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* (E° = +0.34 volts) • O₂ + 4e -> 2H₂O (E° = +0.82 volts) If you balance and combine the reactions so that 34 moles of NH4* are oxidized to NO₂, how many moles of H* will be produced?

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* (E° = +0.34 volts) • O₂ + 4e -> 2H₂O (E° = +0.82 volts) If you balance and combine the reactions so that 34 moles of NH4* are oxidized to NO₂, how many moles of H* will be produced?

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter17: Electrochemistry

Section: Chapter Questions

Problem 83AP

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