Chapter 5 – Oxidation and Reduction
Consider:
+6
2-
0
SO4 + 2CH2O +
2H+
-2
+4
→ H2S + 2CO2 + 2H2O
reduction
oxidation
A redox reaction is the outcome of a reduction and an oxidation half
reaction
SO42- + 8e + 10H+ → H2S + 4H2O
CH2O + H2O → CO2 + 4e + 4H+
By convention, redox reactions are written as REDUCTION reactions
SO42- + 8e + 10H+ → H2S + 4H2O
(2)
CO2 + 4e + 4H+ → CH2O + H2O
(3)
SO42- + 2CH2O + 2H+ → H2S + 2CO2 + 2H2O
(1)
The overall equation is the difference of the chemical equations for the two half
reactions; but number of electrons gained and lost must be the same
(1) = (2) + 2 x (-3)
 Gibbs energy for reaction = difference of the standard Gibbs
energies for the half reactions
Because half reactions must always occur in pairs, only differences in G have any
physical meaning
Chose one half reaction as a standard, and measure all Gibbs energies relative to
this half reaction
2H+(aq) + 2e → H2(g)
Go = 0
Measure G for each half reaction by coupling to a half reaction of known
G (such as the H+|H2 couple) in a galvanic cell
Fig. 5.1
For the reaction
cAA + cBB → cCC + cDD
let the reaction advance by an infinitesimally small amount d.
Greek - Xi
Then
dnA
dnB
dnC
dnD
=
=
=
=
-cAd
-cBd
+cCd
+cDd
The chemical potential  of a pure substance is defined as
 G 
 

 n T , p
For the reaction
cAA + cBB → cCC + cDD
dG = CdnC + DdnD + AdnA + BdnB
= CcCd + DcDd - AcAd - BcBd
= (CcC + DcD - AcA - BcB)d
In general


dG    cJ  J d
 J



dG    cJ  J d
 J

 G 

   cJ  J  G
J
  T , p
Hence
dG = G d
and the maximum work that the reaction can do as it proceeds through an
extent d at constant T and p is
dw = G d
As the reaction advances by d then nd electrons travel
from the anode to the cathode.
e
cathode
anode
The total charge transferred is therefore
-neNAd
= -nFd
where F = eNA is the Faraday constant
Since
Work done
Potential Difference, E =
charge transferred
dw
E=
-nFd
dw   nEFd
and since we had that dw = G d
G = -nFE
and under standard conditions
Go = -nFEo
Since by definition
2H+(aq) + 2e → H2(g)
Go = 0
it follows immediately that the standard reduction potential is
2H+(aq) + 2e → H2(g)
Eo = 0.00 V
Given
SO42- + 8e + 10H+ → H2S + 4H2O
Eo = -0.221 V
CO2 + 4e + 4H+ → CH2O + H2O
Eo = -0.072 V
SO42- + 2CH2O + 2H+ → H2S + 2CO2 + 2H2O
SO42- + 8e + 10H+ → H2S + 4H2O
Eo = -0.221 V
2CH2O + 2H2O → 2CO2 + 8e + 8H+
Eo = 0.072 V
Remember that you
don’t double Eo!
Because G = -nFEo, doubling n
will have doubled G already
SO42- + 2CH2O + 2H+ → H2S + 2CO2 + 2H2O
Eo = -0.221 V + 0.072 V = -0.149 V
Go
= -nFEo = -8 x 96 485 C mol-1 x 0.072 J C-1
= 115 kJ mol-1
G > 0  reaction is not spontaneous
If Eo < 0, the reaction is not spontaneous
For a spontaneous reaction, Eo > 0
The electrochemical series lists half reactions in order of Eo.
The more positive Eo the stronger the species is as an oxidising agent
See Table 5.2
Example
Calculate Eocell and G for the oxidation
of copper by nitric acid
Recall that for the general reaction
aA + bB → cC + dD
aCc aDd
Q a b
aA aB
where
aIi   i [I]i
and  i is the activity coefficient (=1 for dilute solutions)
Q is called the reaction quotient and Q = K, the equilibrium constant, only
when a system is at equilibrium
G = Go + RT ln Q
G = Go + RT ln Q
-nFE = -nFEo + RT ln Q
RT
E = E 
ln Q
nF
o
E = Eo 
2.303 RT
log Q
nF
Nernst Equation
or
0.0592 V
E = E 
log Q (T= 298 K)
n
o
At equilibrium, Q = K, and E = 0 because G = 0
So
E = Eo 
RT
ln Q
nF
RT
0 = E 
ln K
nF
nFEo
ln K =
RT
o
Examples
How does the reaction
O2 + 4H+ + 4e → 2H2O
Eo = 1.23 V
depend on pH? Hence determine Eo for the reaction
O2 + 2H2O + 4e → 4OH–
Example 2
Given:
Br2 + 2e → 2Br–
Cl2 + 2e → 2Cl–
Eo = 1.065 V
Eo = 1.359 V
calculate the equilibrium constant at standard thermodynamic conditions for the reaction
Br2 + 2Cl– → Cl2 + 2Br–
OVERPOTENTIALS
Since
Co2+ + 2e → Co
2H+ + 2e → H2
Eo = -0.277 V
Eo = 0.000 V
cobalt metal in contact with acid should release hydrogen vigorously
Co + 2H+ → Co2+ + H2
Eocell = 0.277 V
The reaction does occur but is slow.
Thermodynamics tells us nothing about the speed (kinetics) of a reaction.
Empirical observation: reduction of one couple by another occurs at a
significant rate only when the difference in potentials of the two couples
exceeds a characteristic value known as the overpotential.
To reduce H+ to H2: need an overpotential of about -0.6 V
To oxidise H2O to O2: need an overpotential of about +0.6 V
Example
Assuming that we can maintain an exposed surface, will Al undergo rapid oxidation in
water at pH 7?
Al3+ + 3e → Al
2H+ + 2e → H2
Eo = -1.66 V
Eo = 0.00 V
Example 2
What minimum potential must a couple have to be able to oxidise water at pH 7?
Xn+ + ne → X
O2 + 4H+ + 4e → 2H2O
Eo = x V
Eo = 1.23 V
Redox stability in aqueous solution
Let X be a species that is dissolved in water
X + e → X-
EoX
Now since:
O2 + 4H+ + 4e → 2H2O
Eo1 = 1.23V
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
and:
what are the permissible values of EoX for X to survive at a given pH?
O2 + 4H+ + 4e → 2H2O
Eo1 = 1.23V
0.0592 V
E = E 
log Q
n
o
1
E = 1.23 V  0.0592 V x pH
as we have seen before
Similarly for
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
E = Eo 2 
0.0592 V
log Q
n
E = 0.00 V  0.0592 V x pH
Let’s look specifically at pH 7
O2 + 4H+ + 4e → 2H2O
Eo1 = 1.23V
E = 1.23 V  0.0592 V x pH
EpH7 = 1.23 - 0.0592 x 7 = 0.82 V
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
E = 0.00 V  0.0592 V x pH
EpH7 = 0.00 - 0.0592 x 7 = -0.41 V
O2 + 4H+ + 4e → 2H2O
Eo1 = 1.23V
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
EpH7 = 0.82 V
EpH7 = -0.41 V
X + e → X-
EoX
Suppose EoX < -0.41 V. For example, let EoX = -0.50 V
2X- → 2X + 2e
2H3O+ + 2e → H2 + 2H2O
2H3O+ + 2X- → 2X + 2H2O + H2
0.50 V
-0.41 V
0.09 V
The reaction is spontaneous and X- will not survive. It will be oxidised by water.
O2 + 4H+ + 4e → 2H2O
Eo1 = 1.23V
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
EpH7 = 0.82 V
EpH7 = -0.41 V
X + e → X-
EoX
Suppose EoX > 0.82 V. For example, let EoX = 0.90 V
4X + 4e → 4X2H2O → O2 + 4H+ + 4e
2H2O + 4X → 4X- + O2 + 4H+
0.90 V
-0.82 V
0.08 V
The reaction is spontaneous and X will not survive. It will by reduced by water.
Finally suppose -0.41 V < EoX < 0.82 V. For example, let EoX = 0.10 V
4X + 4e → 4X2H2O → O2 + 4H+ + 4e
2H2O + 4X → 4X- + O2 + 4H+
0.10 V
-0.82 V
-0.72 V
The reaction is not spontaneous.
2X- → 2X + 2e
2H3O+ + 2e → H2 + 2H2O
2H3O+ + 2X- → 2X + 2H2O + H2
The reaction is not spontaneous.
-0.10 V
-0.41 V
-0.51 V
O2 + 4H+ + 4e → 2H2O
Eo1 = 1.23V
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
EpH7 = 0.82 V
EpH7 = -0.41 V
0.82 V
For a species to be stable in water, its redox potential
must lie between these values at pH 7.
This is sometimes referred to as
the Electrochemical Window
-0.41 V
Recall, however, that the overpotential may make species with Eo values
outside the electrochemical window kinetically stable – i.e., they can exist in
solution because the kinetics of their reaction with water are slow
By applying the Nernst equation, we can get the stability field of water at
any pH.
E = 1.23 V  0.0592 V x pH
E = 0.00 V  0.0592 V x pH
Figure 5.3
Example
Is Co(III) (a) thermodynamically and (b) kinetically stable in water at pH 0
and at pH 7?
O2|H2O
Co(III) + e → Co(II)
Eo
= 1.92V
1.23 V
0.82 V
pH 0
pH 7
0.00 V
-0.41 V
H+|H2
Aluminium should react spontaneously
with water at pH 7:
Al3+ + 3e → Al
O2|H2O
Eo = -1.66 V
H+|H2
2Al → 2Al3+ + 6e
6H+ + 6e → 3H2
Eo = 1.66 V
EpH7 = -0.41
2Al + 6H+ → 2Al3+ + 3H2
Eo = 1.25 V
pH 7
0.82 V
-0.41 V
It doesn’t because the surface is passivated by the formation of a tough coating
of Al2O3 which sticks tightly to the bulk Al metal and protects it.
This can be enhanced by anodising the metal.
The metal is made the anode in an electrolytical cell and oxidised to form the
protective film of the oxide.
Nitric acid can be used to passivate some metals such as stainless steels.
Some couples have very positive potentials but the species can still exist is
water. Examples include:
O2|H2O
pH 7
Cr2O72-|Cr3+
Eo = 1.38 V
0.82 V
2+
o
MnO4 |Mn
E = 1.51 V
Hence at pH 7 the Ecell for oxidation of
water to produce O2 is 0.56 V and 0.69 V,
respectively.
H+|H2
Reason why Cr2O72- and MnO4- can still exist:
Requires a 3e and a 5e transfer, respectively.
Multiple electron transfer reactions are usually very slow.
These species are under kinetic control
-0.41 V
Natural waters
potential controlled by
amount of dissolved O2
and by the presence of
reducing couples from
organic matter
Fig.5.12
pH controlled by CO2/H2CO3/HCO3-/CO32- equilibrium
Disproportionation reactions
Since
Cu+|Cu
Cu2+|Cu+
Eo = 0.52V
Eo = 0.16 V
Cu+ lies in the stability field of water
But Cu+ solutions are unstable because they can undergo
disproportionation
2Cu+ → Cu2+ + Cu(s)
A disproportionation reaction is one where the oxidation state of an
element is simultaneously increased and decreased
Disproportionation occurs because:
Cu+ + e → Cu
Cu+ → Cu2+ + e
Eo = 0.52
Eo = -0.16
2Cu+ → Cu + Cu2+
Eo = 0.36 V
In a comproportionation reaction two species of an element in two different
oxidation states form a product in which the element is in an intermediate
oxidation state
Ag2+ + Ag → 2Ag+
Eo = 1.18 V
Representing potential data
Latimer diagrams summarise the standard potential (in V)
between species of an element.
+1.20
ClO4+7
+1.18
+1.65
ClO3-
HClO2
+5
+3
+1.67
HClO
+1
+1.36
Cl2
Cl-
0
-1
in acid
So we can immediately write down the half reaction and the
potential connecting any two species.
For example, ClO3- and HClO2
ClO3-
HClO2
- Balance O by adding H2O
ClO3-
HClO2 + H2O
- In acid, balance H by adding H+. (In base, add H2O to
the side of the equation requiring H, and OH- to the other
side.)
ClO3- + 3H+
HClO2 + H2O
- Balance charges by adding e
ClO3- + 3H+ + 2e
HClO2 + H2O
Eo = +1.18V
+1.20
ClO4+7
+1.18
+1.65
ClO3-
HClO2
+5
+3
+1.67
HClO
+1
+1.36
Cl2
Cl-
0
-1
in acid
Suppose we are interested in the potential between two
non-adjacent couples.
Goverall =  Gindividual steps
(First Law of Thermodynamics)
-noverall FEooverall = -n1FEo1 + -n2FEo2 + -n3FEo3
+1.20
ClO4+7
+1.18
+1.65
ClO3-
HClO2
+5
+3
+1.67
HClO
+1
+1.36
Cl2
Cl-
0
-1
in acid
o + -n FEo
o
-n
FE
+
-n
FE
1
1
2
2
3
3
Eooverall =
-noverallF
+ n3Eo3
= n1Eo1 + n2Eo2
n1 + n2 + n3
=
2 x 1.65 + 1 x 1.67
2+1+1
= 1.65 V
+ 1 x 1.36
+1.20
ClO4+7
+1.18
+1.65
ClO3-
HClO2
+5
+3
+1.67
HClO
+1
+1.36
Cl2
Cl-
0
-1
and this refers to the reaction
HClO2 + 3H+ + 4e
Cl-
+ 2H2O
in acid
Frost diagrams are plots of X(N)/X(0) (in units of nEo) against
the oxidation number N (or n)
X(N) + Ne-
X(0)
Since G = -nFEo, G  -nEo
Eo
Example
Construct a Frost diagram from the following Latimer
diagram (in aq acidic solution)
Tl3+
+1.25
Tl+
-0.34
+0.72
Tl+ + e = Tl0
Eo = -0.34 V
n=1
nEo = -0.34 V
Tl3+ + 3e = Tl0
Eo = +0.72 V
n=3
nEo = +2.16 V
Tl
Tl+ + e = Tl0
Eo = -0.34 V
n=1
nEo = -0.34 V
Tl3+ + 3e = Tl0
Eo = +0.72 V
n=3
nEo = +2.16 V
3
nEo
2
+2.16 V
1
0
-0.34 V
-1
0
1
2
3
4
N
Note that in general
n2Eo2
nEo
n2Eo2 – n1Eo1
Eo =
n2 – n1
n1Eo1
n2
n1
N
Tl+ is the most stable oxidation state
Tl3+ is a strong oxidising agent because slopes of the lines to the
other two states are large and positive
See SA pp. 155-158
Pourbaix diagrams
Plots of E vs pH
We will, as an example, derive the Pourbaix diagram for iron
Two Latimer diagrams pertain
In acid ([H+] = 1 M):
0.77 V
Fe3+
-0.44 V
Fe(OH)2
Fe
In alkali ([OH-] = 1 M)
-0.56 V
Fe3+
-0.887 V
Fe(OH)2
Fe
Pourbaix diagrams:
•correlate Latimer diagrams at pH 0 and pH 14
•take into account speciation or oxidation state of the element
0.77 V
Fe3+
-0.44 V
Fe(OH)2
The half reaction
Fe3+ + e → Fe2+
Eo = 0.77 V
does not involve a proton so Eo is independent of pH
Fe
Fe3+ + e → Fe2+
Fe3+ will precipitate out of solution as pH is increased. We can calculate the pH
at which this will occur from the KSP for Fe(OH)3.
Fe(OH)3(s)  Fe3+ + 3OH–
KSP = 4.11 x 10-37 M4
At what pH will [Fe3+] = 1.00 M?
KSP = 4.11 x 10-37 M4 = [Fe3+][OH–]3
[OH–] = (4.11 x 10-37/1)0.333
= 7.43 x 10-13 M
So [H+] = 10-14/7.43 x 10-13 = 1.35 x 10-2 M
hence pH = 1.87
Fe(OH)3  Fe3+ + 3OH-
Vertical lines in a Pourbaix
diagram indicate where two
species of an element in the
same oxidation state are in
equilibrium
To calculate the
Fe(OH)3|Fe2+
line...
Fe3+ + e → Fe2+
Eo = 0.77 V
3OH- + 3H+ → 3H2O
Go = -74.3 kJ mol-1
-239.7 kJ mol-1
Fe(OH)3 → Fe3+ + 3OH-
Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O
207.6 kJ mol-1
Go = -nFEo
-106.4
kJ mol-1
= -1 x 96485
x 0.77
Go = -RT ln KSP
= -8.315 x 298 x ln (4.11 x 10-37)
Go = -nFEo
Eo = -Go /nF
= 106400/1 x 96485
= 1.10 V
Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O
Eo = 1.10 V
E = Eo – RT/nF ln Q
E = 1.10 – 3 x 0.0592 x pH
This must cross the Fe3+/Fe(OH)3 line when
0.77 = 1.10 – 3(0.0592)pH
or pH = 1.87
which confirms the result we got from the KSP calclation
0.77 V
Fe3+
-0.44 V
Fe(OH)2
Fe
1.1
Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O
1.1
Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O
From the KSP for Fe(OH)2
Fe(OH)2  Fe2+ + 2OH–
KSP = 1.61 x 10-15 M3
At what pH will [Fe2+] = 1.00 M?
KSP = 1.61 x 10-15 M3 = [Fe2+][OH–]2
[OH–] = (1.61 x 10-15/1)0.5
= 4.01 x 10-8 M
So [H+] = 10-14/4.01 x 10-8 = 2.49 x 10-7 M
hence pH = 6.61
1.1
Fe(OH)2  Fe2+ + 2OH-
The half reaction
Fe2+ + 2e → Fe
Eo = -0.44 V
does not involve a proton so Eo is independent of pH
1.1
Fe2+ + 2e → Fe
An expression for the potential for the Fe(OH)3|Fe(OH)2 couple can be derived
from the following data
Fe(OH)3 + 3H+ + e → Fe2+ + 2H2O
Eo 1.10 V
Go -106.4 kJ mol-1
3H2O → 3H+ + 3OH-
239.7 kJ mol-1
Fe2+ + 2OH- → Fe(OH)2
-84.4 kJ mol-1
Fe(OH)3 + e → Fe(OH)2 + OH-
Eo -0.51 V
Go 48.9 kJ mol-1
E = Eo – RT/nF ln Q
E = -0.51 + 0.0592 x pOH
E = -0.51 + 0.0592 x (14 – pH)
E = 0.316 – 0.0592 x pH
1.1
0.316
Fe(OH)3 + e → Fe(OH)2 + OH-
...and finally the value of Fe(OH)2|Fe couple can be found by
similar considerations, and the Nernst equation applied.
E = -0.060 – 0.0592 x pH
Overlaying Pourbaix diagrams
The feasibility of a reaction can be predicted by overlaying the relevant
Pourbaix diagrams
stability field for As(V)
stability field for As(III)
At pH < 5.5 and at
pH > 9, Fe3+ has the
potential to oxidise
As3+ to As5+
5.5
9
For example
0.65
0.45
Fe(OH)3 + e + 3H+ → Fe2+ + 3H2O
E = 0.65
2
As3+ → As5+ + 2e
E = -0.45
As3+ + 2Fe(OH)3 + 6H+ → 2Fe2+ + 6H2O + As5+ E = 0.20 V
For 5.5 < pH < 9
As5+ will oxidise
Fe2+ to Fe3+
The effect of complex formation on Eo values
The Eo value of a metal ion is very dependent on the ligands of the ion
Example, for the Fe3+|Fe2+ couple
Ligand
phenanthroline
H 2O
CN-
Eo /V
1.14
0.77
0.36
Ligand
phenanthroline
H2O
CN-
N
N
N
N
N
N
N
Fe
Fe
N
N
N
 back bonding from
metal to phen ligand
stabilises Fe(II)
Eo /V
1.14
0.77
0.36
Ligand
phenanthroline
H2O
CN-
Eo /V
1.14
0.77
0.36
-
CN
-
-
NC
CN
Fe
-
-
NC
CN
-
CN
Negatively charged ligands favour the
higher positive charge of Fe(III)
NH3
H 3N
Co3+|Co2+
NH3
Co
H 3N
0.11 V
NH3
NH3 is a better 
donor ligand than
H2O and so stablises
Co(III)
NH3
OH2
H2 O
OH2
1.84 V
Co
H 2O
OH2
OH2