Given the information below, which is more favorable energetically, the oxidation of succinate to fumarate by NAD+ or by FAD? Fumarate + 2H+ + 2e- → Succinate E°´ = 0.031 V NAD+ + 2H+ + 2e- → NADH + H+ E°´ = -0.320 FAD + 2H+ + 2e- → FADH2 E°´ = -0.219

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+

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