Further to your question. Below is an example of a comproportionation red-ox reaction with nitrogen. You can run it in phreeqc using the standard database and examine the output. You will see that both N(5) andN(-3) were consumed and the solution now has a new species N(0) (which is N2) which is the product of the red-ox reaction. The reaction that happens here is the difference of the following half reactions specified in phreeqc database:
1) 2 NO3- + 12 H+ + 10 e- = N2 + 6 H2O
2) NO3- + 10 H+ + 8 e- = NH4+ + 3 H2O
1 - 2 = NO3- + 2H+ + 2e + NH4+ = N2 + 3H2O
The only thing I am worried about here is that the resulting pH of the solution is 1.5 (starting is 5) while the reaction is supposed to consume H+. I hope you will be able to figure it out.
SOLUTION 1
temp 25
pH 5
pe 4
redox pe
units mmol/kgw
density 1
-water 1 # kg
Na 250 charge
N(5) 250
N(-3) 100
REACTION 1 # this is needed to start red-ox reaction