1. Electrochemistry
Ch 13 pg 225 Princeton Review

2. 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-)
19.1

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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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4. Galvanic Cell Flow of e- Flow of current Voltmeter e– e– Zinc anode
Copper
cathode
Cl– K+
Salt bridge
Cotton
plugs SO 4 2–
Cu2+
Zn2+ SO 4 2–
ZnSO4 solution
CuSO4 solution
2e–
2e–
Cu2+ Zn
Zn2+ Cu
Zn is oxidized
to Zn2+ at anode.
Cu2+ is reduced
to Cu at cathode.
Zn(s)
Zn2+(aq) + 2e–
2e– + Cu2+(aq)
Cu(s)
Net reaction
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)

5. Anode Reaction
Cathode Reaction

6. SO 4 2–
Cu2+
Zn2+ SO 4 2–
ZnSO4 solution
CuSO4 solution

7. Voltmeter e– e–
Zinc
anode
Copper
cathode
Cl– K+
Salt bridge
Cotton
plugs SO 4 2–
Cu2+
Zn2+ SO 4 2–
ZnSO4 solution
CuSO4 solution
Salt bridge provides electrical neutrality by providing negative anions to equal the positive cations being created at the Zn anode during oxidation. And cations ions (K+) to replace Cu 2+ being used up at reduction.

8. Voltmeter e– e–
Zinc
anode
Copper
cathode
Cl– K+
Salt bridge
Cotton
plugs SO 4 2–
Cu2+
Zn2+ SO 4 2–
ZnSO4 solution
CuSO4 solution Zn

9. Voltmeter e– e–
Zinc
anode
Copper
cathode
Cl– K+
Salt bridge
Cotton
plugs SO 4 2–
Cu2+
Zn2+ SO 4 2–
ZnSO4 solution
CuSO4 solution
2e– Zn
Zn2+
Zn is oxidized
to Zn2+ at anode.
Zn(s)
Zn2+(aq) + 2e–

10. Voltmeter e– e–
Zinc
anode
Copper
cathode
Cl– K+
Salt bridge
Cotton
plugs SO 4 2–
Cu2+
Zn2+ SO 4 2–
ZnSO4 solution
CuSO4 solution
2e– Zn
Zn2+ Cu
Zn is oxidized
to Zn2+ at anode.
Zn(s)
Zn2+(aq) + 2e–

11. Voltmeter e– e–
Zinc
anode
Copper
cathode
Cl– K+
Salt bridge
Cotton
plugs SO 4 2–
Cu2+
Zn2+ SO 4 2–
ZnSO4 solution
CuSO4 solution
2e–
2e–
Cu2+ Zn
Zn2+ Cu
Zn is oxidized
to Zn2+ at anode.
Zn(s)
Zn2+(aq) + 2e–

12. Voltmeter e– e–
Zinc
anode
Copper
cathode
Cl– K+
Salt bridge
Cotton
plugs SO 4 2–
Cu2+
Zn2+ SO 4 2–
ZnSO4 solution
CuSO4 solution
2e–
2e–
Cu2+ Zn
Zn2+ Cu
Zn is oxidized
to Zn2+ at anode.
Cu2+ is reduced
to Cu at cathode.
Zn(s)
Zn2+(aq) + 2e–
2e– + Cu2+(aq)
Cu(s)

13. Voltmeter e– e–
Zinc
anode
Copper
cathode
Cl– K+
Salt bridge
Cotton
plugs SO 4 2–
Cu2+
Zn2+ SO 4 2–
ZnSO4 solution
CuSO4 solution
2e–
2e–
Cu2+ Zn
Zn2+ Cu
Zn is oxidized
to Zn2+ at anode.
Cu2+ is reduced
to Cu at cathode.
Zn(s)
Zn2+(aq) + 2e–
2e– + Cu2+(aq)
Cu(s)
Net reaction
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)

14. Galvanic Cells anode oxidation cathode reduction spontaneous
redox reaction
19.2

15. Current The flow of positive charge.
Current is always in the opposite directions from electron flow.
Flow of e-
Flow of current

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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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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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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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19. Electrolytic Cells
The difference in electrical potential between the anode and cathode is called:
cell voltage
electromotive force (emf)
cell potential
Used in electro plating
Cell Diagram
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M & [Zn2+] = 1 M
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
Anode OX
Cathode
RED
19.2

20. Reduction Potentials
reduction potential (E0): is the voltage associated with a reduction reaction
The more positive E0 the greater the tendency for the substance to be reduced
On the AP test you will be given a chart of reduction potentials. You can reverse them and change the sign on the voltage to get oxidation potentials.
19.3

21. 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
19.3

22. Using the Table
TOP:
F2 + 2e-  2F- = E° = +2.87
Large reduction potential = more likely to be reduced and thus a strong oxidizing agent.
Bottom:
Li+ + e-  Li = E° = -3.05
Li Li+ + e- = E° = +3.05
Large oxidation potential = more likely to lose an electron and become oxidized thus it’s a good reducing agent.

23. Spontaneity of Redox Reactions
Redox rxns will occur spontaneously if its cell potential has a positive value. Or if its free energy (G) is negative.
DG0
=-nFEcell
19.4

24. Example Look at the spontaneous rxn: Zn + Ag+  Zn2+ + Ag E0 = + 1.56
Zn  Zn2+ + 2e- (LEO)
E0 = reverse = +0.76
Ag+ + e- Ag (GER)
E0 =
E = Eox + Ered
= =
E is positive so G is negative and the rxn is spontaneous

25. Example with coefficients
The number of electrons lost must equal the number of electrons gained. In order to insure this we must multiply the half reaction by integers to balance. This does not change the Cell potential E°.

26. Example Fe3+ + Cu  Cu2+ + Fe 2+ ½ rxns
GER Fe3+ + e-  Fe E° = 0.77 V
LEO Cu  Cu e E° = V
The reduction must occue 2x for every 1 OX rxn
Note the voltage of the OX is reversed because our chart is of reduction potentials
E° cell = V E° = 0.43 V

27. ∆G° = - nFE° G = Gibbs free energy kJ/mol n = moles of e- exchanged
Under standard conditions the maximum cell potential (E°) is directly related to the free energy (G) difference between the reactants and the products.
This equation allows us to test for ∆G° in the lab.
∆G° = - nFE°
G = Gibbs free energy kJ/mol
n = moles of e- exchanged
F = Faraday’s constant 96,500 coulombs/mole
(how much charge is produced for every 1 mole of e- )
E° = Standard reduction potential

28. Example Calculate ∆G° for the reaction Fe + Cu2+  Cu + Fe 2+
Write ½ rxns to ID red.ox atoms
Fe +  Fe 2+
LEO (rev potential)
Cu2+  Cu
GER
Use chart to find reduction potentials
Fe +  Fe E = = v
Cu2+  Cu E = v

29. E cell = = 0.78
F = 96,500
n = 2 mole (2 e- per atom)
3. Plug in
G = -2(96,500)(0.78)
= x 105 J
4. Conclude
G = negative so rxn is spontaneous
E cell = positive spontaneous
** if E is negative rxn will not occur and it is not spontaneous

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Nernst Equation
The relationship between cell potential and concentrations of cell components
At 25°C: or
(at equilibrium)
K and Q = [P]/[R]
n = # of electrons
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CONCEPT CHECK!
Explain the difference between E and E°.
When is E equal to zero?
ε is the cell potential at any condition
When is E° equal to zero?
ε is the cell potential under standard conditions (1.0M, or 1 atm and 25C)
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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32. Q = reaction quotient
Q = concentration of products [ ] raised to its coefficient divided by [ ] reactants raised to its coefficient

33. Example Calculate E for the following reaction given
[Al3+] = 1.5M [Mn2+] = 0.5 M Ecell = 0.48
2 Al + 3 Mn2+  2Al Mn
Q = = 18
0.53
n = 2 Al  2Al Mn2+  3 Mn
= 3(2) = = 2(3)= 6
n = 6

34. Cont.
Plug it in:
E = 0.48 – (0.0591/6) log 18
= 0.47 V

35. Electrolysis and Mass Changes
charge (C) = current (A) x time (s)
1 mole e- = 96,500 C
19.8

36. 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