Question

Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. NO2- + 6e- -> NH4+ (-0.41 volts) O2 + 4e- -> 2H2O (+0.82 volts) Using the information given, calculate the ΔE for this reaction, balance the full reaction to determine the n, the number of electrons transfered when 3 molecules of NH4+ are oxidized. Finally, use the simplified Nernst Equation ΔG = -nFΔE, where F = 96.48 kJ (mol e-) to determine the Gibbs Free energy available to do work! Report your answer in kJ rounded to two decimal places. Report only the numeric portion of your answer e.g. 1.01, not 1.01 kj per mole. Answers should ALWAYS be negative since this is a spontaneous reaction..

Answer 1

Answer;

Nitrification is a process of nitrogen compound oxidation (effectively, loss of electrons from the nitrogen atom to the oxygen atoms), and is catalyzed step-wise by a series of enzymes.

{2NH4+ + 3O2 -> 2NO2- + 4H+ + 2H2O}}} (Nitrosomonas, Comammox)

?????

∆E=E cathode-Eanode. Cathode=reduction

= 0.82-(-0.41) Anode=oxidation

=. + 1.23V

And ,

∆G=-nF∆E

∆G=-3×96.48×1.23

= -356.0112