Lecture No. 8. E_h-pH Diagrams II.

In the As+S diagram, notice there is no field for native arsenic, which does occur from place to place. Would it be stable in water, in the absence of sulfur? Yes. If we look at the next overhead, we see the E_h-pH diagram for As in the absence of sulfur, and there is native arsenic.

Repeating what I said in the last lecture, in solution, arsenous acid (sometimes called arsenious acid), H₃AsO₃, is stable as a very weak acid. Under basic conditions it can form arsenite salts, such as sodium arsenite, Na AsO₂. Note that the acid loses one molecule of water in forming the salt.

H₂AsO₃^– + Na⁺ ¨ NaAsO₂ + H₂O

There probably is also a monohydrogen arsenite ion in very basic solutions, HAsO₃^2–, but there is no tribasic arsenite ion, AsO₃^3–. (Note that the arsenic oxyanions are shown in the As-no S diagram.) Arsenic acid, on the other hand, has 3 separate ionizations and forms salts analogous to phosphates, e.g., NaH₂AsO₄, Na₂HasO₄, and Na₃AsO₄.

Take the antimony E_h-pH diagram with several grains of salt; antimony chemistry is not all that well characterized.

Antimony is superficially similar to arsenic, but its higher atomic number makes it slightly more metallic than arsenic. The Sb₂O₅ stability field also includes dissolved species collectively known as antimonic acid. There is no evidence for the existence of an H₃SbO₄ molecule or of H₂SbO₄^– anions. There is an antimonate anion with the formula Sb(OH)₆^–, which can be prepared in the laboratory under drastic conditions by adding dilute alkali hydroxide to SbCl₅ but which probably does not occur in nature. It has been observed in certain antimonate salts such as NaSb(OH)₆ prepared in the laboratory. Most antimonates are prepared by heating oxides, and they are best regarded as mixed oxides, such as CaOáSb₂O₅.

In the +3 oxidation state, antimony tends to form the antimonyl cation, SbO⁺ (more evidence of its metallic character). The acid shown in this diagram, Sb(OH)₃, is a hydrated form of the oxide, Sb₂O₃. Antimonite salts, MH₂AsO₃, do exist.

The +4 oxidation state does not exist. The compound shown as Sb₂O₄ is a mixed oxide of the formula Sb^VSb^IIIO₄. Note that there are an antimony (III) sulfide and a thioantimonite.

The next overhead shows stability fields for zinc compounds in the presence of sulfur and in the absence of carbonate. Note the stability fields for the zinc sulfide monohydrate and sphalerite. A mixed oxide-sulfide is stable in the pH range where zinc oxide is a solid.

In the case of iron with sulfur and without carbonate, note that the iron monosulfide (mineral form not specified) has only a very small stability field within the stability field of water. Pyrite is much more stable, probably because of its negative free energy of crystallization.

Reviewing the aluminum diagram (see handout), we see that from 4 redox and acid-base relationships for aluminum and two redox equations for water we get a diagram that covers the entire stability field for water (merely two vertical lines) and the relationships among Al, Al³⁺, Al(OH)₃, and Al(OH)₄^– below the water instability line. There is one exception: the boundary between Al metal and Al(OH)₄^–. What do we do here?

There are two answers to this. The first is to note the equation:

Al(OH)₄^– + 3 e^- ¨ Al + 4 OH^–

The pe/pH slope of the line that starts at the intersection of the vertical line (Al(OH)₃/Al(OH)₄^–) and the slanted line (Al/Al(OH)₃) in the lower right corner should therefore be –4/3. This is rigorously true as long as we can assume that Al(OH)₄^– is the only soluble species.

The second answer is to say that this part of the diagram has little meaning, since water is unstable in this E_h region. Also, at high pH, Al(OH)₄^– is only one of several solution species. Other species include Al₃(OH)₁₁^2– and Al₂O(OH)₆^2–.

Let’s consider how the diagram would change if we changed our initial assumptions:

1. Suppose the total dissolved Al was 0.001 M instead of 1.0 M.

The boundary line between Al³⁺ and Al(OH)₃ would be farther to the right (higher pH). Remember that it is governed by the solubility product constant,

K_sp = {Al³⁺}{OH^–}³ = 5« 10^-33

Making Al³⁺ activity lower by a factor or 10³ means that {OH^–}³ can be larger by 10³, or {OH^–} can be larger by a factor of 10. In other words, the boundary would be at pH 4.2 rather than 3.2.

The dividing line between Al(OH)₃ and Al(OH)₄ will move to a lower pH, for similar reasons.

2. Suppose we stipulate that 1 mM carbon dioxide (or carbonate or bicarbonate, depending on pH) is present.

It will have no effect on the aluminum phases, because aluminum does not form a carbonate at any pH. However, if we let the carbonate be redox-active, we will have a methane region at low E_h.

3. Suppose we add 1 mM phosphate to the system.

We will now have to consider the presence of a variscite (AlPO₄) phase, which will exist instead of Al(OH)₃ at some pH values.

4. Suppose we add 1 mM sulfate to the system.

We will find that there are several aluminum sulfate mineral phases that exist at various pH values above the sulfate reduction line. We will also find that, at E_h values where sulfate is unstable, Al(OH)₃ will take over.

The point of all this is that the E_h-pH diagram is good only for the set of assumptions with which we start. If we change the assumptions and throw in a different mix of substances, the diagram can change a lot.

There will almost certainly be one problem on the midterm exam involving construction of an E_h-pH diagram. Make sure you are familiar with the principles we used to make this diagram for aluminum.

The midterm exam will be a take-home exam. You are free to use your notes and any inanimate references, but you must work alone. The exam will be due one week after it is passed out.

Now let’s consider redox ladders. In natural systems there may be several potential oxidizing agents present. Depending on the pH, nitrate, MnO₂, or FeOOH may be the strongest oxidizer after oxygen. Near pH 7, Mn (IV) is reduced first, then Fe (III), then nitrate, and then nitrite. It is only after these stronger oxidizers are depleted that it is possible for sulfate reduction to take place. Why? First, because the stronger oxidizers are often poisonous to the sulfate-reducing bacteria. Second, because any sulfide that is generated in the presence of nitrate, nitrite, Mn (IV), or Fe (III) would be immediately reoxidized to sulfate. Once sulfate becomes depleted, it is possible for bicarbonate to be reduced to methane. There are also a number of fermentation reactions, in which organic carbon disproportionates to methane and carbon dioxide under bacterial action. For example, sugars can be fermented as follows:

2 CH₂O + H₂O ¨ CH₄ + HCO₃^– + H⁺

These fermentation reactions usually begin in the same E_h range as sulfate reduction. They usually do not directly influence the inorganic redox equilibria, but their indirect influence is large. Sulfate reducing bacteria need certain organic compounds that are produced by other heterotrophic anaerobes including the fermenters. Without their bacterial symbionts or some other source of these organics, the sulfate reducers go inactive.

A simple redox ladder for neutral pH is the following:

The ladder is for the case where there is fairly abundant bicarbonate in solution. It could be elaborated further by adding more of the environmentally significant organic redox couples. Note in this table that the bicarbonate/acetate redox couple is slightly lower on the ladder than sulfate reduction. Some sulfate reducing bacteria need acetate in order to function; the acetate is usually produced by a symbiotic bacterium. Also note that the formate reduction is below the water-reduction potential. This means that water could oxidize carbohydrates to formate with the release of hydrogen. (This is similar to the reactions many anaerobic bacteria use to make hydrogen.)

Here is another way to look at the redox ladder:

To previous lecture: Lecture No7. Eh-pH Diagrams

To next lecture: Lecture No. 9. Classes of Geochemical Environments

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