Updated: November 28, 2001. Copyright © 2001 byWalt W. McNab, Concord, CA, U.S.A. All Rights Reserved.
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MULTISPECIES REACTIVE TRANSPORT IN GROUNDWATER |
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Senior Scientist Exponent-Failure Analysis Associates Oakland, California. |
TOPIC A: BASIC PROCESSES AND EQUATIONS
In the previous lecture, we learned that mass balance and mass action expressions can be used to determine how the masses of different chemical elements in an aqueous system, represented as components, are distributed among various species that may be present. The mass balance equation for one element, hydrogen, is replaced by a charge balance equation to solve problems that involve the exchange of protons, H+.
For many groundwater systems, particularly those subject to contamination by metals and/or organic compounds, this system of equations is inadequate as it does not explicitly account for a separate class of chemical interactions: oxidation-reduction reactions, or
redox reactions. Redox reactions involve the exchange of valence electrons between reactants and products. An understanding of redox processes is essential for predicting the chemical evolution of an aquifer because the different redox states exhibited by an element (e.g., carbon, nitrogen, sulfur, common metals such as iron and manganese), can behave very differently from one another. However, from a geochemical speciation modeling standpoint, redox processes are problematic because
(1) shifts in the redox state
of a system can result, mathematically, in shifts in species concentrations that
span many orders of magnitude, and
(2) as explained below, redox
calculations involve working with a theoretical entity, the hypothetical aqueous
electron, a species which does not actually exist in measurable quantities.
Also, from a practical point of view,
redox reactions are problematic in that, more than most other classes of
reactions, they exhibit the greatest disparity between what the models predict
should happen and what is actually observed. The explanation for this touches
upon reaction kinetics, a topic we will consider in a future lecture.
As stated, redox reactions involve the exchange of electrons between species.
Oxidation reactions are those reactions where an electron is liberated (this may or may not directly involve oxygen itself, as we shall see). For example, the conversion of ferrous iron to ferric iron is an oxidation reaction:.gif){kind=small}
(Eq.-1)
Here, Fe2+ is the species that is being oxidized.
Reduction reactions, on the other hand, consume electrons. We can easily write the reverse of Eq.-1 as a reduction reaction:.gif){kind=small}
(Eq.-2)
Notice in both of these reactions there appears a species, e-, that we have not dealt with so far, e-. This is the
hypothetical aqueous electron, a species that is not observable in natural systems because it does not actually exist! Rather, electrons that are transferred between species in redox reactions exist as part of the structure of the species that holds them at anyone time. In other words, unlike other aqueous species, the e- does not "swim around" by itself in the aqueous medium. Redox reactions, by their nature, imply electron transfer. Each oxidation reaction must be coupled with a corresponding reduction reaction, and vice-versa. Thus, Eqs.1-2 are not really chemical reactions at all, but are instead merely half-reactions. A complete redox reaction, therefore, is a linear combination of an oxidation and reduction reaction so that the net release or consumption of electrons in the aqueous solution is always equal to zero. For example, the half-reaction of the oxidation of ferrous iron (Eq.-1) can be combined in this way with the half-reaction of the reduction of oxygen,.gif){kind=small}
(Eq.-3)
to form a balanced coupled redox reaction (multiplying both sides of Eq.-1 by 4 to balance e-):
.gif){kind=small}
(Eq.-4)
A complete description of the oxidation of Fe2+ to Fe3+, coupled with the reduction of O2 to H2O, is provided by Eq.-4. So why described redox reactions using half-reactions? Why not just use the coupled redox reactions? There are two principal reasons why we call upon half-reactions. First, the concept of a half-reaction is central to understanding the subject of electrochemistry, which deals with the
electric currents and voltage differences associated with electrochemical cells fueled by redox reactions.
This is a broad subject that is addressed by
most introductory chemistry textbooks. In the context of geochemical speciation
modeling, however, many of the details of electrochemical theory need not be
mentioned and hence will not be explored further.
Second, from a computational standpoint which is of more immediate interest to us, half-reactions are useful for calculating redox equilibrium. If we continue our assumption of thermodynamic equilibrium for all species in the system, an assumption we began using in the previous lecture, then things become complicated when fully-coupled redox reactions are employed. This is because many coupled redox reactions can be written between the species in a system that can participate in electron-transfer reactions. Some (but not necessarily all) of these reactions will involve the transfer of oxygen atoms (e.g., Eq.-4). However, oxygen, like hydrogen, is a constituent of the water phase itself, so a corresponding mass balance equation is problematic. As a result, a mathematical description of a system characterized by multiple coupled redox reactions may not be properly constrained.
To get around this problem, geochemical speciation modeling usually makes use of the hypothetical aqueous electron. We recognize that the e- species is not really a species at all (itís not included in mass balance or charge balance equations, for example), but rather serves as a measure of redox potential. This potential, and the related concept of operational valence, provides us with the necessary mathematical constraints to solve the equilibrium problem in the aqueous phase.
In an aqueous system in which all of the species are in a state of thermdynamic equilibrium, we begin the process of expanding our basic set of equations from Lecture 2 to include redox processes simply by adding redox half-reactions to the list of mass action expressions. For example, based on Eq.-1, we may write,
.gif)
(Eq.-5)
Recall from Lecture 2 that we use a charge balance equation as a replacement for a mass balance expression for the element hydrogen because the high concentration of H2O in water, in comparison to other aqueous species, makes this mass balance problematic. Oxygen, of course, presents a similar problem and suggests an analogous solution. Indeed, the means out of this conundrum is the concept of operational valence. Operational valences are assigned to individual components or species in the system; they reflect the capacity of such species to gain or donate electrons to other species. A very simple set of rules exists to assign operational valences to redox active species:
1. Starting with an assumed operational valence of zero, subtract 1 for every hydrogen atom present in the molecule. CH4(aq) (dissolved methane) is assigned an operational valence of ñ4, for example.
2. Now, for each oxygen atom present, add 2. Thus, for example, O2(aq) is assigned an operational valence of +4.
3. Finally, adjust the operational valence upward or downward by the actual charge of the aqueous species. Thus, H2CO3, HCO3-, and CO32- are each
The operational valence concept is a book-keeping device that is used to assure conservation of valence electrons in modeled systems. Just as in the case of a reaction that is mass and charged balanced, the total operational valence among reactants and among products must also balance. For example, consider again the oxidation of ferrous iron by oxygen (operational valences shown in blue):
(Eq.-6)
The total operational valence on both sides of the reaction (reactants and products) is 12, assuring us that electrons have indeed been conserved.
So, ultimately, how is the concept of operational valence put into practice? Suppose that you start out with a water that is out of redox equilibrium (i.e, a water where two species are likely to react with one another, such as Fe2+ and O2 in by Eq.-6). The water will have a finite number of valence electrons available for transfer in redox reactions. This number is fixed by the concentrations of redox active species, multiplied by their operational valence. Finding the final equilibrium state is equivalent to adjusting the quantity of e- in the final solution composition, along with solving the mass balance, charge balance, and mass action equations, in such a way that operational valence is conserved.
Solving problems involving multiple redox reactions can be quite complex, and occasionally very difficult, mathematically. Simple example problems outlining the complete solution of a set of equations describing the system, as in the carbonate equilibria example of Lecture 2, are not so easy to come by. Solving these types of problems is the business of specialized geochemical speciation models employing sophisticated numerical solution methods, a subject we will focus much of our attention on later in this course
Although redox half-reactions are largely
an artificial construct, than can be very useful for gaining insights into the
solution chemistry of natural waters beyond being a mere modeling convenience.
The half-reaction introduces two important related concepts in the study of
natural waters, the
The question of which member of the redox couple is dominant in a given
environment depends on the activity of
e-, as suggested by Eq.-5. Because of
the analogy between operational balance and
mass action relationship for
associated with the reaction given by Eq.-3 is:
.gif)
(Eq.-8)
With a little mathematical manipulation, and the definitions of pH (see Question #2, Lecture 2) and pE (Eq.-7), we see that,
.gif)
(Eq.-9)
Assuming a neutral pH of 7.0, and an activity of dissolved oxygen equal to its molar concentration in equilibrium with the atmosphere, about 2.8 x 10-4 mol/L, we can calculate that the solution pE is approximately 13.6.
The other end of the redox range is encountered when methane, CH4, becomes the dominant species of carbon, in comparison to CO2, when conditions become very reducing,such as in a peat bog or in organic rich sediments found under a marsh. Calculating the pE in such an environment is left as an exercise (see Question #2 below).
A very critical assumption central to all of our reasoning above is that of redox equilibrium. That is, we have assumed that all redox couples are in mutual equilibrium, implying that a single value of pE describes all redox conditions in the given system. While this makes affairs neat and tidy from a modeling perspective, it rarely reflect reality.
In the real world, the rates of redox
reactions arenotoriously slow. Typically, this is because redox reactions
involve breaking relatively strong covalent bonds. Microbial agents often
participate in such reactions, extracting energy from them. However, the rates
of such processes are necessarily slow, as microbial populations must adjust to
changes in water chemistry and must also come in contact with the redox-active
species themselves. In subsequent lectures, we will explore some reaction
kinetic models for redox reactions and will also look at some examples of the
role that reaction kinetics of redox reactions play in influencing reactive
transport processes.
1. Groundwater samples that are to be
analyzed for dissolved ferrous iron must be handled very carefully to assure
accurate analytical results. Can you guess what the problem is?
2. A reasonable lower limit for pE values
encountered in groundwater environments is given by the CO2/CH4
redox couple. If the pH is near neutral (pH = 7), what is the pE when the
activities of these two species are equivalent? Note:
.gif)
(Eq.-9a)
You are now ready to continue to LECTURE 4: Phase Equilibria.
You may e-mail me questions and comments.
Walt W. McNab
E-mail address: Walt McNab <WaltMcNab@prodigy.net>
Contamination of a soil sample by
atmospheric oxygen is to be avoided when analyzing for ferrous iron.
Oxygen, O2, readily
reacts with Fe2+ to form
Fe3+, which combines with
OH- in solution to
no headspace (i.e.,
a pocket of air) trapped within the sample vial.
