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?
O2 + 4H+ + 4e → 2H2O
Eo1 = 1.23V
0.0592 V
E = E −
log Q
n
E = 1.23 V − 0.0592 V x pH
o
1
as we have seen before
Similarly for
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
E = E o2 −
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
E pH7 = 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
E pH7 = 0.00 -0.0592 x 7 = -0.41 V
O2 + 4H+ + 4e → 2H2O
Eo1 = 1.23V
E pH7 = 0.82 V
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
E pH7 = -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
E pH7 = 0.82 V
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
E pH7 = -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
E pH7 = 0.82 V
2H3O+ + 2e → 2H2O + H2
Eo2 = 0.00 V
E pH7 = -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.
H2O oxidised
above this line
E = 1.23 V − 0.0592 V x pH
E = 0.00 V − 0.0592 V x pH
Figure 5.3
H2O reduced
below this line
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 = 0
Eo
= 1.92V
1.23 V
0.82 V
pH 0
pH 7
0.00 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
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 6e 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
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
Example
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
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
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
+1.36
Cl2
Cl-
0
-1
+1
in acid
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
+1.20
ClO4+7
+1.18
+1.65
ClO3-
HClO2
+5
+3
+1.67
HClO
+1
+1.36
Cl2
Cl-
0
-1
in acid
+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