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) If you balance and combine the reactions so that 28 molecules of NH4 are oxidized to NO2-, how many molecules of O2 will be reduced to H2O

The molecules of oxygen gas that will be reduced to water are 42 molecules

Explanation:

We are given:

The substance having highest positive potential will always get reduced and will undergo reduction reaction. Here, oxygen will undergo reduction reaction will get reduced.

will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

The half reactions follows:

Oxidation half reaction:       ( × 4)

Reduction half reaction:     ( × 6)

Overall reaction:

We are given:

Molecules of = 28

By Stoichiometry of the reaction:

4 molecules of reacts with 6 molecules of oxygen gas

So, 28 molecules of will react with = molecules of oxygen gas

Hence, the molecules of oxygen gas that will be reduced to water are 42 molecules