Question

Calculate the standard free energy change for the reduction of O2 by NADH. The relevant standard reduction potentials for the two half-reactions are: e°' (V) ½ O2 + 2 H+ + 2 e- ? H2O 0.815 NAD+ + H+ + 2 e- ? NADH -0.315 The Faraday constant is 96,485 JV?¹mol?¹. Extra Credit: Is the reaction spontaneous? Where in an organism does this reaction occur? (Hint: this reaction does not occur in a single step.)

          Calculate the standard free energy change for the reduction of O2 by NADH. The relevant standard reduction potentials for the two half-reactions are:

e°' (V)
½ O2 + 2 H+ + 2 e- ? H2O  0.815
NAD+ + H+ + 2 e- ? NADH  -0.315

The Faraday constant is 96,485 J*V?¹*mol?¹.

Extra Credit: Is the reaction spontaneous? Where in an organism does this reaction occur? (Hint: this reaction does not occur in a single step.)

Calculate the standard free energy change for the reduction of O2 by NADH. The relevant standard reduction potentials for the two half-reactions are:

e°' (V)
½ O2 + 2 H+ + 2 e- ? H2O 0.815
NAD+ + H+ + 2 e- ? NADH -0.315

The Faraday constant is 96,485 J*V?¹*mol?¹.

Extra Credit: Is the reaction spontaneous? Where in an organism does this reaction occur? (Hint: this reaction does not occur in a single step.)

Added by Elizabeth M.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

02/08/2023

Step 1

½ O2 + 2 H+ + 2 e- → H2O (E° = 0.815 V) NAD+ + H+ + 2 e- → NADH (E° = -0.315 V) Show more…

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Calculate the standard free energy change for the reduction of O2 by NADH: The relevant standard reduction potentials for the two half-reactions are: ½ O2 + 2 H+ + 2 e- → H2O 0.815 NAD+ + H+ + 2 e- → NADH -0.315 The Faraday constant is 96,485 J*V^-1*mol^-1 Extra Credit: Is the reaction spontaneous? Where in an organism does this reaction occur? (Hint: this reaction does not occur in a single step:)