Below are the reduction half reactions for chemolithoautotrophic denitrification, where hydrogen is a source of electrons...

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Below are the reduction half reactions for chemolithoautotrophic denitrification, where hydrogen is a source of electrons...

Below are the reduction half reactions for chemolithoautotrophic denitrification, where hydrogen is a source of electrons and energy and nitrate is the terminal electron acceptor.

NO₃^- + 10e^- -> N₂ (E⁰ = +0.74 V)

H⁺ + 2e^- -> H₂ (E⁰ = -0.42 V)

If you balance and combine the reactions so that 155 molecules of H2 gas are oxidized to H⁺, how many molecules of NO₃^- will be reduced to dinitrogen gas?

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Answer 1

Answer #1

The combined reaction is

5 H₂ + 2 NO₃^- 10 H⁺ + N₂

molecules of H₂ oxidized = 155

molecules of NO₃^- reduced = (2/5) * (molecules of H₂ oxidized)

molecules of NO₃^- reduced = (2/5) * (155 molecules)

molecules of NO₃^- reduced = 62 molecules

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Answer 2

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Below are the reduction half reactions for chemolithoautotrophic denitrification, where hydrogen is a source of electrons...

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