Redox Geochemistry
WHY?
• Redox gradients drive life processes!
– The transfer of electrons between oxidants and
reactants is harnessed as the battery, the source
of metabolic energy for organisms
• Metal mobility  redox state of metals and
ligands that may complex them is the critical
factor in the solubility of many metals
– Contaminant transport
– Ore deposit formation
REDOX CLASSIFICATION OF
NATURAL WATERS
Oxic waters - waters that contain
measurable dissolved oxygen.
Suboxic waters - waters that lack
measurable oxygen or sulfide, but do
contain significant dissolved iron (> ~0.1
mg L-1).
Reducing waters (anoxic) - waters that
contain both dissolved iron and sulfide.
The Redox ladder
O2
Oxic
Aerobes
H2O
NO3- Dinitrofiers
Post - oxic
N2
MnO2
Mn2+
Maganese reducers
Fe(OH)3
Fe2+
Sulfidic
Iron reducers
SO42H2S
Methanic
Sulfate reducers
CO2
CH4
Methanogens
H2O
H2
The redox-couples are shown on each stair-step, where the
most energy is gained at the top step and the least at the
bottom step. (Gibb’s free energy becomes more positive
going down the steps)
Oxidation – Reduction Reactions
•
•
•
•
Oxidation - a process involving loss of electrons.
Reduction - a process involving gain of electrons.
Reductant - a species that loses electrons.
Oxidant - a species that gains electrons.
• Free electrons do not exist in solution. Any
electron lost from one species in solution must be
immediately gained by another.
Ox1 + Red2  Red1 + Ox2
Half Reactions
• Often split redox reactions in two:
– oxidation half rxn 
• Fe2+  Fe3+ + e-
– Reduction half rxn 
• O2 + 4 e- + 4 H+  2 H2O
• SUM of the half reactions yields the total
redox reaction
4 Fe2+  4 Fe3+ + 4 eO2 + 4 e- + 4 H+ 2 H2O
4 Fe2+ + O2 + 4 H+  4 Fe3+ + 2 H2O
Redox Couples
• For any half reaction, the oxidized/reduced
pair is the redox couple:
– Fe2+  Fe3+ + e– Couple: Fe2+/Fe3+
– H2S + 4 H2O  SO42- + 10 H+ + 8 e– Couple: H2S/SO42-
ELECTRON ACTIVITY
• Although no free electrons exist in solution, it is useful
to define a quantity called the electron activity:
pe   log ae
• The pe indicates the tendency of a solution to donate or
accept a proton.
• If pe is low, there is a strong tendency for the solution to
donate protons - the solution is reducing.
• If pe is high, there is a strong tendency for the solution to
accept protons - the solution is oxidizing.
THE pe OF A HALF REACTION - I
Consider the half reaction
MnO2(s) + 4H+ + 2e-  Mn2+ + 2H2O(l)
The equilibrium constant is
K
aMn2
4
H
a a
2
e
Solving for the electron activity
 aMn2
ae    4
 Ka 
 H




1
2
THE pe OF A HALF REACTION - II
Taking the logarithm of both sides of the above
equation and multiplying by -1 we obtain:
or
 aMn2
 log ae   1 2 log  4
 a 
 H

  1 2 log K


 aMn2
pe   1 2 log  4
 a 
 H

  1 2 log K


THE pe OF A HALF REACTION - III
We can calculate K from:
 Gro
log K 
2.303RT
 ( G of  Mn2   2G of  H 2O  G of  MnO2 )

2.303RT
 ( 228.1  2( 237.1)  ( 453.1))

 43.65
3
2.303(8.314  10 )( 298.15)
so
 aMn2 
pe   1 2 log  4
 a 
 H

  21.83


WE NEED A REFERENCE
POINT!
Values of pe are meaningless without a point of
reference with which to compare. Such a point
is provided by the following reaction:
½H2(g)  H+ + eBy convention
G of  H   G of  H 2  G of e  0
so K = 1.
K
a H  ae 
1
pH22
1
THE STANDARD HYDROGEN
ELECTRODE
If a cell were set up in the laboratory based on the
half reaction
½H2(g)  H+ + eand the conditions a H+ = 1 (pH = 0) and p H2 = 1, it
would be called the standard hydrogen electrode
(SHE).
If conditions are constant in the SHE, no reaction
occurs, but if we connect it to another cell
containing a different solution, electrons may
flow and a reaction may occur.
STANDARD HYDROGEN
ELECTRODE
H2 = 1 atm
Platinum
electrode
½H2(g)  H+ + e-
aH+ = 1
ELECTROCHEMICAL CELL
H2 = 1 atm
V
Platinum
electrode
Platinum
electrode
Salt Bridge
3+
Fe
aH+ = 1
½H2(g) 
H+
2+
Fe
+
e-
Fe3+ + e-  Fe2+
ELECTROCHEMICAL CELL
We can calculate the pe of the cell on the right with
respect to SHE using:
 aFe2
pe   log 
 a 3
 Fe

  12.8


If the activities of both iron species are equal, pe =
12.8. If a Fe2+/a Fe3+ = 0.05, then
pe   log 0.05  12.8  14.1
The electrochemical cell shown gives us a method
of measuring the redox potential of an unknown
solution vs. SHE.
DEFINITION OF Eh
Eh - the potential of a solution relative to the SHE.
Both pe and Eh measure essentially the same thing.
They may be converted via the relationship:

pe 
Eh
2.303RT
Where  = 96.42 kJ volt-1 eq-1 (Faraday’s constant).
At 25°C, this becomes
pe  16.9 Eh
or
Eh  0.059 pe
Eh – Measurement and meaning
• Eh is the driving force for a redox reaction
• No exposed live wires in natural systems
(usually…)  where does Eh come from?
• From Nernst  redox couples exist at some
Eh (Fe2+/Fe3+=1, Eh = +0.77V)
• When two redox species (like Fe2+ and O2)
come together, they should react towards
equilibrium
• Total Eh of a solution is measure of that
equilibrium
FIELD APPARATUS FOR Eh
MEASUREMENTS
CALIBRATION OF ELECTRODES
• The indicator electrode is usually platinum.
• In practice, the SHE is not a convenient field reference
electrode.
• More convenient reference electrodes include saturated
calomel (SCE - mercury in mercurous chloride solution)
or silver-silver chloride electrodes.
• A standard solution is employed to calibrate the
electrode.
• Zobell’s solution - solution of potassium ferric-ferro
cyanide of known Eh.
CONVERTING ELECTRODE
READING TO Eh
Once a stable potential has been obtained, the reading can
be converted to Eh using the equation
Ehsys = Eobs + EhZobell - EhZobell-observed
Ehsys = the Eh of the water sample.
Eobs = the measured potential of the water sample relative
to the reference electrode.
EhZobell = the theoretical Eh of the Zobell solution
EhZobell = 0.428 - 0.0022 (t - 25)
EhZobell-observed = the measured potential of the Zobell
solution relative to the reference electrode.
Figure 5-6 from Kehew (2001). Plot of Eh values computed from the
Nernst equation vs. field-measured Eh values.
PROBLEMS WITH Eh MEASUREMENTS
• Natural waters contain many redox couples NOT at
equilibrium; it is not always clear to which couple (if
any) the Eh electrode is responding.
• Eh values calculated from redox couples often do not
correlate with each other or directly measured Eh values.
• Eh can change during sampling and measurement if
caution is not exercised.
• Electrode material (Pt usually used, others also used)
– Many species are not electroactive (do NOT react electrode)
• Many species of O, N, C, As, Se, and S are not electroactive at Pt
– electrode can become poisoned by sulfide, etc.
Other methods of determining the
redox state of natural systems
• For some, we can directly measure the redox
couple (such as Fe2+ and Fe3+)
• Techniques to directly measure redox SPECIES:
–
–
–
–
–
–
Amperometry (ion specific electrodes)
Voltammetry
Chromatography
Spectrophotometry/ colorimetry
EPR, NMR
Synchrotron based XANES, EXAFS, etc.
Free Energy and Electropotential
• Talked about electropotential (aka emf, Eh) 
driving force for e- transfer
• How does this relate to driving force for any
reaction defined by Gr ??
Gr = nE
or G0r = nE0
– Where n is the # of e-’s in the rxn,  is Faraday’s
constant (23.06 cal V-1), and E is electropotential (V)
• pe for an electron transfer between a redox
couple analagous to pK between conjugate acidbase pair
Electromotive Series
• When we put two redox species together, they will
react towards equilibrium, i.e., e- will move 
which ones move electrons from others better is the
electromotive series
• Measurement of this is through the electropotential
for half-reactions of any redox couple (like Fe2+ and
Fe3+)
– Because Gr = nE, combining two half reactions in a
certain way will yield either a + or – electropotential
(additive, remember to switch sign when reversing a rxn)
-E  - Gr, therefore  spontaneous
• In order of decreasing strength as a reducing agent
 strong reducing agents are better e- donors
• Fe3+ + e-  Fe2+
E0= +0.77
• 2 H2O  4 H+ + O2 + 4 eE0= +1.23
----------------------------------------------------------4 Fe3+ + H2O  4 Fe2+ + 4 H+ + O2
+2.00 V
Reaction direction!!
4 Fe2+ + 4 H+ + O2  4 Fe3+ + H2O
-2.00 V