Direction: The standard redox potential values for two half-reactions are given below. The value for Faraday’s constant is 96.48 kJ V–1 mol–1 and Gas constant R is 8.31 JK–1 mol–1.
NAD+ + H+ + 2e- ? NADH – 0.315 V
FAD + 2H+ + 2e– ? FADH2 – 0.219 V
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The ?G for the oxidation of NADH by FAD is
-
- – 9.25 kJ mol–1
- – 103.04 kJ mol–1
- + 51.52 kJ mol–1
- – 18.5 kJ mol–1
Correct Option: D
We have, from Gibbs free energy equation:-
ΔG = –nFΔE
Where,
ΔG = change in Gibbs free energy,
n = number of electrons per mole product,
F = Faraday’s constant,
ΔE = change in electrode potential of the reaction.
Now,
ΔGOXIDATION = –nFΔ(EREDUCTION – EOXIDATION)
Given, from equation: n = 2 (2e– are exchanged)
EOXIDATION = –0.315 V [Since NADH is oxidized to NAD+ by loss of 2e–] EREDUCTION = –0.219 V [Since FAD is reduced to FADH2 by accepting 2e–]
Therefore,
ΔGOXIDATION = – 2 * 96.48 * {(–0.219) – (–0.315)}
= – 18.524 kJ/mol
Therefore ΔG for oxidation of NADH by FAD is:
–18.524 kJ/mol
