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) 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

Answer 1

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

Explanation:

We are given:

$E^o_{(NO_2^-/NH_4)}=-0.41V\\E^o_{(O_2/H_2O)}=0.82V$

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

$NH_4$ 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:  $NH_4\rightarrow NO_2^-+6e^-$     ( × 4)

Reduction half reaction:  $O_2+4e^-\rightarrow 2H_2O$   ( × 6)

Overall reaction: $4NH_4+6O_2\rightarrow 4NO_2^-+12H_2O$

We are given:

Molecules of $NH_4$ = 28

By Stoichiometry of the reaction:

4 molecules of $NH_4$ reacts with 6 molecules of oxygen gas

So, 28 molecules of $NH_4$ will react with = $\frac{6}{4}\times 28=42$ molecules of oxygen gas

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