Transcript Document

Electrochemistry
Ch 13 pg 225
Princeton Review
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
0
0
2+ 2-
2Mg (s) + O2 (g)
2Mg
O2 + 4e-
2MgO (s)
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2O2-
Reduction half-reaction (gain e-)
19.1
Flow of e-
Galvanic Cell
Flow of current
Voltmeter
e–
Zinc
anode
e–
Cotton
plugs
SO42–
Cu2+
SO42–
Zn2+
ZnSO4 solution
2e–
Copper
cathode
Cl–
K+
Salt bridge
CuSO4 solution
2e–
Cu2+
Zn
Zn2+
Cu
Zn is oxidized
to Zn2+ at anode.
Zn(s)
Cu2+ is reduced
to Cu at cathode.
Zn2+(aq) + 2e–
2e– + Cu2+(aq)
Net reaction
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Cu(s)
SO42–
Cu2+
Zn2+
ZnSO4 solution
SO42–
CuSO4 solution
Voltmeter
e–
Zinc
anode
e–
Copper
cathode
Cl–
K+
Salt bridge
Cotton
plugs
SO42–
Zn2+
ZnSO4 solution
Cu2+
SO42–
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.
Voltmeter
e–
Zinc
anode
e–
Cotton
plugs
SO42–
Zn2+
ZnSO4 solution
Zn
Copper
cathode
Cl–
K+
Salt bridge
Cu2+
SO42–
CuSO4 solution
Voltmeter
e–
Zinc
anode
e–
Cotton
plugs
SO42–
Zn2+
ZnSO4 solution
2e–
Zn
Zn2+
Zn is oxidized
to Zn2+ at anode.
Zn(s)
Zn2+(aq) + 2e–
Copper
cathode
Cl–
K+
Salt bridge
Cu2+
SO42–
CuSO4 solution
Voltmeter
e–
Zinc
anode
e–
Cotton
plugs
SO42–
Zn2+
ZnSO4 solution
2e–
Zn
Copper
cathode
Cl–
K+
Salt bridge
Cu2+
SO42–
CuSO4 solution
Zn2+
Cu
Zn is oxidized
to Zn2+ at anode.
Zn(s)
Zn2+(aq) + 2e–
Voltmeter
e–
Zinc
anode
e–
Cotton
plugs
SO42–
Zn2+
ZnSO4 solution
2e–
Copper
cathode
Cl–
K+
Salt bridge
Cu2+
SO42–
CuSO4 solution
2e–
Cu2+
Zn
Zn2+
Cu
Zn is oxidized
to Zn2+ at anode.
Zn(s)
Zn2+(aq) + 2e–
Voltmeter
e–
Zinc
anode
e–
Cotton
plugs
SO42–
Zn2+
ZnSO4 solution
2e–
Copper
cathode
Cl–
K+
Salt bridge
Cu2+
SO42–
CuSO4 solution
2e–
Cu2+
Zn
Zn2+
Cu
Zn is oxidized
to Zn2+ at anode.
Zn(s)
Zn2+(aq) + 2e–
Cu2+ is reduced
to Cu at cathode.
2e– + Cu2+(aq)
Cu(s)
Voltmeter
e–
Zinc
anode
e–
Cotton
plugs
SO42–
Cu2+
SO42–
Zn2+
ZnSO4 solution
2e–
Copper
cathode
Cl–
K+
Salt bridge
CuSO4 solution
2e–
Cu2+
Zn
Zn2+
Cu
Zn is oxidized
to Zn2+ at anode.
Zn(s)
Cu2+ is reduced
to Cu at cathode.
Zn2+(aq) + 2e–
2e– + Cu2+(aq)
Net reaction
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Cu(s)
Galvanic Cells
anode
oxidation
cathode
reduction
spontaneous
redox reaction
19.2
Current
• The flow of positive charge.
• Current is always in the opposite directions
from electron flow.
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
Cathode
OX
RED
19.2
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
•
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
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.
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
0
=-nFEcell
19.4
Example
Look at the spontaneous rxn:
Zn + Ag+  Zn2+ + Ag
E0 = + 1.56
Zn  Zn2+ + 2e- (LEO)
E0 = -0.76 reverse = +0.76
Ag+ + e- Ag (GER)
E0 = + 0.80
E = Eox + Ered
= .76 + .80 = + 1.56
E is positive so G is negative and the rxn is
spontaneous
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°.
Example
Fe3+ + Cu  Cu2+ + Fe 2+
½ rxns
GER Fe3+ + e-  Fe 2+
LEO Cu  Cu2+ + 2 e1.
2.
E° = 0.77 V
E° = -0.34 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 = - 0. 34 V + 0.77 E° = 0.43 V
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
Example
Calculate ∆G° for the reaction
Fe + Cu2+  Cu + Fe 2+
1. Write ½ rxns to ID red.ox atoms
Fe +  Fe 2+
0
+2
LEO (rev potential)
Cu2+  Cu
+2
0
GER
2. Use chart to find reduction potentials
Fe +  Fe 2+
E = -0.44 = +0.44 v
Cu2+  Cu
E = +0.34 v
E cell = 0.44+0.34 = 0.78
F = 96,500
n = 2 mole (2 e- per atom)
3. Plug in
G = -2(96,500)(0.78)
= - 1.5 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
Nernst equation
Gives the relationship between cell potential
and concentrations of cell components.
E = E°cell - (0.0591/n) log Q
Q = reaction quotient
Q = concentration of products [ ] raised to its
coefficient divided by [ ] reactants raised to
its coefficient
Example
Calculate E for the following reaction given
[Al3+] = 1.5M [Mn2+] = 0.5 M
Ecell = 0.48
2 Al + 3 Mn2+  2Al 3+ + 3 Mn
Q = 1.52
= 18
0.53
n = 2 Al  2Al 3+
0
+3 = 3(2) = 6
n=6
3 Mn2+  3 Mn
+2
0 = 2(3)= 6
Cont.
Plug it in:
E = 0.48 – (0.0591/6) log 18
= 0.47 V
Electrolysis is the process in which electrical energy is used
to cause a nonspontaneous chemical reaction to occur.
19.8
Electrolysis of Water
19.8
Electrolysis and Mass Changes
charge (C) = current (A) x time (s)
1 mole e- = 96,500 C
19.8
How much Ca will be produced in an electrolytic cell of
molten CaCl2 if a current of 0.452 A is passed through the
cell for 1.5 hours?
Anode:
Cathode:
2Cl- (l)
Ca2+ (l) + 2eCa2+ (l) + 2Cl- (l)
Cl2 (g) + 2eCa (s)
Ca (s) + Cl2 (g)
2 mole e- = 1 mole Ca
C
s 1 mol e- 1 mol Ca
mol Ca = 0.452
x 1.5 hr x 3600 x
x
s
hr 96,500 C 2 mol e= 0.0126 mol Ca
= 0.50 g Ca
19.8
Chemistry In Action: Dental Filling Discomfort
2+
Hg2 /Ag2Hg3 0.85 V
2+
Sn /Ag3Sn -0.05 V
2+
Sn /Ag3Sn -0.05 V