Electrochemistry:
Oxidation-Reduction Reactions
gaining to 2e-
Zn(s) + Cu+2 (aq)  Zn2+(aq) + Cu(s)
loss of 2e-
Copper is reduced because it goes down
in charge and is the oxidizing agent
Zinc is oxidized - it goes up in charge and
is the reducing agent
Electrochemistry
Messing with electrons
What is the charge on each atom
+1
+1
+7
+7
-8
+3
-2
+3
KMnO4
-3
-2
+1
Cr(OH)3
+3
-2
+3
-1
=1
Fe(OH)2+1
Electrochemistry: Oxidation-Reduction Reactions
By writing the oxidation number of each element under the
reaction equation, we can easily see the oxidation state
changes that occur
0
+1
+2
0
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(aq)
In any oxidation-reduction reaction (redox),
both oxidation and reduction must occur.
Electrochemistry:
Balancing Oxidation-Reduction Reactions
Oxidation Number Method
-3e’s X 4 = -12
4 Al(l)
0
+ 3 MnO2  2Al2O3 + 3 Mn
+4 -2
+4e’s X 3 = +12
+3 -2
0
Electrochemistry:
Balancing Oxidation-Reduction Reactions
Sample problem:
+2
-2e’s X 5 = -10
+4
I2O5(s) +5 CO(g)  I2(s) + 5 CO2(g)
+5
+5e’s X 2 = 10
0
Half-Reaction
Method
:
MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (acid)
Half-Reaction Method
(in acid)
:
MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g)
Step 1: Divide into two half-reactions,
MnO4-  Mn2+
C2O42-  CO2
Step 2: Balance main element
MnO4-  Mn2+
C2O42-  2CO2
Step 3: Balance the O: atoms by adding H2O
MnO4-  Mn2+ + 4H2O
C2O42-  2CO2
Step 4: Balance the H atoms by adding H+
8H+
-  Mn2+ + 4H O
MnO
+
4
2
C2O42-  2CO2
Step 5: Balance charge with electrons
+7
+2
5e- + 8H++ MnO4-  Mn2+ + 4H2O
-2
0
C2O42-  2CO2 + 2e-
Step 6: Make the electrons
balance
:
10
16
2
2
8
5e- + 8H++ MnO4-  Mn2+ + 4H2O
5
10
10
C2O42-  2CO2 + 2e-
Step 7: Cancel and add
10e- + 16H++ 2MnO4-  2Mn2+ + 8H2O
+
5C2O42-  10CO2 +10e-
16H+ + 2MnO4- + 5C2O42-  2Mn2+ + 10CO2 + 8H2O
Balancing in base
CN- + MnO4-  CNO- + MnO2
-1
+1
3
6 + 6 3 - 3
H2O + CN  CNO + 2H + 2e
6
3e-+
2
8
4H+ +
+3
2
MnO4  MnO2
4
+ 2H2O
0
2H+ + 3CN- + 2MnO4-  3CNO- + 2MnO2 + H2O + 2OH2OH2H2O
1
Balancing in base
Cr(OH)3 + ClO  CrO4-2 + Cl2
Have at it
4Cr(OH)3 + 6ClO + 8OH- 4CrO4-2 + 3Cl2 + 10H2O
Note that the activity
series is simply the
oxidation halfreactions of the
metals ordered from
the highest to lowest
Using the hydrogen electrode as a reference
2H+ (1 M) + 2e-  H2 (1 atm, 25 °C) E°red = 0 V
Cell EMF or Voltage
From the table:
Flip Zn+2 + 2e-  Zn red= -.76V
Cu+2 + 2e-  Cu red= .34V
Zn  Zn+2 + 2e- ox = .76V
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
E°cell = 1.10 V
E ° ox + E ° red = E°cell
oxidation
+
reduction
Voltaic or Galvanic Cells
Sample Problem. Draw a voltaic cell
using the following equation and label
all parts
Cr2O72-(aq) + I-(aq)  Cr3+(aq) + I2(s)
Spontaneity and Extent of Redox Reactions
A positive emf indicates a spontaneous process, and
a negative emf indicates a nonspontaneous one.
Sample Problem: Using the standard electrode potentials,
determine whether the following reaction are spontaneous
A. Cu(s) + 2H+(aq)  Cu2+(aq) + H2(g)
Cu(s)  Cu2+(aq) + 2eEox = -.34V
2H+(aq) + 2e-  H2(g)
Ered = 0.0V
Ecell = -.34V
B. Cl2(g) + 2I-(aq)  2Cl-(aq) + I2(s)
Cl2(g) + 2e-  2Cl-(aq)
Ered = 1.36V
2I-(aq)  2e- + I2(s)
Eox = -.54V
Ecell = .82V
EMF and Free Energy Change
Any redox reaction involves free energy change
(G) which also may be used as a measure of
spontaneity or work (max or min.)
G° = -nE °
n = # of moles of e-s transferred
 = 96500 C the charge of one mole of e-s
or 96,500 J/V-mol eNote that because n and  are both positive values, a
positive value in E leads to a negative value of G.
Use the standard electrode potentials to
calculate the standard free energy
change, G°, for the following reaction
2Br-(aq) + F2(g) Br2(l) + 2F-(aq)
2Br-(aq)  Br2(l) + 2e E°ox = -1.06
F2(g) + 2e-  2F-(aq) E°red = 2.87
2Br-(aq) + F2(g) Br2(l) + 2F-(aq) E°cell = 1.81 V
G = -nE
G = -(2 mol e- )(96,500 J/volt-mol e-)( 1.81 V)
G° = -3.49 x 105 J = -349 kJ
EMF and Equilibrium Constant
“Remember in Chapter 19, we related G° to
the equilibrium constant, K?”
G° = -RT lnK
G ° = -nE °
-nE° = -RT lnK
Simplify
E ° = 0.0591 logK
n
Calculate the K for the following reaction
O2 (g) + 4H+ (aq) + 4Fe2+ (aq)  4Fe3+ (aq) + 2H2O (l)
O2 (g) + 4H+ (aq) + 4e-  2H2O (l)
E°red = 1.23 V
4Fe2+ (aq)  4Fe3+ (aq) + 4e-
E°ox = -0.77 V
O2 (g) + 4H+ (aq) + 4Fe2+ (aq)  4Fe3+ (aq) + 2H2O (l)
E°cell = 0.46 V
log K = n E °
= 4(0.46V) = 31.1
0.0591V
0.0591V
K = 1.36 x 1031
Calculate the equilibrium constant for the reaction
IO3- (aq) + 5Cu (s) + 12H+ (aq) I2 (s) + 5Cu2+
(aq)
+ 6H2O (l)
2IO3- (aq) + 12H+ (aq) + 10e-  I2 (s) + 6H2O (l) red= 1.20V
5Cu (s)  5Cu2+ (aq) + 10e-
ox= -.34V
 = .0591 log k
.86V = .0591 log k
n
10
Log K = 145.5
K = 10145.5
cell= .86V
Standard Conditions
G° = -RT lnK
G = G° + RT lnQ
Non- Standard Conditions
G = G° + 2.30 RT log Q
-nE = -nE ° + 2.30 RT log Q
Nernst Equation
When T=298K
E = E ° - 2.30 RT log Q
n
E = E ° - 0.0591 log Q
n
Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s) E °= 1.10 V
2e- + Cu2+ (aq)  Cu (s)
Zn (s)  Zn2+ (aq) + 2e[Zn2+]
Therefore E = 1.10 ° - 0.0591 log
[Cu2+]
2
If [Zn2+]=0.05M, and [Cu2+] = 0.50 M
[0.05]
E = 1.10 ° - 0.0591 log
= 1.13V
[.50]
2
Calculate the emf generated by the the following reaction
Cr2O72- + 14 H + + 6I-  2Cr3+ + 3I2 + 7H2O
red=
1.33V
ox= -.54V
[Cr2O72-] = 2.0 M
[H + ]
= .05 M
[I-]
[Cr3+]
= .25 M
= 1.0 x 10-5M cell= .79V
 = .79V - .0591 log
 = .68V
6
(1 x 10-5)2
(2.0)(.05)14(.25)6
Concentration Cells
What happens when both cells are identical
except for concentration. Nature will try to
equalize the two cells.
E= Eº - .0591 log Q
n
E = 0 - .591 log .01 = .591V
.1
Lead Storage Battery
anode: Pb (s) + SO42- (aq)  PbSO4 (s) + 2eE °= 0.356 V
cathode:PbO2(s)+ SO42-(aq)+ H+(aq)+ 2e-  PbSO4(s)+ 2H2O(l)
E °= 1.685 V
Pb(s)+ PbO2(s)+ 4H+ +2SO4(aq) 2PbSO4(s)+
2H2O(l)
E °= 2.041 V
Note that one advantage of the lead
storage battery is that it can be
recharged because the PbSO4
produced during discharge adheres
to the electrodes
Dry Cell alkaline
anode: Zn (s)  Zn 2+ (aq) + 2ecathode:2NH4+(aq)+2MnO2(s)+ 2e Mn2O3(s) + 2NH3(aq) + H2O (l)
Fuel Cells
anode: 2H2(g)+ 4OH-(aq) 4H2O(l) + 4ecathode: 4e- + O2(g) + H2O(l)  4OH-(aq)
2H2(g) + O2(g)  2H2O(l)
Electrolytic Cells:
Notice 
(+ ox.)
These electrodes
are inert
Electrolysis
voltage source acts
like an electron
pump
(- red)
Eº = ( )
Electrolysis is driven by an outside energy source
Electrolysis of Aqueous Solutions
Sodium cannot be prepared by electrolysis of
aqueous solutions of NaCl, because water
is more easily reduced than Na+:
Possible Cathode reactions
(Note: Not in table on AP exam)
2H2O + 2e-  H2 + 2OH- E°red = -0.83
Na+ + e-  Na(s)
E°red = -2.71
The possible Anode reactions are:
2Cl-  Cl2
2H2O  4H+ + O2 + 4e-
E°ox = -1.36
E°ox = -1.23
Therefore
2Cl-(aq)  Cl2(g)
E°ox = -1.36
2H2O + 2e-  H2 + 2OH- E°red = -0.83
E°cell > -2.19
Electrolysis of With Active Electrodes
Possibilities
Ni(s)  Ni2+(aq) + 2e-
E°ox = 0.28
2H2O  4H+ + O2 + 4e- E°ox = -1.23
anode:
Ni(s)  Ni2+(aq) + 2e-
cathode:
Ni2+(aq) + 2e-  Ni(s)
Electroplating creates a silver lining
Quantitative Aspects of Electrolysis
Calculate the mass of aluminum produced in 1.00
hr. by the electrolysis of molten AlCl3 if the current
is 10.0 A. (C = amperes x seconds)
(10.0A)(1.00 hr) (3600 sec)(1C) (1) (1molAl) (27.0 g Al)
(1 hr) (1 A-s)(96,500C) (3) (1 mol Al)
= 3.36 g of Al
Quantitative Aspects of Electrolysis
The half-reaction for the formation of magnesium metal
upon electrolysis of molten MgCl2 is
Mg2+ + 2e-  Mg. Calculate the mass of magnesium
formed upon passage of 60.0 A for a period of 4000 s
(60.0A)(4000s)(1C)(1)
(mole Mg)(24.3g)
(A•s)(96,500C) (2) (mole Mg)
30.2g Mg
Electrical Work
since G= wmax
and G = - nE
then wmax= - nE
The unit employed by electric
utilities is kilowatt-hour
(kWh =3.6 x 106J)
Corrosion reactions are redox reactions in which a metal is
attacked by some substance in its environment and converted
to an unwanted compound.
All metals except gold and platinum are thermodynamically
capable of undergoing oxidation in air at room temperature
The rusting of iron is known to require oxygen; iron does not rust in water unless O2
is present.
E°red = 1.23 V
E°red = 0.44 V
Note that as the pH increases, the reduction of O2 becomes less favorable
The corrosion of Iron:Protecting the surface with tin
Tin has lower E°ox so less likely to oxidize until the
surface is broken then it accelerates as it makes a
voltaic cell with iron.
Fe(s)  Fe2+ + 2e- E°ox = 0.44 V
Sn(s)  Sn2+ + 2e- E°ox = 0.14 V
The corrosion of Iron:Cathodic Protection
Fe(s)  Fe2+(aq) + 2e- E°ox = 0.44 V
2+(aq) + 2e- E° = 0.76V
Zn(s)

Zn
ox
Zinc is more
positive and
goes first and has
an oxide coat
that seals.
sacrificial
anode
Galvanization
The corrosion of Iron:Cathodic Protection