Conceptual Models > Example input solutions

Three redox potential in the same example. What does each one do?

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Juan Sebastian:
Hello everyone,

In the first example of the manual (speciation example), the solution has the following composition:

SOLUTION 1  SEAWATER FROM NORDSTROM AND OTHERS (1979)
        units   ppm
        pH      8.22
        pe      8.451
        density 1.023
        temp    25.0
        redox   O(0)/O(-2)
        Ca              412.3
        Mg              1291.8
        Na              10768.0
        K               399.1
        Fe              0.002
        Mn              0.0002  pe
        Si              4.28
        Cl              19353.0
        Alkalinity      141.682 as HCO3
        S(6)            2712.0
        N(5)            0.29    gfw   62.0
        N(-3)           0.03    as    NH4
        U               3.3     ppb   N(5)/N(-3)   
        O(0)            1.0     O2(g) -0.7
   END

My question is: In this example, there are three types of redox potentials (i.e., pe, redox O(0)/O(-2), and N(5)/N(-3)). What does each one do?

Thanks for your support!

dlparkhurst:
The answer is not much.

The pe is used to determine the distribution of Mn among Mn redox states.

The pe calculated from O2(aq) is used to determine the distribution of Fe among its redox states.

The pe calculated from N(-3)/N(5) is used to determine the distribution of U among its redox states.

Juan Sebastian:
Thank you for your quick response Mr. Parkhurst,

I still have the following question: is the pe 8.451 used to define the chemical speciation of the rest of the elements (e.g., S(6), C(4), etc.), or is the one calculated from O2(aq) used to define the chemical speciation of the rest of the elements?

Thank you so much!

Juan Sebastian:
I read more and understood that the pe calculated from O2(aq) is used to determine the distribution of the rest of the elements.

Thank you!

dlparkhurst:
All of the other elements are either not redox active, or you have defined individual redox states, in which case there is no need for a pe.

Note that if the solution is used in a reaction calculation, all of the redox elements and element valence states will react to redox equilibrium.

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