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zhenek [66]
3 years ago
10

Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energ

y 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
Chemistry
1 answer:
SSSSS [86.1K]3 years ago
4 0

<u>Answer:</u> The molecules of oxygen gas that will be reduced to water are 42 molecules

<u>Explanation:</u>

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:

<u>Oxidation half reaction:</u>  NH_4\rightarrow NO_2^-+6e^-     ( × 4)

<u>Reduction half reaction:</u>  O_2+4e^-\rightarrow 2H_2O   ( × 6)

<u>Overall reaction:</u> 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

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