Answer:
Oxidation with FAD is more favorable energetically.
Explanation:
With FAD as the electron acceptor;
∆E'° = (E'° from reduction reaction) – (E'° from oxidation reaction)
∆E'° = (-0.219) - 0.031 = -0.25 V
The standard free energy change, ∆G'° = -nF∆E'° ; where n is number of moles of electrons = 2, F is Faraday's constant = 96.5 KJ/V.mol
∆G'° = -2 × 96.5 KJ/V.mol × (-0.25)
∆G'° = 48.25 KJ/mol
With NAD+ as electron acceptor
∆E'° = (E'° from reduction reaction) – (E'° from oxidation reaction)
∆E'° = (-0.320) - 0.031 = -0.351 V
The standard free energy change, ∆G'° = -nF∆E'° ; n = 2, F = 96.5 KJ/V.mol
∆G'° = -2 × 96.5 KJ/V.mol × (-0.351)
∆G'° = 67.74 KJ/mol
From the above values obtained, the oxidation by FAD is more favorable energetically as the free energy is less positive than with NAD+
