The Geochemistry of Rocks and
Natural Waters
Course no. 210301
Introduction to
Electrochemistry
A. Koschinsky
Electrochemistry - Redox reactions
Electrochemistry translates the chemical energy of a reduction–oxidation reaction into
electrical energy. Reduction–oxidation reactions involve the transfer of electrons.
Electrical energy comes from the movement of electrons through a wire.
When the substances involved in oxidation and reduction half–reactions are physically
separated, it is called an electrochemical cell. Each half reaction occurs on the
surface of an electrode. Each electrode is immersed in a solution containing ions
needed for the half–reaction. The solutions are connected by a salt bridge so that
ions can move between solutions. The force moving electrons through the wire is
called the electromotive force, electrochemical potential, or voltage (E). The unit for
this force is volts (V).
.
Since electrons are negative, a
negative charge might build up with
the reduction and a positive charge
with the oxidation. The salt bridge
moves ions into those solutions to
maintain electric neutrality without
mixing of solutions
A- --> A + e-
B + e- --> B-
Electrochemistry - Redox reactions
Since electrochemical potential (E) measures the driving force of a reaction, it is similar in
concept to Gibbs free energy (G). The relationship is
G = –nFE
where n is the number of electrons transferred in the redox reaction and F is Faraday's
constant, 9.65 x 104 C/mol e–. Thus a positive potential represents a spontaneous
reaction and a negative potential denotes a non-spontaneous reaction.
Like free energy, electrochemical potential is related to the equilibrium constant (K), where
E0 = E at standard conditions
Using the actual values for the constants at 25 °C and rearranging, the equation becomes
Electrochemistry - Redox reactions
Cell potentials (E) depend primarily on the identity of the reaction. Since cells separate the
half–reactions, it is often convenient to talk about the potential of the half–reaction or
the electrode potential. Only potential differences can be measured, but no single
half-cell reaction. So that electrode potentials could be tabulated, one half–cell was
defined as having an electrode potential of exactly 0 V. This half–cell is called the
standard hydrogen electrode (SHE). It uses an inert platinum electrode with 1 M H+
and 1 atm H2 for the following reaction
2 H+ + 2 e– <--> H2
Tables list potentials for reduction half–reactions at standard state (1 M, 25 °C and 105 Pa)
with SHE as the oxidation half–reaction. These are called standard reduction
potentials. To obtain standard oxidation potentials, the opposite sign is used. These
tabulated values can be used to determine the standard cell potential for any
electrochemical cell. The standard cell potential is the difference between the
reduction potential of the cathode and the reduction potential of the anode.
E°cell = E°c – E°a
Electrochemistry - Redox reactions
To correct for non-standard conditions, the Nernst equation is used to determine the cell
potential (Ecell).
where R is the gas constant, T is temperature, n is the moles of electrons in the reaction, F
is Faraday's constant, and Q is the reactant quotient, where all solutions are in units
of molarity and concentration of gases are expressed as the partial pressure in
atmospheres.
An alternative representation of redox state: pe or pe
A "p" unit is a log10-based quantity. We can report the activity of any chemical species in
p units, such as pH.
The pe is the -log10 of the electron activity:
pe = -log ae-
The pe is related to the reduction potential by the FRT factor of the Nernst equation
pe = E / 0.059
Electrochemistry - Redox reactions
The Nernst equation may be recast using pe formalism:
pe = pe0 - 1/n log Q
Example:
n
Electrochemistry Redox reactions
Half cell potential is given by:
E = E0 - (RT/nF) ln{[Red]/[Ox]}
where E0 is the thermodynamic driving force for
the reaction when the other 1⁄2 cell reaction is
SHE (standard hydrogen electrode, E0 = 0 V).
NOTE: All of these E0 values are based on
writing the equation as a reduction (this is also
known as the standard reduction potential):
Cl2 + 2e- ---> 2Cl- E0 = 1.36 V
Strong tendency to occur as a reduction (i.e. as
written)
Na+ + e- ---> Na
E0 = -2.71 V
Strong tendency to occur as an oxidation (i.e.
opposite direction as written)
Electrochemistry - Redox reactions
Examples:
E = E0 - (RT/nF) ln{[Red]/[Ox]}
1. Calculate the half cell potential for a Zn electrode in 10-2 M Zn(NO3)2
reaction: Zn2+ + 2e- ---> Zn
E = E0 - (0.059/n) log 1/[Zn2+]
E = -0.76 - 0.059/2 log {1/0.01} = -0.82 V
The reaction proceeds as an oxidation, therefore Zn is the anode.
Any half-cell potential calculated using the Nernst Equation is an E for a hypothetical cell
which has the half cell of interest as the cathode and the SHE as the anode. However, the
negative sign on our example indicates that if assembled, the Zn electrode would be the
anode.
2. Consider half-cell SHE, but p(H2) and [H+] are different: [p{H2} = 0.5 atm and [H+] = 0.01 M]
reaction: 2H+ + 2e- ---> H2
E = E0 - 0.0059/2 log {pH2/[H+]2} = 0 - 0.059/2 log [0.5/(0.01)2]
E = -0.1093 V
Electrochemistry - Redox reactions
Exercise 13
What is the pe (or redox potential E) of the following solution:
Water (pH = 7) in equilibrium with atmospheric oxygen (pO2 = 0.2);
K = 10 20.75
1/4 O2(g) + H+ + e- = 1/2 H2O
K = [H2O]1/2 / [H+] [e-] pO21/4 = 10 20.75
log K = 20.75 = 0 - log H+ - log e- - 1/4 log pO2 = pe + pH - 1/4 log pO2
pe = 20.75 - 7 - 1/4(-0.7) = 13.58
E = 0.80 V
Electrochemistry - Redox reactions
Exercise 14
What is the pe (or redox potential E) of the following solution:
Lake water (pH = 7) in equilibrium with MnO2 (manganate) and 10-5 M Mn2+
MnO2 + 4 H+ + 2 e- = Mn2+ + 2 H2O
G0 = -228.0 + 2 x (-237,18) - (-453.1) = -249.26 kJ mol-1
Log K = -G0 / 2.3 RT = 249.26 / 5.7066 = 43.6
Log K = 43.6 = log[Mn2+] + 4 pH + 2 pe
pe = 1/2 (43.6 + 5 - 28) = 10.3
E = 0.61 V
Eh-pH diagrams
Learn to construct and use Eh-pH (or pe-pH) diagrams
T = 25oC
pH = 1 atm
20
16
Fe
3+
HO
2
12
•
2
O
pO = 1 atm
2
2
•
pe
8
Fe
4
2+
Fe(OH)3(s)
•
0
HO
2
H
-4
2
Fe
-8
)2 (s
Fe = 10-6 mol L-1
-12
0
2
4
6
•
(OH
8
10
12
)
14
pH
pe-pH diagram for the Fe - O2 - H20 system
Diagrams that display
relationships between oxidized
and reduced species and
phases.
They are a type of activityactivity diagram!
Useful to depict general
relationships, but difficulties of
using field-measured pe (Eh)
values should be kept in mind.
Constructed by writing half
reactions representing the
boundaries between
species/phases.
UPPER STABILITY LIMIT OF WATER (pe-pH)
The following half reaction defines the conditions under
which water is oxidized to oxygen:
1/2O2(g) + 2e- + 2H+  H2O
The equilibrium constant for this reaction is given by
1
K = 12 2 2
pO2 ae- aH 
log K = - 1 2 log pO2 - 2 log ae- - 2 log aH 
log K = - 1 2 log pO2  2 pe  2 pH
Solving for pe we get
pe = 1 2 log K  1 4 log pO2 - pH
This equation contains three variables, so it cannot be
plotted on a two-dimensional diagram without making
some assumption about pO2. We assume that pO2 = 1
atm. This results in
1
pe =
2
log K - pH
We next calculate log K using
Gr° = -237.1 kJ mol-1
237,100 J mol -1
log K =
= 41.53
-1
-1
2.303(8.314 J K mol )(298.15 K )
pe = 20.77 - pH
LOWER STABILITY LIMIT OF WATER (pe-pH)
At some low pe, water will be reduced to hydrogen by the
reaction
H+ + e-  1/2H2(g)
1
K=
pH22
ae - a H 
log K = - 1 2 log pH 2  pe  pH
We set pH2 = 1 atm. Also, Gr° = 0, so log K = 0.
pe = - pH
T = 25oC
pH = 1 atm
20
A pe-pH diagram showing the
stability limits of water. At
conditions above the top
dashed line, water is oxidized
to O2; at conditions below the
bottom dashed line, water is
reduced to H2. No natural
water can persist outside
these stability limits for any
length of time.
2
16
O
HO
2
12
pO = 1 atm
2
2
pe
8
Water stable
4
0
HO
2
H
-4
2
-8
-12
0
2
4
6
8
pH
10
12
14
UPPER STABILITY LIMIT OF WATER (Eh-pH)
To determine the upper limit on an Eh-pH diagram, we
start with the same reaction
1/2O2(g) + 2e- + 2H+  H2O
but now we employ the Nernst equation
0.0592
1
Eh = E log 12 2
n
pO2 aH 
0
0.0592
1
Eh = E log 12 2
2
pO2 aH 
0
- G
- ( -237.1)
E =
=
= 1.23 volts
n
(2)( 96.42)
0
0
r
Eh = 1.23  0.0296 log pO22 aH2 
1
Eh = 1.23  0.0148 log pO2 - 0.0592 pH
As for the pe-pH diagram, we assume that pO2 = 1 atm.
This results in
Eh = 1.23 - 0.0592 pH
This yields a line with slope of -0.0592.
LOWER STABILITY LIMIT OF WATER (Eh-pH)
Starting with
H+ + e-  1/2H2(g)
1
we write the Nernst equation
2
p
0
.
0592
H2
0
Eh = E log
1
aH 
We set pH2 = 1 atm. Also, Gr° = 0, so E0 = 0. Thus, we
have
Eh = -0.0592 pH
1.4
Eh-pH diagram showing the
stability limits of water at
25°C and 1 bar. Note the
similarity to the pe-pH
diagram.
T = 25oC
pH = 1 atm
1.2
2
1.0
pO = 1 atm
O
2
2
HO
2
0.8
0.4
0.2
20
0.0
16
T = 25oC
pH = 1 atm
2
HO
2
H
-0.2
-0.4
O
HO
2
12
pO = 1 atm
2
2
8
2
pe
Eh (volts)
0.6
-0.6
4
0
-0.8
HO
2
H
-4
-1.0
0
2
4
6
8
10
12
14
2
-8
pH
-12
0
2
4
6
8
pH
10
12
14
Range of Eh-pH conditions in
natural environments based on
data of Baas-Becking et al.
(1960) Jour. Geol. 68: 243284.
Fe-O2-H2O SYSTEM
Fe
A preliminary
mapping of the
species and phases in
pe-pH space.
3+
Fe
Fe(OH)3(s)
2+
Fe(OH)2(s)
pH
Fe(OH)3/Fe(OH)2 BOUNDARY
First we write a reaction with one phase on each side, and
using only H2O, H+ and e- to balance, as necessary
Fe(OH)3(s) + e- + H+  Fe(OH)2(s) + H2O(l)
Next we write the mass-action expression for the reaction
1
K=
ae - a H 
Taking the logarithms of both sides and rearranging we get
log K = - log ae- - log aH  = pe  pH
And then
pe = log K - pH
Next, we calculate rG° and log K.
Gr° = Gf°Fe(OH)2 + Gf°H2O - Gf°Fe(OH)3
Gr° = (-486.5) + (-237.1) - (-696.5)
Gr° = -27.1 kJ mol-1
27,100 J mol -1
log K =
= 4.75
-1
-1
2.303(8.314 J K mol )(298.15 K)
So now we have
pe = 4.75 - pH
This is a line with slope -1 and intercept 4.75.
T = 25oC
pH = 1 atm
20
pe-pH diagram showing the
first Fe boundary. This
boundary will surely intersect
another boundary and be
truncated, but at this point
we don’t know where this
intersection will occur. So for
now, the boundary is shown
stretching across the entire
Eh-pH diagram.
2
16
O
HO
2
12
pO = 1 atm
2
2
pe
8
4
0
Fe
HO
2
H
-4
Fe
(OH
)3 (s
)
)2 (s
)
(OH
2
-8
-12
0
2
4
6
8
pH
10
12
14
Fe(OH)2/Fe2+ BOUNDARY
Again we write a balanced reaction
Fe(OH)2(s) + 2H+  Fe2+ + 2H2O(l)
No electrons are required to balance this reaction. The
mass-action expression is:
K=
aFe2 
a
2
H
log K = - log aFe2  2 pH
pH = 1 2 log K - 1 2 log aFe2
Gr° = Gf°Fe2+ + 2Gf°H2O - Gf°Fe(OH)2
Gr° = (-90.0) + 2(-237.1) - (-486.5)
Gr° = -77.7 kJ mol-1
77,700 J mol -1
log K =
= 13.61
-1
-1
2.303(8.314 J K mol )(298.15 K)
pH = 1 2 (13.61) - 1 2 log aFe2 = 6.81 - 1 2 log aFe2
To plot this boundary, we need to assume a value for Fe 
a Fe2+  m Fe2+. This choice is arbitrary - here we choose
Fe =10-6 mol L-1. Now we have
pH = 6.81 - 1 2 ( -6) = 9.81
T = 25oC
pH = 1 atm
20
This diagram illustrates the
plotting of the second
boundary required for this
diagram. Note that the portion
of the Fe(OH)3(s) /Fe(OH)2(s)
boundary from about pH 10 to
pH 0 was erased as it is
metastable. Also, the portion of
the Fe2+ /Fe(OH)2 boundary at
high pe is also metastable and
has been erased. The next
boundary to be calculated is
the Fe(OH)3(s) /Fe2+ boundary.
2
16
O
HO
2
12
pO = 1 atm
2
2
pe
8
4
0
HO
2
H
-4
Fe
Fe2+
2
-8
Fe = 10 mol L
-6
-12
0
2
4
)3 (s
Fe
-1
6
8
pH
(OH
10
(OH
)2 (s
12
)
)
14
Fe(OH)3/Fe2+ BOUNDARY
Again we write a balanced reaction
Fe(OH)3(s) + 3H+ + e-  Fe2+ + 3H2O(l)
The mass-action expression is:
K=
aFe2 
ae - a
3
H
log K = log aFe2  pe  3 pH
pe = log K - log aFe2 - 3 pH
Gr° = Gf°Fe2+ + 3Gf°H2O - Gf°Fe(OH)3
Gr° = (-90.0) + 3(-237.1) - (-696.5)
Gr° = -104.8 kJ mol-1
-1
104,800 J mol
log K =
= 18.36
-1
-1
2.303(8.314 J K mol )(298.15 K )
pe = 18.36 - 1 2 log aFe2 - 3 pH
To plot this boundary, we again need to assume a value for
Fe  a Fe2+  m Fe2+. We must now stick with the choice
made earlier, i.e., Fe =10-6 mol L-1. Now we have
pe = 18.36 - ( -6) - 3 pH = 24.36 - 3 pH
T = 25oC
pH = 1 atm
20
The third boundary is now
plotted on the diagram. This
boundary will probably
intersect the Fe2+/Fe3+
boundary, but at this point, we
do not yet know where the
intersection will be. Thus, the
line is shown extending
throughout the diagram.
2
16
O
HO
2
12
pO = 1 atm
2
2
pe
8
4
Fe
2+
0
HO
2
H
-4
Fe
)3 (s
2
Fe
-8
Fe = 10 mol L
-6
-12
0
2
4
(OH
-1
6
8
pH
10
(OH
)2 (s
12
)
)
14
Fe3+/Fe2+ BOUNDARY
We write
Fe3+ + e-  Fe2+
Note that this boundary will be pH-independent.
K=
aFe2 
a Fe2 
ae- aFe3
a Fe3
=1
pe = log K
Gr° = Gf°Fe2+ - Gf°Fe3+
Gr° = (-90.0) - (-16.7) = -73.3 kJ mol-1
-1
73,300 J mol
log K =
= 12.84
-1
-1
2.303(8.314 J K mol )(298.15 K )
pe = 12.8
The Fe2+/Fe3+ boundary now
truncates the Fe2+/Fe(OH)3
boundary as shown. There
remains just one boundary to
calculate - the Fe(OH)3(s) /Fe3+
boundary. Because the reaction
for the Fe2+/Fe3+ boundary
does not include any protons,
this boundary is horizontal, i.e.,
pH-independent.
o
T = 25 C
pH = 1 atm
20
16
2
O
Fe
3+
HO
2
12
pO = 1 atm
2
2
pe
8
Fe
4
2+
Fe(OH)3(s)
0
HO
2
H
-4
2
Fe
-8
Fe = 10 mol L
-6
-12
0
2
4
)2 (s
)
-1
6
8
pH
(OH
10
12
14
Fe(OH)3/Fe3+ BOUNDARY
Fe(OH)3(s) + 3H+  Fe3+ + 3H2O(l)
K=
aFe3
a
3
H
log K = log aFe3  3 pH
pH = 1 3 log K - 1 3 log aFe3
Gr° = Gf°Fe3+ + 3Gf°H2O - Gf°Fe(OH)3
Gr° = (-16.7) + 3(-237.1) - (-696.5) = -31.5 kJ mol-1
-1
31,500 J mol
log K =
= 5.52
-1
-1
2.303(8.314 J K mol )(298.15 K)
pH = 1 3 (5.52) - 1 3 ( -6) = 3.84
Final pe-pH diagram for the FeO2-H2O system. Note that the
solubility of iron phases is greater
when the dissolved iron species
is the reduced Fe2+. In other
words, Fe is more soluble under
reducing conditions. Because
most natural waters have pH
values in the range 5.5-8.5, they
will not contain much iron unless
redox conditions are relatively
reducing.
T = 25oC
pH = 1 atm
20
16
2
O
Fe
3+
HO
2
12
pO = 1 atm
2
2
pe
8
Fe
4
2+
Fe(OH)3(s)
0
HO
2
H
-4
2
Fe
-8
Fe = 10 mol L
-6
-12
0
2
4
)2 (s
)
-1
6
8
pH
(OH
10
12
14
Eh-pe and Eh-pH diagrams for
the Fe-O2-H2O system at two
different Fe concentrations
T = 25oC
pH = 1 atm
20
16
2
O
Fe
3+
HO
2
12
pO = 1 atm
2
2
pe
8
Fe
4
2+
Fe(OH)3(s)
0
HO
2
H
-4
2
Fe
-8
Fe = 10 mol L
-6
-12
0
2
4
)2 (s
)
-1
6
8
pH
(OH
10
12
14
pe-pH diagram for the Fe-OH-C-H2O system
pe-pH diagram for the FeO2-C-H2O system involving
other Fe minerals
Hematite = Fe3+2O3 (Fe2O3)
Magnetite = Fe2+Fe3+2O4 (Fe3O4)
Siderite = FeCO3
Pyrite = FeS2
Another type of stability (activity-activity) diagram:
Complexation of Hg in solution depending on the concentrations of bromide and
iodide as ligands