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 8.3 J K-1 mol-1.
NAD+ + H+ + 2e– → NADH E° = −0.315 V
FAD + 2H+ + 2e– → FADH2 E° = −0.219 V
Q.52
The ΔG° for the oxidation of NADH by FAD is
Options:
(A) −9.25 kJ mol-1
(B) −103.04 kJ mol-1
(C) +51.52 kJ mol-1
(D) −18.5 kJ mol-1
ΔG° for Oxidation of NADH by FAD
Redox reactions involving biological molecules such as NADH and
FAD are fundamental in bioenergetics. This problem evaluates the
relationship between standard redox potential (E°) and Gibbs free energy
change (ΔG°).
Given Data
Standard Reduction Half-Reactions:
NAD+ + H+ + 2e− → NADH
E° = −0.315 V
FAD + 2H+ + 2e− → FADH2
E° = −0.219 V
Constants:
Faraday constant (F) = 96.48 kJ V−1 mol−1
Number of electrons transferred (n) = 2
Step 1: Identify Oxidation and Reduction
- NADH is oxidized to NAD+
- FAD is reduced to FADH2
Step 2: Calculate Cell Potential (E°cell)
E°cell = E°cathode − E°anode
E°cell = (−0.219) − (−0.315) = +0.096 V
Step 3: Calculate ΔG°
ΔG° = −nFE°cell
ΔG° = −(2)(96.48)(0.096)
ΔG° = −18.5 kJ mol−1
Correct Answer
Option (D): −18.5 kJ mol−1
Conclusion
The oxidation of NADH by FAD is a spontaneous process with a negative Gibbs free
energy change:
ΔG° = −18.5 kJ mol−1
