1. Electrochemistry
Interchange of
electrical and
chemical energy.

2. Electron Transfer Reactions
Electron transfer reactions are oxidation-reduction or redox reactions.
the energy released by a spontaneous reaction is converted to electricity or
electrical energy is used to cause a nonspontaneous reaction to occur
Therefore, this field of chemistry is often called ELECTROCHEMISTRY.

3. Electrochemistry Terminology #1
Oxidation – A process in which an element attains a more positive oxidation state
Na(s)  Na+ + e-
Reduction – A process in which an element attains a more negative oxidation state
Cl2 + 2e-  2Cl-

4. Electrochemistry Terminology #2
An old memory device for oxidation and reduction goes like this…
LEO says GER
Lose Electrons = Oxidation
Gain Electrons = Reduction

5. Electrochemistry Terminology #3
Oxidizing agent
The substance that is reduced is the oxidizing agent
Reducing agent
The substance that is oxidized is the reducing agent

6. 2Mg (s) + O2 (g) 2MgO (s) O2 + 4e- 2O2-2+ 2-
2Mg (s) + O2 (g) MgO (s)
2Mg Mg2+ + 4e-
Oxidation half-reaction (lose e-)
O2 + 4e O2-
Reduction half-reaction (gain e-)
19.1

7. Electrochemistry Terminology #4
Anode
The electrode where oxidation occurs
Cathode
The electrode where reduction occurs
Memory device:
An ox
Red Cat

8. 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.

9. Method of Half Reactions
Cr2O72-(aq)  Cr3+(aq)  
SO32-(aq)  SO42-(aq)
Now, balance all elements except O and H.

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

11. Method of Half Reactions (continued)
Cr2O72-(aq)  Cr3+(aq) + 7H2O 
H2O +SO32-(aq)  + SO42-(aq)
How can we balance the hydrogen atoms?

12. Method of Half Reactions (continued)
This reaction occurs in an acidic solution.
14H+ + Cr2O72-  2Cr3+ + 7H2O 
H2O +SO32-  SO H+
How can we write the electrons?

13. Method of Half Reactions (continued)
This reaction occurs in an acidic solution.
6e- + 14H+ + Cr2O72-  2Cr3+ + 7H2O 
H2O +SO32-  SO H+ + 2e-
How can we balance the electrons?

14. Method of Half Reactions (continued)
14H+ + 6e- + Cr2O72-  2Cr3+ + 7H2O
3[H2O +SO32-  SO e- + 2H+]
Final Balanced Equation:
Cr2O SO H+  2Cr3+ + 3SO H2O

15. 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)

16. Half–Reactions
The overall reaction is split into two half–reactions, one involving oxidation and one reduction.
8H+ + MnO4- + 5Fe2+  Mn2+ + 5Fe3+ + 4H2O
Reduction: 8H+ + MnO4- + 5e-  Mn2+ + 4H2O
Oxidation: 5Fe2+  5Fe3+ + 5e-

17. How do we harness the energy from the reaction?
If they are in the same solution;
energy is released as heat and can do no work.
Separate the solution;
MnO4- in one beaker and Fe2+ in another requiring the e- to be transferred through a wire producing a current.

18. Figure 17.1: Schematic of a method to separate the oxidizing and reducing agents of a redox reaction. (The solutions also contain counterions to balance the charge.)
Current
stop
flowing due to separation of charge.

19. Keeps the net charge at zero; allows
Figure 17.2: Galvanic cells can contain a salt bridge as in (a) or a porous-disk connection as in (b).
Keeps the net charge at zero; allows
the ions to flow w/o mixing the solutions.

20. Galvanic Cell
A device in which chemical energy is changed to electrical energy.

21. - + anode oxidation cathode reduction spontaneous redox reaction 19.2
19.2

22. 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 (º), 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. (volt= Joule/coulomb)

23. Table of Reduction Potentials
Measured against the Standard Hydrogen Electrode

24. Figure 17.5: (a) A galvanic cell involving the reactions Zn → Zn2+ + 2e- (at the anode/oxidation) and 2H+ + 2e-→ H2 (at the cathode/reduction) has a potential of 0.76 V. (b) 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.

25. Standard Reduction Potentials
Potential of half-reactions
are summed to obtain cell potential.
2H+(aq) + Zn(s) → Zn2+(aq) + H2(g)
The anode compartment?
The cathode compartment?
Cathode consists of a platinum electrode (inert conduction) in contact w/1M H+ and bathed by H gas at 1atm.
Standard Hydrogen Electrode.
Zn → Zn2+ + 2e- oxidation
2H+ + 2e- → H2 reduction

26. Total cell potential = 0.76V; must be divided to get the half-potentials
Assign this reaction 2H+ + 2e- → H2 reduction
Where:
[H+]= 1M and PH2=1 atm
A potential of exactly 0 V, then the reaction
will have a potential of 0.76 V because
Zn → Zn2+ + 2e- oxidation

27. 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 calculate the cell potential, we must reverse reaction 2.
Cu → Cu2+ + 2e– – E ° = – 0.34 V

28. Overall Balanced Cell Reaction
2Fe3+ + 2e– → 2Fe E ° = 0.77 V (cathode)
Cu → Cu2+ + 2e– – E ° = – 0.34 V (anode)
Cell Potential:
E °cell = E °(cathode) – E °(anode)
**not on formula sheet! *** the – in the equation automatically flips the anode
E °cell = 0.77 V – 0.34 V = 0.43 V

29. Manipulations to obtain a balanced oxidation-reduction reaction
One half-reaction must be reversed, which means the sign must be reversed.
Since the # of e- gained must equal the # of e- lost, the half-reactions must be multiplied by integers. But the value of º is not changed since it is an intensive property.
A galvanic cell runs spontaneously in the direction that gives a positive value for E °cell.

30. Line Notation
Anode components are listed on the left and the cathode on the right.
Mg(s) Mg2+(aq) Al3+(aq) Al(s)
Mg → Mg2+ + 2e– (anode)
Al3+ + 3e– → Al (cathode)
Phase difference- single line
Salt bridge- double line

31. The relationship between thermodynamics and electrochemistry

32. Cell Potential, Electrical Work, and Free Energy
Work that can be done by a cell depends on the “push” driving force. (emf)
Defined in terms of potential difference (V) between two points in a circuit.
Copyright©2000 by Houghton Mifflin Company. All rights reserved.

33. Volt – J of work/C of charge transferred
emf = potential difference(V) = work (J) charge(C)
1J work is produced or needed, when 1C of charge is transferred between two points in the circuit that differ by a potential of 1 volt.

34. Work is from the point of view of the system. Work flowing out is (-)
Cell produces current, ξ is (+), current is used to do work (-).
Thus, ξ and w have opposite signs
ξ= -w/q (charge)
-w= ξq
Max work is obtained at max ξ (cell potential)
-wmax=q ξmax wmax= -q ξmax

35. Problem? Wmax will never be reached Why??
*in any real spontaneous process some E is always lost . Actual work is less that max work.
Entropy??? What is the only type of process in which you can reach wmax ?

36. Although we may never achieve wmax, we can calculate max potential
Although we may never achieve wmax, we can calculate max potential.We use wmax to calculate efficiency.
Suppose we have a galvanic cell wmax= 2.50V
And 1.33 mol of e- were passed through the cell and the actual potential of 2.10V is registered.
The actual work done is:
w= - ξq
ξ = actual potential difference at which the current flowed (2.10V or 2.10J/C)
q = quantity of charge in coulombs transferred

37. Faraday (F) The charge on 1mol of e- = faraday(F)
96,485 C of charge/mol e-
q=nF= (1.33mol)(96,485C/mol)
w= -q ξ = -(1.33mol)(96,485C/mol)(2.10J/C) =
-2.96x105J
wmax= -(1.33mol)(96,485C)(2.50J/C) =
-3.21x105J
Efficiency = w/wmax x 100% = 83.8%

38. Relate potential of cell to free energy at constant T and P : wmax=ΔG
wmax= -qξmax = ΔG
q=nF
ΔG=-nF ξ
Assume at this point any potential is max.

39. For standard conditions:
ΔG = nFE
The max cell potential is directly related to the free energy difference between the reactants and the products in the cell. (experimental means!)
A positive cell potential yeilds a -ΔG

40. Calculate ΔG° for the reaction Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq) Is this reaction spontaneous?
Half reaction: Cu2+ + 2e- → Cu ξ° = 0.34 V
Fe(s) → Fe2+ + 2e ξ° = 0.44 V
Cu2+ + Fe → Fe2+ + Cu
ξ° = 0.78 V

41. Copyright©2000 by Houghton Mifflin Company. All rights reserved.
Predict whether 1M HNO3 will dissolve gold metal to form a 1 M Au3+ solution.
Copyright©2000 by Houghton Mifflin Company. All rights reserved.

42. Concentration Cell
. . . a cell in which both compartments have the same components but at different concentrations.

43. Cell Potential and concentration
Standard conditions (1M)
Cu(s) + 2Ce4+(aq) → Cu2+ + 2Ce3+(aq)
ξ= 1.36V
What will the potential be if [Ce4+] is greater than 1M?
LeChatelier? Increase or decrease cell potential?
Increase driving force on the e-

44. For the reaction a. ξ°< 0.48V b. ξ°> 0.48V
2Al(s) + 3Mn2+(aq) → 2Al3+(aq) + 3Mn(s) ξ°= 0.48 V
Will ξcell be larger or smaller than ξ° in the following cases?
a. [Al3+] = 2.0M [Mn2+] = 1.0M
b. [Al3+] = 1.0M [Mn2+] = 3.0M
a. ξ°< 0.48V
b. ξ°> 0.48V

45. Cell potential and concentration
An increase in [reactant] caused an increase in cell potential, and vice versa.
Therefore we can construct [cells] of the same components.

46. Which way will electron flow to equilibrium? Ag+ + e- → Ag ξ°=0.80V
Figure 17.9: A concentration cell that contains a silver electrode and aqueous silver nitrate in both compartments
Ag+ + e- → Ag
ξ°=0.80V
e- 
Which way
will electron
flow to
equilibrium?
anode cathode

47. Nernst Equation
Dependence of cell potential on concentration, results from the dependence ΔG° on concentration.
ΔG= ΔG° + RTln(Q) Q= reaction quotient
ΔG = -nF ξ
-nFξ =-nF ξ° + RTln(Q)
ξ = ξ ° -RT/nF ln(Q)

48. The Nernst Equation R = 8.31 J/(molK) T = Temperature in K
Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M.
R = 8.31 J/(molK)
T = Temperature in K
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol e-

49. Nernst Equation Simplified
At 25 C (298 K) the Nernst Equation is simplified this way:

50. Both sides have the same components but at different concentrations.
???
Concentration Cell
Both sides have the same components but at different concentrations.
Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.

51. Both sides have the same components but at different concentrations.
???
Concentration Cell
Both sides have the same components but at different concentrations.
Anode
Cathode
The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentration
Zn2+ (1.0M) + 2e-  Zn (reduction)
Zn  Zn2+ (0.10M) + 2e- (oxidation)
Zn2+ (1.0M)  Zn2+ (0.10M)

52. Both sides have the same components but at different concentrations.
Concentration Cell
???
Concentration Cell
Both sides have the same components but at different concentrations.
Anode
Cathode
Step 2: Calculate cell potential using the Nernst
Equation (assuming 25 C).
Zn2+ (1.0M)  Zn2+ (0.10M)

53. Nernst Calculations
Zn2+ (1.0M)  Zn2+ (0.10M)

54. ξ°cell =0.48V for a cell based on the reaction:
2Al(s) + 3Mn2+(aq)→2Al3+(aq) + 3Mn(s)
[Mn2+] = 0.50M
[Al3+] = 1.50M

55. 2Al(s) + 3Mn2+(aq)→2Al3+(aq) + 3Mn(s)
[Mn2+] = 0.50M [Al3+] = 1.50M
Half reactions
2Al →2Al3+ + 6e-
3Mn3+ + 6e- →3Mn
n= 6
Voltage reduced (Le Chatelier) since the [reactant] is lower
than 1.0M and the [products] is higher than 1.0M,
ξcell is less than ξ°cell

56. *The cell will spontaneously discharge until it reaches equilibrium.
at that point…
Q=K (the equilibrium constant) and ξcell=0.

57. Dead battery is at equilibrium.
At equilibrium the components in the two cell compartments have the same free energy, and ΔG=0.
The cell can no longer do work.

58. Calculation of Equilibrium Constants for Redox Reactions
At equilibrium, Ecell = 0 and Q = K.

59. Spontaneity of Redox Reactions
19.4

60. Calculating an Equilibrium Constant from a Cell Potential
Zn + Cu2+  Zn2+ + Cu E0 = V

61. One of the Six Cells in a 12–V Lead Storage Battery
Anode rxn: Pb + HSO4-  PbSO4 + H+ + 2e-
Cathode rxn: PbO2 + HSO H+ + 2e- + PbSO4 + 2H2O
Cell rxn: Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq)  2PbSO4(s) + 2H2O(l)
Discharge lowers the concentration
of sulfuric acid, and the density of
the solution.
PbSO4(s) forms on the grids (electrodes)
and can be recharged.

62. Charging a Battery
When you charge a battery, you are forcing the electrons backwards (from the + to the -). To do this, you will need a higher voltage backwards than forwards. This is why the ammeter in your car often goes slightly higher while your battery is charging, and then returns to normal.
In your car, the battery charger is called an alternator. If you have a dead battery, it could be the battery needs to be replaced OR the alternator is not charging the battery properly.

63. A Common Dry Cell Battery
Acidic version
1.5V

64. A Mercury Battery Basic version: last longer, less corrosion
your calculator

65. Schematic of the Hydrogen-Oxygen Fuel Cell
galvanic cells for which the reactants are continuously supplied.
2H2(g) + O2(g)  2H2O(l)
anode:
2H2 + 4OH  4H2O + 4e
cathode:
4e + O2 + 2H2O  4OH

66. Chemistry In Action: Dental Filling Discomfort
Sn /Ag3Sn V 2+
Hg2 /Ag2Hg V 2+
Sn /Ag3Sn V 2+

67. Corrosion:
Process of returning metals to their natural state – the ores from which they were originally obtained.
Involves oxidation of the metal.
Some metals, such as copper, gold, silver and platinum, are relatively difficult to oxidize. These are often called noble metals.

68. The Electrochemical Corrosion of Iron
Steel is not homogeneous and has nonuniformities where
oxidation occurs easily (anodic regions)
E- flow through the moisture on the surface of the steel (salt bridge)
to the cathodic region and react with O2

69. Corrosion Prevention
1. Application of a coating (like paint or metal plating)
Chromium and Tin form durable oxide coating
Galvanizing with Zinc
2. Alloying: Stainless steel contains chromium and nickel (changes the reduction potential)
3. Cathodic Protection:
Protects steel in buried fuel tanks and pipelines.

70. Cathodic Protection
Mg: active metal furnish the e- instead of the iron, preventing the iron from oxidizing

71. Producing aluminum metal Chrome plating
Electrolysis:
Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
Electrolytic cell
Charging a battery
Producing aluminum metal
Chrome plating

72. Figure 17.19: (a) A standard galvanic cell based on the spontaneous reaction (b) A standard electrolytic cell. A power source forces the opposite reaction

73. Electrolytic Processes
Electrolytic processes are NOT spontaneous. They have:
A negative
cell potential, (-E0)
A positive free energy change, (+G)

74. 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

75. Stoichiometry of Electrolysis
current and time  quantity of charge
1 ampere = 1 coulomb/second
1 mole of e- = 96,485 coulombs
Coulombs of charge = amps (C/s) × seconds(s)
quantity of charge  moles of electrons

76. How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of A is passed through the cell for 1.5 hours?
Anode:
2Cl- (l) Cl2 (g) + 2e-
Cathode:
2 mole e- = 1 mole Ca
Ca2+ (l) + 2e Ca (s)
Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)
mol Ca = 0.452 C s
x 1.5 hr x 3600 s hr
96,500 C
1 mol e- x
2 mol e-
1 mol Ca x
= mol Ca
= 0.50 g Ca
19.8

77. Electrolysis of Water
19.8

78. Electrolysis of Water In acidic solution Anode rxn: Cathode rxn:
-1.23 V
Cathode rxn:
-0.83 V
-2.06 V

79. Electroplating a Spoon

80. Electroplating of Silver
Anode reaction:
Ag  Ag+ + e-
Cathode reaction:
Ag+ + e-  Ag
Electroplating requirements:
1. Solution of the plating metal
2. Anode made of the plating metal
3. Cathode with the object to be plated
4. Source of current

81. Solving an Electroplating Problem
Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO3 using a 20.0 Ampere current?
Ag+ + e-  Ag
5.0 g
1 mol Ag
1 mol e- C
1 s
1 mol e-
20.0 C g
1 mol Ag
= 2.2 x 102 s

82. Ag+ + e- → Ag ξ°= 0.80V Cu2+ + 2e- → Cu ξ°= 0.34V
Suppose a solution in an electrolytic cell contains the inos Cu2+, Ag+, and Zn2+. If the voltage is very low and is gradually turned up, in which order will the metals be plated out onto the cathode?
Ag+ + e- → Ag ξ°= 0.80V
Cu2+ + 2e- → Cu ξ°= 0.34V
Zn2+ + 2e- → Zn ξ°= -0.76V

83. Zn - Cu Galvanic Cell Zn  Zn2+ + 2e- E = +0.76V
The less positive, or more negative reduction potential becomes the oxidation…
Cu2+ + 2e-  Cu E = +0.34V
Zn  Zn2+ + 2e E = +0.76V
Zn + Cu2+  Zn2+ + Cu E0 = V

84. Line Notation Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) | || |
An abbreviated representation of an electrochemical cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Anode
material
Anode
solution
Cathode
solution
Cathode
material | || |