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+ (+0.34 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 transferred when 440 moles of NH4+ are oxidized. Finally, use the simplified Nernst Equation



ΔG = -nFΔE, where F = 96.5 kJ (mol e- × V)-1



to determine the Gibbs Free energy available to do work!

Report your answer in kJ rounded to two decimal places.

Answer 1

To calculate the ΔE for the reaction, we need to subtract the reduction potential of the anode (oxidation half-reaction) from the reduction potential of the cathode (reduction half-reaction):

ΔE = E(cathode) - E(anode)

= (+0.82 V) - (+0.34 V)

= +0.48 V

The balanced full reaction can be obtained by adding the two half-reactions together:

4NO2- + O2 + 8H+ -> 4NH4+ + 2H2O

From the balanced equation, we can see that 4 moles of NO2- are consumed for every mole of NH4+ oxidized. Therefore, the number of electrons transferred (n) is 6 * 4 = 24 electrons per mole of NH4+.

To calculate the Gibbs free energy change (ΔG) in kilojoules, we can use the equation:

ΔG = -nFΔE

where F is the Faraday constant (96.5 kJ/(mol e- × V)).

Let's calculate ΔG:

ΔG = - (24 mol e-) * (96.5 kJ/(mol e- × V)) * (0.48 V)

= -1174.08 kJ

Therefore, the Gibbs free energy available to do work in this reaction is approximately -1174.08 kJ (rounded to two decimal places).