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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. +10e-> N₂ (E° = +0.74 V) • NO3 + • H*+2e -> H₂ (E° = -0.42 V) If you balance and combine the reactions so that 420 moles of H₂ gas are oxidized to H*, how many moles of N₂ gas will be produced?

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. +10e-> N₂ (E° = +0.74 V) • NO3 + • H*+2e -> H₂ (E° = -0.42 V) If you balance and combine the reactions so that 420 moles of H₂ gas are oxidized to H*, how many moles of N₂ gas will be produced?

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Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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 (E° = +0.74 V) • H* + 2e -> H₂ (E° = -0.42 V) If you balance and combine the reactions so that 420. moles of H₂ gas are oxidized to H*, how many moles of N₂ gas will be produced?
Transcribed Image Text: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 (E° = +0.74 V) • H* + 2e -> H₂ (E° = -0.42 V) If you balance and combine the reactions so that 420. moles of H₂ gas are oxidized to H*, how many moles of N₂ gas will be produced?
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