pe sensitivity to calcite equilibrium

[davidsegura]

Hello,

I wish to equilibrate an almost saturated solution with calcite and to use the output as an input for another 3D reactive-transport program. I am using the phreeqc.dat database. I have no data on redox conditions but the rest of the data about solution 1 comes from laboratory analysis. I have two questions:

1) Why does the pe even change after calcite equilibrium in this simple model? It went from 4 to 13.39. Why is it so sensible even if calcite is almost at equilibrium? If my pe is very low (less than -4) it does not change after equilibration, why?

2) In one of my scenarios, I would like that this fluid stayed at whichever pe i want after calcite equilibration. How can I find such a solution?

Code: [Select]

SOLUTION 1
    temp      13
    pH        7
    pe        4
    units     mol/kgw
    density   1
    C         0.00609658
    Ca        0.00318539
    Cl        0.000606249 charge
    K         6.59271e-05
    Mg        0.000155064
    N(5)      0.000452186
    Na        0.000504207
    S(6)      0.00053622
    -water    1 # kg
SAVE SOLUTION 1

EQUILIBRIUM_PHASES 1
    Calcite   0 10

END




[dlparkhurst]

1) Why does the pe even change after calcite equilibrium in this simple model? It went from 4 to 13.39. Why is it so sensible even if calcite is almost at equilibrium?

Initial solution may defined with redox disequilibrium. In your case, the concentrations of H2(aq), O2(aq), C(4), and C(-4) are determined by the pe, and all will be in redox equilibrium. However, S(6) and N(5) are defined individually, and will not be in redox equilibrium with the other redox couples. It is useful to separate your calculation into two steps: one to bring your initial solution to redox equilibrium, and one to equilibrate with calcite.

Code: [Select]

SOLUTION 1
    temp      13
    pH        7
    pe        4
    units     mol/kgw
    density   1
    C         0.00609658
    Ca        0.00318539
    Cl        0.000606249 charge
    K         6.59271e-05
    Mg        0.000155064
    N(5)      0.000452186
    Na        0.000504207
    S(6)      0.00053622
    -water    1 # kg
END
RUN_CELLS
-cell 1
END
USE solution 1
EQUILIBRIUM_PHASES 1
    Calcite   0 10
END

When the pe is 4, C is distributed predominantly to C(4), and the presence of N(5) generates a high pe when the solution is brought to redox equilibrium.

When the pe is -r, C is distributed predominantly to C(-4), which is sufficient to reduce all of the N(5), resulting in a low pe consistent with the presence of methane.

The equilibration with calcite has a minor effect on pe in the final step for initial pe 4. It has a greater effect when initial pe is -4 because the final pe is determined by the ratio of C(4)/C(-4), and you are adding C(4) to solution, which changes that ratio significantly.

[davidsegura]

Dr. Parkhurst,

Thank you for your swift reply. I have another question.

In order to give more redox liberty I supressed the oxydation states for N and S. Prior to calcite equilibrium I have the pe I wish with the corresponding H2(aq), O2(aq), C, N and S speciation.

The equilibration with calcite now decreases the pe from 4 to 0. I wonder why this decrease occurs if we are adding C(4), and if there is a way to compensate this and keeping the pe by adding some reducing species? for example H2(aq)?

Code: [Select]

SOLUTION 1
    temp      13
    pH        7
    pe        4
    units     mol/kgw
    density   1
    C         0.00609658
    Ca        0.00318539
    Cl        0.000606249 charge
    K         6.59271e-05
    Mg        0.000155064
    N         0.000452186
    Na        0.000504207
    S         0.00053622
    -water    1 # kg
END
RUN_CELLS
-cell 1
END
USE solution 1
EQUILIBRIUM_PHASES 1
    Calcite   0 10
END


[dlparkhurst]

I don't really understand your concern about the pe. To me, the redox states for the redox elements are the important criteria. For example, now your nitrogen is all N2(aq). Is that what you want?

The reason the pe is changing is that there is no redox buffer in your solution when you specify pe 4 the way that you have. A redox buffer occurs when two valence states of a redox element are present in significant quantities. If you look at the distribution of species in your initial solution, no two valence states of the same element are present in concentrations above, say 1e-14, which is a rough estimate of roundoff error in solving the system of equations. When you react with calcite, the pe basically drifts to a point where some redox couple exists. In this case, the point reached (for me) had a little bit of N(5) in addition to N(0) and a pe of 10. However, with the small numbers involved and roundoff errors in solving the equations, it could have reached a point with some N(-3) and a pe of -1 instead. Another way to say it is that the concentrations of the redox states other than the dominant redeox state for each element in your system are all less than 1e-14 in the pe range of -1 to 10.

If you really, really want to stabilize the pe, you can add an artificial redox buffer. With the following definition, there will be a species H2O-0.01 that is always present in a concentration of around 1e-9. It provides a redox couple for which two redox states (H2O and H2O-0.01) exist in significant concentrations the intermediate pe ranges. It will be insignificant if other redox processes occur in concentrations greater than 1e-9.

Code: [Select]

SOLUTION_SPECIES
H2O + 0.01e- = H2O-0.01
log_k -9
Here is another way to fix the pe. I do not recommend it because you can try to set a pe that is outside the stability field of water and, in general, you should let the reactions determine the pe.

Code: [Select]

PHASES
Fix_e-
e- = e-
log_k 0

SOLUTION 1
    temp      13
    pH        7
    pe        4
    units     mol/kgw
    density   1
    C         0.00609658
    Ca        0.00318539
    Cl        0.000606249 charge
    K         6.59271e-05
    Mg        0.000155064
    N         0.000452186
    Na        0.000504207
    S         0.00053622
    -water    1 # kg
END
RUN_CELLS
-cell 1
END
USE solution 1
EQUILIBRIUM_PHASES 1
    Calcite   0 10
    Fix_e- -4 O2 10
END

[davidsegura]

Dr Parkhurst,

Thank you for your reply. I am sorry I was not clear from the beggining.

My intention is to have a solution that is in equilibrium with calcite for a given pe.

Then, I want to mix this water with another very similar water, but enriched with CH4 and CO2 (known concentrations, pe = -2.55). The idea is to discuss how the insitu redox conditions of the aquifer influence the CH4 oxydation rate and how the chances of detecting would decrease it in a hypothetical water monitoring program.

I saw the changes in pe after equilibration with calcite as a problem, because I would like to start from typical groundwater pe values, between -4 and 4. Of course what matters is the redox state of the species more than the actual pe, which is only a consequence of the redox states.

The aquifer itself is composed of almost pure calcite. Do you think the artificial redox buffer would help? Please excuse me if there are issues with my idea, I am new to redox modeling.

[dlparkhurst]

The pe is simply a ratio of redox state species, for instance

Code: [Select]

pe = log([Fe+3]/[Fe+2]) - logK

If there were complete redox equilibrium, the pe would be the same when calculated from each redox couple. Two problems: (1) redox equilibrium often is not attained, and (2) concentrations of one side of the couple are often unmeasureable.

You are trying to specify a pe, presumably measured by a platinum electrode, in a range where there is no measureable couple. In my opinion, the corresponding pe is meaningless.

I suggest that you consider redox environments as oxic (O2 present), nitric (Nitrate present), ferric (iron-oxyhydroxide equilibrium), sulfidic (sulfide present), or methanic (methane present), without too much regard for the pe. If you are mixing an oxic with a methanic water, I think it will take some time for the biota to cause the reactions that approach redox equilibrium, but PHREEQC will calculate redox equilibrium and the corresponding pe.

[davidsegura]

Thank you very much @dlparkhurst,

I will try mixing methanic and oxic waters then, without worrying much about the pe.

Best regards

[MichaelZ20]

Hi David!
Can you please explain why just pe of 4 was chosen as the default value?

[dlparkhurst]

It is pretty arbitrary; pe 4 is not oxidizing and it is not reducing under most conditions. It is near the center of water stability a 25C and 1 atm. However, with high temperature and pressure, it is possible the pe 4 is outside the stability field of water creating large concentrations of O2(aq).

As I have been saying, in general, pe is overrated. It only affects total concentrations of redox elements in SOLUTION definitions, like As, Se, Fe, U, others. Usually, the major elements are analyzed and defined by redox state--O2(aq), S(6), C(4) or Alkalinity, N(5), N(-3), and the pe does not affect these definitions. In reaction calculations (all calculations except SOLUTION), the pe is a calculated value.

[MichaelZ20]

David, thank you for the explanation!