Lecture No. 10. Some Handy Computational Tools

We spent some time recently examining and interpreting E_h-pH (or pe-pH) diagrams. These diagrams (see illustration) show stable chemical species at various pe and pH values while element concentrations are fixed. In that sense we are looking at constant-c planes in the 3-dimensional log c-pH-pe space.

Now we are going into diagrams that let us make predictions in the other dimensions of this space. Our first example will involve pe-log C planes where the pH is constant.

Before we construct the first diagram, here are a few formulas we covered in past lectures.

Equilibrium Constants

For a reaction aA + bB ¨ cC + dD, the equilibrium constant is

Where {I} is the activity of species I and i is the number of I molecules in the reaction. K is related to the free energy of reaction by the formula

D G⁰ = –RT ln K

Furthermore, the electrochemical potential E⁰ of a reaction is related to the free energy by the formula

Therefore, E⁰ and K are related by the equation

or

We have found that pe is a linear function of log K! We are finding that pe, which you will recall is a fictitious but very useful measure of reducing potential, can be very useful. We are about to discover another use for pe. The pe value, a dimensionless number, is directly proportional to E_h and represents the availability of electrons as if there were an actual electron activity in solution:

pe = -log₁₀{e^-}

Now suppose we wrote down the reduction potential of a half-reaction in terms of an equilibrium constant. All of the reactants, including the electrons in the half-reaction, would be in the equilibrium constant expression.

Reduction potential, or E_h, is easily converted to log K by the equation shown above. If you look at the list of reduction potentials I handed out previously, you will notice that the E⁰ values have already been converted to log K values.

All we need is to have the E_h value under one set of conditions, so E⁰ is usable. By its very nature, an equilibrium constant is valid under any set of concentrations.

Let’s look at a few of the equations for sulfur:

SO₄^2– + 9 H⁺ + 8 e^- ó HS^– + 4 H₂O
log K = 34.0

H⁺ + HS^– ó H₂S
log K = 7

S + 2 H⁺ + 2 e^- ó H₂S
log K = 4.8

SO₄^2– + 8 H⁺ + 6 e^- ó S + 4 H₂O
log K = 36.2

S + H⁺ + 2 e^- ó HS^–
log K = -2.2

From these equations we can construct a set of log c-pe diagrams. The first one will be at pH 10, for a total sulfur activity (È concentration) of 10^–3 M.

For the first sulfate half-reaction,

Therefore

log K = 34 = log {HS^–} – log {SO₄^2–} – 9 log {H⁺} – 8 log {e^-}

34 = log {HS^–} – log {SO₄^2–} + 9 pH + 8 pe

Since we have set the pH to 10,

34 = log {HS^–} – log {SO₄^2–} = 90 + 8 pe

–56 = log {HS^–} – log {SO₄^2–} + 8 pe

We can turn the equation around by solving for pe:

pe = –7 –1/8 log {HS^–} + 1/8 log {SO₄^–}

On the plot, we set pe as the horizontal axis and log c as the vertical axis. We set a horizontal limit line at log c = –3.

Since the concentration scale is logarithmic, whichever species is at a visibly higher concentration on the diagram is the predominant one.

First let’s see what happens when the sulfate concentration and the bisulfide (HS^–) concentrations are equal:

pe = -7

Since sulfate and bisulfide are at equal concentrations, the concentration of each will be 1« 10^–3 / 2. The logarithm of 0.5« 10^–3 is –3 + log 0.5 = –3.301.

Looking at the equation again, we can rewrite it so that the logs of the concentrations are in terms of pe:

log {HS^–} = –56 + log {SO₄^2–} – 8 pe

and

log {SO₄^–} = 56 + log {HS^–} + 8 pe

The slope of the log sulfate line with respect to pe is +8, and the slope of the log bisulfide line is –8. We can now fill in that part of the log c-pe diagram for pH 10.

Moving on to the second equation,

log K = 7 = log {H₂S} – log {HS^–} – log {H⁺}

log K = 7 = log {H₂S} – log {HS^–} + pH

log K = 7 = log {H₂S} – log {HS^–} + 10

log {H₂S} = log {HS^–} – 3

At this pH, the H₂S concentration is always smaller by the HS^– concentration by a factor of 1000. We can plot it if we wish, or we can just consider it negligible.

We have now accounted for all the soluble species. We can draw smooth curves from the equal concentration point for sulfate and bisulfide, up to the limit line. Below a pe of just under –7, log {HS^–} will be –3; above a pe that is just a little over –7, log {SO₄^2–} will be –3. If we need exact numbers between pe values of –7.2 and –6.8, we can solve 2 equations in 2 unknowns (the equilibrium constant equation and the sum of the two concentrations).

Next we skip the third equation (H₂S is negligible) and go on to the fourth one:

log K = 36.2 = log {S} + 4 log {H₂O} – log {SO₄^2–} + 8 pH + 6 pe

36.2 = log {S} – log {SO₄^2–} + 80 + 6 pe

log {S} = log {SO₄^2–} – 43.8 – 6 pe

But what does {S} mean? Since sulfur is a solid at 25¡ C, its activity should be 1.

What it means is that, if the calculated activity is less than 1 (i.e., log {S} < 0), solid sulfur cannot exist!

We can calculate the point at which the sulfur activity is equal to that of sulfate. At that point, –6 pe = 43.8, or pe = –7.3. At this pe value, log {SO₄^2–} È – 4. We can plot a few other points.

Similarly, we can plug in the fifth equation:

log K = –2.2 = log {HS^–} – log {S} – log {H⁺} – 2 log {e^-}

log {S} = log {HS^–} + 2.2 + pH + 2 pe

log {S} = log {HS^–} + 12.2 + 2 pe

{S} is equal to {HS^–} at pe = – 12.2/2 = –6.1. We can plot a few more points on the HS^– side of the line:

What we find is that {S} never reaches unity; native sulfur cannot exist at pH 10. If we revisit the E_h-pH diagram for sulfur, we find that is in fact the case; elemental sulfur only exists at low pH.

Now we will construct a log c-pe diagram for sulfur at pH 4. The same equations are used, with one additional one that is derived from the first and second equations:

SO₄^2– + 9 H⁺ + 8 e^- ó HS^– + 4 H₂O
log K = 34.0

H⁺ + HS^– ó H₂S
log K = 7

S + 2 H⁺ + 2 e^- ó H₂S
log K = 4.8

SO₄^2– + 8 H⁺ + 6 e^- ó S + 4 H₂O
log K = 36.2

S + H⁺ + 2 e^- ó HS^–
log K = –2.2

SO₄^2– + 10 H⁺ + 8 e^- ó H₂S + 4 H₂O
log K = 41.0

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Assignment (due next class): Derive Equation 6 and its equilibrium constant from Equations 1 and 2 and their constants._____________________________________________________________________________________

This time, we see from the second equation that H₂S will be the predominant sulfide form; log {HS^–} is always 3 smaller than {H₂S}.

log K = 7 = log {H₂S} – log {H⁺} – log {HS^–}

7 = log {H₂S} + pH – log {HS^–}

7 = log {H₂S} + 4 – log {HS^–}

log {H₂S} = 3 + log {HS^–}

From the sixth equation we see that {SO₄^2–} and {H₂S} are equal when pe = +1/8:

log K = 41 = log {H₂S} + 4 log {H₂O} – log (SO₄^–} – 10 log {H⁺} – 8 log {e^-}

41 = log {H₂S} – log {SO₄^2–} + 10 pH + 8 pe

41 = log {H₂S} – log {SO₄^2–} +40 + 8 pe

log {SO₄^2–} +1 = log {H₂S} + 8 pe

+1 = 8 pe

We can enter the intersection at pe = 0.125, log {SO₄^–} = log {H₂S} = –3.301. Log sulfate will have a slope of +8 vs. pe, and log H₂S will have a slope of –8.

Now to look at sulfur. First we examine hydrogen sulfide:

log K = 4.8 = log {H₂S} – log {S} – 2 log {H⁺} – 2 log {e^-}

4.8 = log {H₂S} – log {S} + 2 pH + 2 pe

4.8 = log {H₂S} – log {S} +8 + 2pe

log {S} = log {H₂S} +8 – 4.8 + 2 pe

log {S} = log {H₂S} + 3.2 + 2 pe

Log sulfur and log {H₂S} are equal at pe = –1.6. We can plot some additional points in the H₂S region:

Now we examine the sulfate equilibrium:

Log K = 36.2 = log {S} + 4 log {H₂O} – log {SO₄^–} – 8 {H⁺} – 6 log {e^-}

36.2 = log {S} – log {SO₄^2–} + 8 pH + 6 pe

36.2 = log {S} – log {SO₄^2–} + 32 + 6 pe

log {S} = log {SO₄^2–} + 4.2 – 6 pe

Log sulfate and log sulfur are equal at pe = +0.7. We can plot some additional points in the sulfate region:

We can fill in the area around pe = 0 by noting that the slope of the line on the positive side is –6 log {S}/pe, and on the negative side it is + 2 log {S}/pe.

We see that at pH 4 there is a very limited area around pe = 0 where elemental sulfur is stable. This compares well with the E_h-pH diagram for sulfur.

Note also that solid sulfur is not bound by the limit of 10^–3 M for total dissolved sulfur, since it is not a dissolved species. It can have an activity > 1 on this diagram. What this really means, however, is that it exists as a solid with activity = 1 whenever the calculated log {S} is greater than zero.

Now let’s ask ourselves what happens if the total dissolved sulfur activity is higher — say 10^–1 M instead of 10^–3 M. We will consider the case at pH 4.

The formulas are the same. We find log {H₂S} = log {SO₄^2–} = –1.301 at pe = +0.125, as before.

On the H₂S side, log {S} = log {H₂S} at pe = –1.6. Calculating some other points, we get

On the sulfate side, log {S} = log {SO₄^2–} when pe = +0.7. We can calculate some other points.

We now have a wider pe range in which S⁰ can exist. If we plot diagrams for S_T = 1 M and for higher pH, we will find sulfur stable at a somewhat higher pH range. I take this as an indication that the Texas and Louisiana sulfur deposits probably formed from a brine, possibly a saturated sulfate brine at high pressure. Higher sulfate and H₂S activities would favor formation of mineral sulfur.

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Reading Assignment (due in about 2 weeks): Johnnie N. Moore, 1994, Contaminant mobilization resulting from redox pumping in a metal-contaminated river-reservoir system, Advances in Chemistry Series 237 (Environmental Chemistry Of Lakes and Reservoirs), pp. 451-71.
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Next: More useful computational tools

Reminder: The take-home midterm exam will be handed out next class. It is due at the beginning of class the following Tuesday.

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

To next lecture: Lecture No 11. More Handy Computational Tools

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