Chapter 20
Electrochemistry
Oxidation States
electron bookkeeping
* NOT really the charge on the species but a
way of describing chemical behavior.
Oxidation: Loss of electrons
Reduction: Gain of electrons
Oxidizer: Oxidizes another species, it gets
reduced. Also called oxidizing agent or
oxidant
Reducing agent or reductant: Reduces another
species, gives up electrons, it is oxidized
Many reactions are oxidation-reduction
(redox) reactions
+5 -2
+2 -2
0
+4 -2
I2O5(s) + 5CO(g)  I2(s) + 5CO2(g)
+4 -2
+1 -2
+1 +5 -2
+2 -2
3NO2(g) + H2O(l)  2HNO3(aq) + NO(g)
Balancing Redox Reactions
Cu(s)  NO
3(aq)
 Cu
2
(aq)
 NO2(g)
Cu (s)  NO
3(aq)
1)
 Cu
Cu  Cu
2
(aq)
 NO2(g)
2
NO3-  NO2
Cu  Cu 2  2e -
2)
2H   NO3-  e -  NO2  H 2 O
Cu  Cu 2  2e -
3)

4H  2NO  2e  2NO2  2H 2 O
4) *
5)
3

-
4H  2NO  Cu  2NO2  2H 2 O  Cu
3
2
(2) 4H, 2N, 6O, Cu  (2) 4H, 2N, 6O, Cu
The following steps summarize the procedure that we use to balance an
oxidation-reduction equation by the method of half-reactions when the reaction
occurs in acid solution:
• Divide the reaction into two complete half-reactions, one for oxidation and the
other for reduction.
• Balance each half-reaction
• First, balance the elements other than H and O
• Next, balance the O atoms by adding H2O.
• Then, balance the H atoms by adding H+
• Finally, balance the charge by adding e- to the side with the greater positive
charge
• Multiply each half-reaction by an integer so that the number of electrons lost in
one half-reaction equals the number gained in the other.
• Add the two half-reactions and simplify where possible by canceling species
appearing on both sides of the equation.
• Check the equation to make sure that there are the same number of atoms of each
kind and the same total charge on both sides.
Voltaic Cells
E released from redox reaction can be
used to perform electrical work.
anode = electrodes where oxidation
occurs
cathode = electrode where reduction
occurs
reduce the cat population
Half Reactions
Zn(s)  Zn2+ + 2eCu2+ + 2e-  Cu(s)
Zn electrode  mass
because Zn(s)  Zn2+ + 2eCu electrode  mass
because Cu2+ + 2e-  Cu(s)
Zn
Cu2+
Cu2+
Zn2+
CuSO4
Zn2+
Cu2+
SO42-
Cell EMF
Electromotive force
E is favorable for e- to flow from anode to
cathode
e-’s at higher potential E in Zn electrode than
in Cu electrode
What do you know about E?
It tends to dissipate.
Lower E state is favorable!
2
Zn (s)  Zn (aq)
 2e 
ClO 3(aq)
 6H (aq)
 6e -  Cl-(aq)  3H 2O (l)
Which reaction at anode?
Which reaction at cathode?
e- 
e- 
V.M.
e- 
(-)
(+)
Salt
bridge
Which electrode is (+) ?
For the reaction
2
2
Zn (s)  Cu(aq)
 Zn (aq)
 Cu(s)
Flow of e- allows the reaction to occur
Zn
anode
e- 
e- 
V.M.
e- 
(-)
(+)
Solution of
ZnSO4
Zn2+
Salt bridge:
can’t just
build up a
charge in the
cell, keep
neutrality by
ions in salt
bridge.
Solution of
CuSO4
Cu2+
The Potential Difference between two
electrodes is measured in volts
1V = 1J/C (energy/unit charge)
Volts = Electrical pressure or electromotive
force
The driving force for reaction to take place.
Potential difference = emf = Ecell = cell
voltage
Measured in volts
Standard emf = standard cell potential,
Ecell
Standard conditions: 1M concentration
for reactants and products and 1atm
pressure when gases are used.
*Standard Electrode Potentials
Ecell = Eox + Ered
Emf and Free Energy
G = free energy
Energy produced by a reaction that can
be used for work
G is a measure of spontaneity
G = -nFE
1F  a faraday
C
J
 96,500
 96,500
mol e
V - mol e -
N = number of moles of e- transferred
E = emf of cell
+E = spontaneous
-G = spontaneous
I 2(s)  5Cu
2
(aq)
 6H 2 O (l)  12H

(aq)
 5Cu (s)  2IO
3(aq)
? G
look in App. E for standard potentials
1
2

I 2  3H 2 O  IO  6H  5e
3
I 2  2IO
Cu
2
3
 Cu
G  -nFE
G  827970J
828kJ
-
E
- 1.195
 0.337
E  -0.858V
J
F  96,500 V -mol
e-
n  10
Concentration and Cell EMF
EMF we have looked at so far has been at
standard condition. i.e. 1 molar solutions
? What if you change concentration
A guy named Nernst worked it out for you!
G = G0 + RT ln Q
Q = reaction quotient
like equilibrium expression but not at
equilibrium (ion product)
Well!
Since G = -nFE
-nFE = -nFE0 + RT ln Q
Solve for E
RT
EE ln Q
nF
in log 10
0
[product ion]
2.30 RT
EE log Q Q 
[reactant ion]
nF
0
NOW you can measure E and determine
concentration of reactant or product
? emf of
at 25°C

2Al (s)  3I2(s)  2Al3(aq)
 6I-(aq)
[Al3+] = 4.0x10-3M
[I-] = 0.010M
step 1) what’s E0
V
Al  Al3+ + 3e-
+1.66
I2  2I- + 2e-
+0.536
E0 = 2.196
step 2) R = 8.314 J/K·mol
F = 96,500
n=6
Find values for all
variables and constants
2.30RT
EE log Q
nF
0
Q = [Al3+]2[I-]6 = (4.0x10-3)2(0.010)6
= 1.6x10-17
log Q = -16.8
step 3) Plug and Chug
2.30(8.3145)(298)
E  2.196 (16.8)
6(96,500)
 2.36V
Equilibrium Constants
When E = 0, Q = Kc
The cell is no longer active
NO NET REACTION
Substitute in Nernst Equation at 298K
0.0592
0E log K c
n
nE 0
log K c 
0.0592
0
Can calculate Kc from E0 for cell. Just assume equilibrium
and replace E with 0 and Q with Kc
amps x seconds = coulombs
coulomb mol of eamp 

sec
sec
ex.
current 60.0A
4.00x103 s
How much Mg can be formed from Mg2+
1) coulombs = (4.00x103s)(60.0A)
= 2.4x105 coulombs


1
mol
e
5

2) moles e  2.4x10 C 
 96,500C 
 2.49 mol e
3) Mg
2
 2e  Mg
-
 1 mol Mg 
mol Mg  2.49 mol e 
- 
 2 mol e 
 1.24 mol Mg
-
24.3g
1.24 mol Mg x
 30.7g Mg
mol Mg
Quantitative Aspects of Electrolysis
Fe2+ + 2e-  Fe
1 mol of Fe2+ needs 2 mol of e- to make
1 mol Fe
1/2 mol Fe2+ need 1 mol e- to make 1/2
mol Fe
Al3+ + 3e-  Al
For 1 mol Al to form 1 mol Al3+ it must give
up 3 mol e-.
For 1/2 mol Al to form 1/2 mol Al3+ it must
give up 3/2 mol e-.
NOW, Can I measure electron flow?
You betcha!
e-