2. What is electrochemistry?
POP QUIZ
What is electrochemistry?
What is a voltaic or galvanic cell?
List 2 types of voltaic (galvanic cell)?

3. 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
Copyright © Cengage Learning. All rights reserved

4. Oxidation half-reaction (lose e-)
Electrochemical processes are oxidation-reduction reactions in which:
the energy released by a spontaneous reaction is converted to electricity or
electrical energy is used to cause a nonspontaneous reaction to occur 2+ 2-
2Mg (s) + O2 (g) MgO (s)
2Mg Mg2+ + 4e-
Oxidation half-reaction (lose e-)
O2 + 4e O2-
Reduction half-reaction (gain e-)

5. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
Free elements (uncombined state) have an oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
In monatomic ions, the oxidation number is equal to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1.

6. HCO3− O = −2 H = +1 3(−2) + 1 + x = −1 C = +4
The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.
Group IA metals are +1, IIA metals are +2 and fluorine is always –1.
6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.
HCO3−
Identify the oxidation numbers of all the atoms in HCO3− ?
O = −2
H = +1
3(−2) x = −1
C = +4

7. Balancing Redox Equations
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
Write the unbalanced equation for the reaction in ionic form.
Fe2+ + Cr2O Fe3+ + Cr3+
Separate the equation into two half-reactions.
Fe Fe3+ +2 +3
Oxidation:
Cr2O Cr3+ +6 +3
Reduction:
Balance the atoms other than O and H in each half-reaction.
Cr2O Cr3+

8. Balancing Redox Equations
For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.
Cr2O Cr3+ + 7H2O
14H+ + Cr2O Cr3+ + 7H2O
Add electrons to one side of each half-reaction to balance the charges on the half-reaction.
Fe Fe3+ + 1e-
6e- + 14H+ + Cr2O Cr3+ + 7H2O
If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.
6Fe Fe3+ + 6e-
6e- + 14H+ + Cr2O Cr3+ + 7H2O

9. Balancing Redox Equations
Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel.
Oxidation:
6Fe Fe3+ + 6e-
Reduction:
6e- + 14H+ + Cr2O Cr3+ + 7H2O
14H+ + Cr2O Fe Fe3+ + 2Cr3+ + 7H2O
Verify that the number of atoms and the charges are balanced.
14x1 – x 2 = 24 = 6 x x 3
For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation.

10. 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–
Copyright © Cengage Learning. All rights reserved

11. 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.
Copyright © Cengage Learning. All rights reserved

12. 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.
Copyright © Cengage Learning. All rights reserved

13. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
Copyright © Cengage Learning. All rights reserved

14. 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.
Copyright © Cengage Learning. All rights reserved

15. Method of Half Reactions
Cr2O72-(aq) 2Cr3+(aq)  
SO32-(aq) SO42-(aq)
How many electrons are involved in each half reaction?
Copyright © Cengage Learning. All rights reserved

16. Method of Half Reactions (continued)
6e- + Cr2O72-(aq) 2Cr3+(aq) 
SO32-(aq) + SO42-(aq) + 2e-
How can we balance the oxygen atoms?
Copyright © Cengage Learning. All rights reserved

17. Method of Half Reactions (continued)
6e- + Cr2O72-(aq) Cr3+(aq) + 7H2O 
H2O +SO32-(aq) SO42-(aq) + 2e-
How can we balance the hydrogen atoms?
Copyright © Cengage Learning. All rights reserved

18. 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?
Copyright © Cengage Learning. All rights reserved

19. 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
Copyright © Cengage Learning. All rights reserved

20. 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)
Copyright © Cengage Learning. All rights reserved

21. 18.1
Write a balanced ionic equation to represent the oxidation of iodide ion (I-) by permanganate ion ( ) in basic solution to yield molecular iodine (I2) and manganese(IV) oxide (MnO2).

22. 18.1
Strategy
We follow the preceding procedure for balancing redox equations. Note that the reaction takes place in a basic medium.
Solution
Step 1: The unbalanced equation is
+ I MnO2 + I2

23. 18.1 Step 2: The two half-reactions are Oxidation: I- I2
Reduction: MnO2 -1 +7 +4
Step 3: We balance each half-reaction for number and type of atoms and charges. Oxidation half-reaction: We first balance the I atoms:
2I I2

24. 18.1
To balance charges, we add two electrons to the right-hand side of the equation:
2I I2 + 2e-
Reduction half-reaction: To balance the O atoms, we add two H2O molecules on the right:
MnO2 + 2H2O
To balance the H atoms, we add four H+ ions on the left:
+ 4H MnO2 + 2H2O

25. 18.1
There are three net positive charges on the left, so we add three electrons to the same side to balance the charges:
+ 4H+ + 3e MnO2 + 2H2O
Step 4: We now add the oxidation and reduction half reactions to give the overall reaction. In order to equalize the number of electrons, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 as follows:
3(2I I2 + 2e-)
2( H+ + 3e MnO2 + 2H2O)
________________________________________
6I H+ + 6e I2 + 2MnO2 + 4H2O + 6e-

26. 6I- + 2 + 8H+ + 8OH- 3I2 + 2MnO2 + 4H2O + 8OH-
18.1
The electrons on both sides cancel, and we are left with the balanced net ionic equation:
6I H I2 + 2MnO2 + 4H2O
This is the balanced equation in an acidic medium. However, because the reaction is carried out in a basic medium, for every H+ ion we need to add equal number of OH- ions to both sides of the equation:
6I H+ + 8OH I2 + 2MnO2 + 4H2O + 8OH-

27. 18.1 Finally, combining the H+ and OH- ions to form water, we obtain
6I H2O I2 + 2MnO2 + 8OH-
Step 5: A final check shows that the equation is balanced in terms of both atoms and charges.

28. 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.)
Copyright © Cengage Learning. All rights reserved

29. 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.
Copyright © Cengage Learning. All rights reserved

30. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
Copyright © Cengage Learning. All rights reserved

31. The next several topics describe battery cells or voltaic cells (galvanic cells).
An electrochemical cell is a system consisting of electrodes that dip into an electrolyte and in which a chemical reaction either uses or generates an electric current. 2

32. 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.
Copyright © Cengage Learning. All rights reserved

33. Galvanic Cells anode oxidation cathode reduction spontaneous
redox reaction

34. A Galvanic Cell
Copyright © Cengage Learning. All rights reserved

35. 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.
Copyright © Cengage Learning. All rights reserved

36. 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.
Copyright © Cengage Learning. All rights reserved

37. Voltaic Cell: Cathode Reaction
To play movie you must be in Slide Show Mode
PC Users: Please wait for content to load, then click to play
Mac Users: CLICK HERE
Copyright © Cengage Learning. All rights reserved

38. Voltaic Cell: Anode Reaction
To play movie you must be in Slide Show Mode
PC Users: Please wait for content to load, then click to play
Mac Users: CLICK HERE
Copyright © Cengage Learning. All rights reserved

39. 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.
Copyright © Cengage Learning. All rights reserved

40. E0 is for the reaction as written
The more positive E0 the greater the tendency for the substance to be reduced
The half-cell reactions are reversible
The sign of E0 changes when the reaction is reversed
Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

41. Comparing Oxidizing Strengths
The oxidizing agent is itself reduced and is the species on the left of the reduction half-reaction.
Consequently, the strongest oxidizing agent is the product of the half-reaction with the largest (most positive) E° value. 2

42. Comparing Reducing Strengths
The reducing agent is itself oxidized and is the species on the right of the reduction half-reaction.
Consequently, the strongest reducing agent is the reactant in the half-reaction with the smallest (most negative) E° value. 2

43. 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
Copyright © Cengage Learning. All rights reserved

44. 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
Copyright © Cengage Learning. All rights reserved

45. 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).

46. 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)
Copyright © Cengage Learning. All rights reserved

47. Galvanic Cells
The difference in electrical potential between the anode and cathode is called:
cell voltage
electromotive force (emf)
cell potential
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M and [Zn2+] = 1 M
Cell Diagram
phase boundary
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode
salt bridge
cathode

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.
Copyright © Cengage Learning. All rights reserved

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.
Copyright © Cengage Learning. All rights reserved

50. Because this is a reduction, the electrode appears on the far right.
When the half-reaction involves a gas, the electrode is an inert material such as platinum, Pt. It is included as a third substance in the half-cell.
For example, the half-reaction of Cl2 being reduced to Cl- is written as follows:
Cl2(g) | Cl-(aq) | Pt
Because this is a reduction, the electrode appears on the far right.
For the oxidation of H2(g) to H+(aq), the notation is
Pt | H2(g) | H+(aq)
In this case, the electrode appears on the far left. 2

51. 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
Example 18.4
Ag electrode in 1.0 M Ag+(aq) and Fe2+ electrode in 1.0 M Fe3+(aq)
Iron is the anode, silver is the cathode (electrons flow from Iron to silver). The cell potential is 0.03 V.
Copper is the anode, silver is the cathode (electrons flow from copper to silver). The cell potential is 0.46 V.
Copyright © Cengage Learning. All rights reserved

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
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.
Copyright © Cengage Learning. All rights reserved

53. 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).
Copyright © Cengage Learning. All rights reserved

54. Just as work is required to pump water from one point to another, so work is required to move electrons.
Water flows from areas of high pressure to areas of low pressure. Similarly, electrons flow from high electric potential to low electric potential. Electric potential can be thought of as electric pressure. 2

55. 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.
Copyright © Cengage Learning. All rights reserved

56. Electrical work = charge x potential difference
Potential difference is the difference in electric potential (electrical pressure) between two points. Potential difference is measured in the SI unit volt (V).
Electrical work = charge x potential difference
The SI units for this are
J = C × V

57. 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–
Copyright © Cengage Learning. All rights reserved

58. A Concentration Cell
Copyright © Cengage Learning. All rights reserved

59. The Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q
DG = -nFE
DG0 = -nFE
-nFE = -nFE0 + RT ln Q
Nernst equation
E = E0 -
ln Q RT nF
At 298 - V n
ln Q E
E = - V n
log Q E
E =
19.5

60. Nernst Equation
The relationship between cell potential and concentrations of cell components
At 25°C: or
(at equilibrium)
Copyright © Cengage Learning. All rights reserved

61. 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.
Copyright © Cengage Learning. All rights reserved

62. 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
Copyright © Cengage Learning. All rights reserved

63. 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.
Copyright © Cengage Learning. All rights reserved

64. One of the Six Cells in a 12–V Lead Storage Battery
Copyright © Cengage Learning. All rights reserved

65. Anode: Pb(s) + HSO4-(aq)  PbSO4(s) + H+(aq) + 2e-
Lead Storage Cell
The electrodes are lead alloy grids: one is packed with a spongy lead to form the anode, and the other is packed with lead dioxide to form the cathode. Both electrodes are in an aqueous solution of H2SO4.
Anode: Pb(s) + HSO4-(aq)
 PbSO4(s) + H+(aq) + 2e-
Cathode: PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e-
 PbSO4(s) + 2H2O(l)
Unlike dry cells, after discharge, lead storage cells can be recharged.
Copyright © Cengage Learning. All rights reserved. 2

66. Each of the six cells in the lead storage cell generates 2 V, yielding 12 V. During discharge, white PbSO4(s) coats each electrode.
To recharge the cell, an external current is used, reversing the previous reactions. Some water decomposes into hydrogen and oxygen gas, so more water may need to be added.
Newer batteries use electrodes with calcium in the lead, which resists decomposition by water. These versions are maintenance free.
Copyright © Cengage Learning. All rights reserved. 2

67. Batteries Lead storage battery Anode:
Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4
Cathode:
PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e PbSO4 (s) + 2H2O (l) 4
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) PbSO4 (s) + 2H2O (l) 4
19.6

68. Batteries Dry cell Leclanché cell Anode: Zn (s) Zn2+ (aq) + 2e-
Cathode:
2NH4 (aq) + 2MnO2 (s) + 2e Mn2O3 (s) + 2NH3 (aq) + H2O (l) +
Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)
19.6

69. Zinc–Carbon Dry Cell: Leclanché Anode: Zn(s)  Zn2+(aq) + 2e-
Cathode: 2NH4+(aq) + 2MnO2(s) + 2e-
 Mn2O3(s) + H2O(l) + 2NH3(aq)
The initial voltage is about 1.5 V, but decreases and deteriorates rapidly in cold weather. 2

70. A Common Dry Cell Battery

71. A Mercury Battery
Copyright © Cengage Learning. All rights reserved

72. Batteries Mercury Battery Anode:
Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-
Cathode:
HgO (s) + H2O (l) + 2e Hg (l) + 2OH- (aq)
Zn(Hg) + HgO (s) ZnO (s) + Hg (l)
19.6

73. Batteries
A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning
Anode:
2H2 (g) + 4OH- (aq) H2O (l) + 4e-
Cathode:
O2 (g) + 2H2O (l) + 4e OH- (aq)
2H2 (g) + O2 (g) H2O (l)
19.6

74. Schematic of the Hydrogen-Oxygen Fuel Cell

75. 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.
Copyright © Cengage Learning. All rights reserved

76. Corrosion Control: Cathodic Protection
Voltaic cells can be used to control corrosion of underground pipelines and tanks. Rusting occurs when water comes in contact with iron. The edge of the water drop, when exposed to air, becomes one pole of a voltaic cell where oxygen is reduced to hydroxide.
Anode: Fe(s)  Fe2+(aq) + 2e-
Cathode: O2(g) + 2H2O(l) + 4e-  4OH-(aq)
Copyright © Cengage Learning. All rights reserved.

77. The Electrochemical Corrosion of Iron
Copyright © Cengage Learning. All rights reserved

78. Corrosion Prevention
Application of a coating (like paint or metal plating)
Galvanizing
Alloying
Cathodic Protection
Protects steel in buried fuel tanks and pipelines.
Copyright © Cengage Learning. All rights reserved

79. This phenomenon is called cathodic protection.
When the buried metal is connected to a more active metal such as magnesium, the magnesium becomes the anode and the iron becomes the cathode. The iron is, therefore, protected from oxidation.
This phenomenon is called cathodic protection.
Copyright © Cengage Learning. All rights reserved.

80. Cathodic Protection
Copyright © Cengage Learning. All rights reserved

81. Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.
19.8

82. Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
Copyright © Cengage Learning. All rights reserved

83. Stoichiometry of Electrolysis
In the 1830s, Michael Faraday showed that the total charge that flows in a circuit is related to the amount of substance released at the electrodes.
One faraday of charge is the charge on one mole of electrons and is equal to 96,485 coulombs.
Copyright © Cengage Learning. All rights reserved.

84. 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
Copyright © Cengage Learning. All rights reserved

85. Stoichiometry of Electrolysis
current and time  quantity of charge
Coulombs of charge = amps (C/s) × seconds (s)
quantity of charge  moles of electrons
Copyright © Cengage Learning. All rights reserved

86. Electric charge = current × time Coulombs = amperes × seconds
For quantitative considerations, we also need to know the magnitude of the current and the time it has flowed.
Electric charge = current × time
Coulombs = amperes × seconds
The ampere, A, is the base unit of current.
The coulomb, C, is equal to an ampere-second.
Copyright © Cengage Learning. All rights reserved.

87. What electric charge is required to plate a piece of automobile molding with 1.00 g of chromium metal using a chromium(III) ion solution?
If the electrolysis current is 2.00 A, how long does the plating take?

88. First, we convert mass to moles Cr; then to moles e-; then, using current, to seconds.
= 2.78 × 103 s = 46.4 min

89. A solution of nickel salt is electrolyzed to nickel metal by a current of 2.43 A. If this current flows for 10.0 min, how many coulombs is this?
How much nickel metal is deposited in the electrolysis?

90. We first find the charge:
Given the half-reaction: Ni2+(aq) + 2e-  Ni(s)
This slide should be followed by the chapter key terms. This was originally the last slide, but I copied the logo slide from another chapter as the final slide.
= g Ni

91. 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.
Copyright © Cengage Learning. All rights reserved

92. 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+
Copyright © Cengage Learning. All rights reserved

93. Production of aluminum Purification of metals Metal plating
Electrolysis of sodium chloride
Production of chlorine and sodium hydroxide
Copyright © Cengage Learning. All rights reserved

94. Producing Aluminum by the Hall-Heroult Process
Copyright © Cengage Learning. All rights reserved

95. Electroplating a Spoon
Copyright © Cengage Learning. All rights reserved

96. Anode: Cl-(l)  ½Cl2(g) + e- Cathode: Na+(l) + e-  Na(l)
Downs Cell
A Downs cell is an electrolytic cell used to obtain sodium metal by electrolysis of sodium chloride. The products must be kept separated or they would react.
Anode: Cl-(l)  ½Cl2(g) + e-
Cathode: Na+(l) + e-  Na(l)
Copyright © Cengage Learning. All rights reserved.

97. The Downs Cell for the Electrolysis of Molten Sodium Chloride
Copyright © Cengage Learning. All rights reserved