PhreeqcUsers Discussion Forum
Redox Equilibria => Oxidation and Reduction => Topic started by: Alexey on September 26, 2015, 07:17:18 PM

Hi. My question is about calculation of pe in solution. As I understand PHREEQC calculates pe for distinct redox couples. Their values may be rather different. How can I understand which of them is right? Or what algorithm is used for their recalculation? So in the example 1 for seawater (http://wwwbrr.cr.usgs.gov/projects/GWC_coupled/phreeqc/html/final70.html) there are 2 redox couples: N(3)/N(5) with pe=4.6750 and O(2)/O(0)  12.3893. I found an equation in the Russian book of Borisov and Shvarov (1992): Eh = SUM(n_i*Eh_i)/SUM(n_i), Eh_i attributed to each redox couple and n_i is a number of electrons in each reaction, should be 8 for N(3)/N(5) and 2 for O(2)/O(0). Therefore, considering that pe and Eh are linearly related parameters, pe of solution should be (4.6750*8+12.3893*2)/10=6.2179. But the example reports pe=8.451. What is wrong here? Or how do you recommend to obtain pe for solution in bulk?

PHREEQC treats initial solutions (SOLUTION definitions) differently from reaction calculations, that is when a solution reacts with anything (EQUILIBRIUM_PHASES, REACTION, MIX, etc).
For initial solutions, each redox state is maintained as defined; redox disequilibrium is allowed. If you define N(3) and N(5), then they are preserved in the initial solution calculations. Thus, you may have a different pe for each redox couple that is defined (N(3)/N(5), O(0)/O(2), and perhaps others). There is also the specified pe, 8.45 in your case. In addition, any redox elements that are defined as total concentrations (Fe, or U) for example will be distributed among their redox state, either by using the pe (default), or by using a specified redox couple (O(0)/O(2) for example).
Once the initial solution calculation is done and the solution reacts with something, then redox equilibrium is established. All redox elements may react with each other such that the pe calculated by each redox couple is the same (pe(O(0)/O(2)) = pe(N(5)/N(3)) = pe(N(5)/N(0)) = pe(S(6)/S(2)) = ...).

Thank you!