 C/mol e-
(charge of 1 mole of electrons)
E = maximum energy potential
-G = +Ecell = spontaneous G unit is J/ mole reaction (coefficient)
ΔG° = –nFE°
Max energy potential is directly related to the energy difference between the reactants and products in a cell.

61. How is everything related? ΔG° = –nFE° = –RT ln K
Free Energy, Redox, and K
How is everything related?
ΔG° = –nFE° = –RT ln K

62. Using Standard Reduction Potentials to Calculate ∆G° and K
(a) Use the standard reduction potentials in Table to calculate the standard free-energy change, ∆G°, and the equilibrium constant, K, at 298 K for the reaction
4 Ag(s) + O2(g) + 4 H+(aq) → 4 Ag+(aq) + 2 H2O(l)
(b) Suppose the reaction in part (a) is written
2 Ag(s) + O2(g) + 2 H+(aq) → 2 Ag+(aq) + H2O(l)
What are the values of E°, ∆G°, and K when the reaction is written in this way?
a)G= -170 kJ/mol k= 9 x1029 b) E= .43V G= -8 3kJ/mol k= 4 x 1014

63. 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.

64. Remember, ΔG = ΔG° + RT ln Q So, –nFE = nFE° + RT ln Q
Nernst Equation
Remember, ΔG = ΔG° + RT ln Q
So, –nFE = nFE° + RT ln Q
Dividing both sides by –nF, we get the Nernst equation:
E = E° – (RT/nF) ln Q
OR E = E° – (2.303 RT/nF) log Q
Using standard thermodynamic temperature and the constants R and F,
E = E° – (0.0592V/n) log Q

65. Nernst Equation
The relationship between cell potential and concentrations of cell components
At 25°C: or
(at equilibrium)
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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. CONCEPT CHECK! a) Ecell <Ecell not b) Ecell> than Ecell not
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68. A Concentration Cell
A cell in which both compartments have the same components at different concentrations.
The electrons will flow in the direction in which to equalize the Ag+ concentration in the two compartments. (Flows to lower the more concentrated ion)
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69. Concentration Cells
The difference in concentration is the only factor that produces cell potential.
E° = 0
For such a cell,
would be 0, but Q would not.
Ecell
Once the concentrations are the same on both sides, Q= 1 and E = 0 (equilibrium).

70. 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. anode over cathode gives positive E (dilute over concentrated)
ε = 0 - (0.0591/2)log(1.0×10-4 / 1.0) = V
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71. EXERCISE!
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74. One of the Six Cells in a 12–V Lead Storage Battery
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75. A Common Dry Cell Battery

76. A Mercury Battery
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77. Schematic of the Hydrogen-Oxygen Fuel Cell
Fuel cells are NOT batteries; the source of energy must be continuously provided.

78. 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.

79. 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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80. Corrosion Corrosion is oxidation. Its common name is rusting.
Water acts like salt bridge. Salt increases the conductivity of the aqueous solution and accelerates corrosion.
Corrosion is oxidation.
Its common name is rusting.

81. Corrosion Prevention
Application of a coating (like paint or metal plating) to protect the metal from oxygen and moisture.
Galvanizing (coating steel with Zn to form a mixed oxide-carbonate coating)
Alloying- mix with a metal with a lower reduction potential.
Cathodic Protection
Protects steel in buried fuel tanks and pipelines.
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82. Preventing Corrosion
Corrosion is prevented by coating iron with a metal that is more readily oxidized.
Zinc is more easily oxidized, so that metal is sacrificed to keep the iron from rusting.

83. Cathodic Protection-Protects steel in buried fuel tanks and pipelines.
Mg oxidized and electrons are given to the iron for reduction.
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84. 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”)

85. Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
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86. Electrons flow in the opposite direction.
Note:
External powers source greater than 1.10V needed
Electrons flow in the opposite direction.
Anode and cathode are reversed.
Ions in the salt bridge flow in the opposite direction.
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87. Production of aluminum Purification of metals Metal plating
Electrolysis of sodium chloride
Production of chlorine and sodium hydroxide
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88. Electrolysis
Molten NaCl
Nonspontaneous reactions can occur in electrochemistry IF outside electricity is used to drive the reaction.
Use of electrical energy to create chemical reactions is called electrolysis.
(+) terminal connected to the anode and the (-) one to the cathode-forces electrons from anode to the cathode.
Use molten NaCl because water solution would see reduction of H and oxidation of O. Use inactive electrodes for NaCl. Active electrodes when plating metal.

89. Electrolysis
Electroplating-use active electrodes during electrolysis to deposit a thin layer of metal onto another metal to improve beauty or resistance to corrosion.
Since using a solution, you must consider the reduction and oxidation of water.

90. Electrolysis
Ni2+ E◦red = -.28V and water E◦red = -.83V so Ni2+ is preferentially reduced and plated on the steel cathode.
At the anode Ni2+ E◦red = -.28V and water E◦red = 1.23V so Ni2+ is preferentially oxidized (more negative E◦red means more easily oxidized).

91. Electroplating a Spoon
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92. Electrolysis and “Stoichiometry”
1 ampere = 1 coulomb
1 Second
I = q t
q = charge (C)
I = current (A)
t = time (s)

93. Electrolysis and “Stoichiometry”
1 coulomb = 1 ampere × 1 second
q = It = nF
q = charge (C)
I = current (A)
t = time (s)
n = moles of electrons that travel through the wire in the given time
F = Faraday’s constant
NOTE: n is different than that for the Nernst equation!

94. Relating Electrical Charge and Quantity of Electrolysis
Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.
3.36 g Al

95. 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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96. 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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