Chapter 5 – Oxidation and Reduction
Consider:
+6
0
-2
+4
SO42- + 2CH2O + 2H+ → H2S + 2CO2 + 2H2O
reduction
oxidation
A redox reaction is the outcome of a reduction and an oxidation half
reaction
SO422- + 8e + 10H+ → H2S + 4H2O
CH2O + H2O → CO2 + 4e + 4H+
1
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
2
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
From physics, we know that:
Potential (E) = - Work (w) / Charge (q)
therefore, w = -qE
Lett us now define
L
d fi a quantity
tit called
ll d a ffaraday
d (F) and
d llett F equall th
the
charge in coulombs per mole of electrons (96,485 C).
Then q = nF
and w= -nFE.
From thermodynamics we know that:
∆G = ∆U -T∆S + ∆(PV)
and U = heat + w.
Therefore, at constant T and P: ∆G = w.
Therefore: ∆G = -nFE
and at standard state: ∆Go = -nFEo
3
Since by definition
2H+(aq) + 2e → H2(g)
Go = 0
it follows immediatelyy that the standard reduction p
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
d ’ double
don’t
d bl Eo!
Because G = -nFEo, doubling n
will have doubled G already
4
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 -0.149
-0 149 J C-1
= 115 kJ mol-1
G > 0  reaction is not spontaneous
If Eo < 0
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
Strong oxidising
agent
See Table 5.2
Weak oxidising
agent
5
Example
Calculate Eocell and G for the oxidation
of copper by nitric acid
Example
Calculate Eocell and G for the oxidation of copper by nitric acid
6
The Nernst Equation
The dependence of the electrode potential on the concentration can be obtained
from the dependence of ∆rG on concentration.
Recall that under non-standard conditions:
∆rG = ∆rG° + RT ln Q
Where Q is the reaction quotient.
For the general reaction:
aA + bB + ne- Ý cC + dD
[C]c [D]d
Q
[[A]]a [B]]b
For systems at equilibrium:
Q = Keq and ∆rG = 0
The Nernst Equation
∆rG = ∆rG° + RT ln Q
(and we know: ∆G = -nFE
∆Go = -nFEo)
-nFE = -nFE° + RT ln Q
E  Eo 
RT
ln Q
nF
Converting to base 10 log: E  E 
o
2.303RT
log Q
nF
Nernst Equation
Or
E  Eo 
0.0592
log Q
n
(at 298 K)
7
At equilibrium, Q = K, and E = 0 because G = 0
Therefore:
0  E0 
RT
ln K
nF
RT
ln K  E o
nF
l K
ln
nFE o
RT
Example
The overall equation for a fuel cell can be written as follows:
2H2(g) + O2(g) Ý 2H2O(l)
Eº = 1.23 V
What cell potential would be produced in a fuel cell operating with O2 and H2 at 5.0
bar and 298 K?
8
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.
E i i l observation:
Empirical
b
ti
reduction
d ti off one couple
l b
by another
th occurs att 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
The Nernst Equation and pH
The Nernst equation for the following half reaction at 25 °C and 1 bar H2(g) pressure
2H+(aq) + 2e- Ý H2(g)
can be written as:
E° = 0.00V
pH2
E = E° - 0.0592 log + 2
[H ]
n
pH2
E = 0 - 0.0592 log + 2
[H ]
2
pH2
E = - 0.0592 log [H+]
1
E = - 0.0592 log [H+]
9
The Nernst Equation and pH
1
E = - 0.0592 log [H+]
+
( 0 0592 log [H ])
0 0592 log(1) - (-0.0592
E = - 0.0592
+
E = - (-0.0592 log [H ])
E = - 0.0592 x pH
Example
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
The Nernst equation for O2 + 4H+ + 4e- Ý 2H2O can be written as:
1
E = E° - 0.0592 log + 4
[H ]
n
10
Example
+
E = 1.23V - (-0.0592 log [H ])
(pH = - log [H+])
E = 1.23V - 0.0592 x pH
E = 1.23V - 0.0592 x 7
E = 0.82 V
Redox stability in aqueous solution
Let X be a species that is dissolved in water
X + e → X-
EoX
Water can act as both a reducing agent:
O2 + 4H+ + 4e → 2H2O
Eo1 = 1.23V
and an oxidising agent:
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
what are the permissible values of EoX for X to survive at a given pH?
11
Eo1 = 1.23V
o
E
=
E
Q
g
o
l
V
2
9
5 n
0
.
0
O2 + 4H+ + 4e → 2H2O
1

H
p
x
V
2
9
5
0
.
0
V
3
2
.
1
=
E

as we have seen before
Similarly for
2H3O+ + 2e → 2H2O + H2
o
E
=
E
2

Q
g
o
l
V
2
9
5 n
0
.
0
Eo2 = 0.00 V
H
p
x
V
2
9
5
0
.
0
V
0
0
.
0
=
E

Let’s look specifically at pH 7
V
2
8
.
0
=
7
x
2
9
5
0
.
0
3
2
.
1
=
E

Eo1 = 1.23V
H
p
x
V
2
9
5
0
.
0
V
3
2
.
1
=
E
O2 + 4H+ + 4e → 2H2O
7
H
p
V
1
4
.
0
=
7
x
2
9
5
0
.
0
0
0
.
0
=
E

Eo2 = 0.00 V
H
p
x
V
2
9
5
0
.
0
V
0
0
.
0
=
E
2H3O+ + 2e → 2H2O + H2
7
H
p
12
Eo1 = 1.23V
V
2
8
.
0
=
E
O2 + 4H+ + 4e → 2H2O
7
H
p
Eo2 = 0.00 V
V
1
4
.
0
=
E
2H3O+ + 2e → 2H2O + H2
7
H
p
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.
Eo1 = 1.23V
V
2
8
.
0
=
E
O2 + 4H+ + 4e → 2H2O
7
H
p
Eo2 = 0.00 V
V
1
4
.
0
=
E
2H3O+ + 2e → 2H2O + H2
7
H
p
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.
13
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
spontaneous.
2X- → 2X + 2e
2H3O+ + 2e → H2 + 2H2O
2H3O+ + 2X- → 2X + 2H2O + H2
-0.10 V
-0.41 V
-0.51 V
The reaction is not spontaneous.
Eo1 = 1.23V
V
2
8
.
0
=
E
O2 + 4H+ + 4e → 2H2O
7
H
p
Eo2 = 0.00 V
V
1
4
.
0
=
E
2H3O+ + 2e → 2H2O + H2
7
H
p
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
14
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.
H2O oxidised
above this line
V
3
2
.
1
=
E
H
p
x
V
2
9
5
0
.
0
V
0
0
.
0
=
E
H
p
x
V
2
9
5
0
.
0


Figure 5.3
H2O reduced
below this line
15
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
pH 7
pH 0
0.00 V
At pH = 0
0.82 V
-0.41 V
H+|H2
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)
( )
At pH = 7
Eo = 1.92V
1.23 V
pH 7
pH 0
0.00 V
0.82 V
-0.41 V
H+|H2
16
O2|H2O
Aluminium should react spontaneously
with water at pH 7:
Al3+ + 3e → Al
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.
17
Some couples have very positive potentials but the species can still exist in
water. Examples include:
O2|H2O
pH 7
Cr2O72-|Cr3+
Eo = 1.38 V
0.82 V
MnO4-|Mn2+
Eo = 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
-0.41 V
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
Natural waters
pH controlled by CO2/H2CO3/HCO3-/CO32- equilibrium
potential controlled by
amount of dissolved O2
and by the presence of
reducing couples from
organic matter
Fig.5.12
18
Disproportionation reactions
Since
Cu+|Cu
Cu2+|Cu+
Eo = 0.52V
Eo = 0.16 V
Both potentials lie within the stability field of water
Cu+ ions neither oxidise nor reduce water
But Cu+ ions in solution 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
19
Example
Fe2+|Fe
Fe3+|Fe2+
E° = -0.44 V
E° = 0.77 V
Do we expect Fe2+ to disproportionate in aqueous solution?
Disproportionation reaction:
3Fe2+ → 2Fe(s) + Fe3+
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
20
Example
Show that the equilibrium constant lies far to the right for the following
comproportionation reaction at 25 °C:
Ag2+ + Ag → 2Ag+
Eo = 1.18 V
 Ag(II) and Ag(0) completely converted to Ag(I) in aqueous soln.
Representing potential data
Latimer diagrams summarise the standard potential (in V)
between species of an element.
ClO4+7
+1.20
ClO3+5
+1.18
HClO2
+3
+1.65
HClO
+1.67
+1
Cl2
0
+1.36
Cl-
in acid
-1
So we can immediately write down the half reaction and the
potential connecting any two species.
21
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-
ClO4+7
+1.20
ClO3-
+1.18
HClO2 + H2O
HClO2
+5
+1.65
HClO
+3
+1.67
Eo = +1.18V
Cl2
+1
+1.36
Cl-
0
in acid
-1
Suppose we are interested in the potential between two
non-adjacent couples.
Cannot just add their standard potentials!
∆rG°overall = ∆rG°individual steps
(First Law of Thermodynamics)
i.e. ∆rG°(a+b) = ∆rG°(a) + ∆rG°(b)
-noverall FEooverall =
-n1FEo1 + -n2FEo2 + -n3FEo3
22
ClO4-
+1.20
+7
ClO3-
+1.18
HClO2
+5
Eo
+1.65
+3
+1.67
HClO
+1
Cl2
+1.36
Cl-
0
in acid
-1
o + -n FEo
o
FE
+
-n
FE
-n
1
2
3
1
2
3
overall =
-noverallF
+ n3Eo3
= n1Eo1 + n2Eo2
n1 + n2 + n3
=
2 x 1.65 + 1 x 1.67
+ 1 x 1.36
2+1+1
= 1.65 V
ClO4+7
+1.20
ClO3-
+1.18
+5
HClO2
+1.65
+3
HClO
+1.67
+1
Cl2
+1.36
Cl-
0
in acid
-1
and this refers to the reaction
HClO2
+ 3H+ + 4e
Cl-
+ 2H2O
23
Frost Diagrams
A plot of NEº for the couple X(N)/X(0) vs. oxidation number (N) of the element:
X(N) + Ne- → X(0)
Eº
NEº = -∆rGº/F
Frost Diagrams
What information can one obtain from the Frost diagram below?
Lower = more thermodynamically stable
Convex = disproportionation
Concave = comproportionation
Upper left = strong oxidising agent in this diagram
Upper right= reducing agent in this diagram
Frost diagram for Mn
24
Frost Diagrams
Can be constructed from Latimer diagrams
Example
Construct a Frost diagram from the following Latimer diagram (in aq acidic solution)
Tl3+
+1.25
Tl+
-0.34
Tl
+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+ + 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
-1
-0.34 V
0
1
2
3
4
N
25
Note that in general
n2Eo2
Eo
nEo
n2Eo2 – n1Eo1
=
n2 – n1
I.e. slope = Eº
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
26
Example
Consider the following half reactions for copper:
Cu3+ + e- → Cu2+
Eo = 1.80 V
Cu+ + e- → Cu
Eo = 0.52 V
Cu2+ + 2e- → Cu
Eo = 0.34 V
(1)
Construct a Latimer diagram for copper
(2)
Construct a Frost diagram for copper
Example
Cu3+ + e- → Cu2+
Eo = 1.80 V
Cu+ + e- → Cu
Eo = 0.52 V
Cu2+ + 2e- → Cu
Eo = 0.34 V
-noverall FEooverall = -n1FEo1 + -n2FEo2
noverall FE0overall - n2FE02
Eo1=
n1F
Eo1=
(2  0.34 V) - (1  0.52 V)
1
Eo1 = 0.16 V
27
Example
3
2.49 V
2.5
Remember:
y = n  E0
nE /V
2
i.e. for Cu2+:
y = 2  0.34 V
= 0.68 V
1.5
1
0.52 V
0.5
0.68 V
0V
0
-0.5
-1
0
1
2
3
Oxidation number
Example
3
Is Cu3+ a strong oxidizing agent?
2.49 V
2.5
nE /V
2
What is the oxidation number of Cu
in the product when Cu3+ is used as
an oxidizing agent?
1.5
1
0.5
0.68 V
0.52 V
0
0V
-0.5
-1
0
1
2
3
Comment on the stability of Cu+ in
acidic aqueous solution.
Oxidation number
28
See SA pp. 164-168
Example
From the Frost diagram for N below, estimate Eº for the NO3- | NO2- couple in
acidic media (red line):
29
Example
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):
Fe3+
0.77 V
Fe(OH)2
-0.44 V
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
30
Fe3+
0.77 V
Fe(OH)2
-0.44 V
Fe
The half reaction
Fe33+ + e → Fe22+
Eo = 0.77
0 77 V
does not involve a proton so Eo is independent of pH
31
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) Ý 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
32
Fe(OH)3 Ý 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...
33
Fe3+ + e → Fe2+
Go = -74.3 kJ mol-1
Eo = 0.77 V
3OH- + 3H+ → 3H2O
-239.7 kJ mol-1
Fe(OH)3 → Fe3+ + 3OHFe(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
Fe3+
0.77 V
Fe(OH)2
-0.44 V
Fe
34
1.1
Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O
1.1
Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O
35
From the KSP for Fe(OH)2
Fe(OH)2 Ý Fe2+ + 2OH–
KSP = 1.61 x 10-15 M3
At what
hat pH will
ill [Fe2+] = 1.00
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 Ý Fe2+ + 2OH-
36
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
0.51
51 + 0
0.0592
0592 x pOH
E = -0.51 + 0.0592 x (14 – pH)
E = 0.316 – 0.0592 x pH
37
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
38
Overlaying Pourbaix diagrams
The feasibility of a reaction can be predicted by overlaying the relevant
Pourbaix diagrams
39
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
40
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+
41