Question

Calculate the standard free energy the reduction of O2 to water by NADH.

Answer 1

NADH + H++ 1/2 O2 -------> NAD+ + H2O

In this equation , NADH is oxidised by O2 to give water.

We know that the standard redox potential of NAD+/NADH is -0.32 V and the redox potential of oxygen is +0.82V.

So standard free energy change of the above reaction is as follows

ΔG⁰ = n x F x ΔE⁰ (where n = no. of electrons, F = Faraday Constant equal to 96.5 kJ/V/mol , E⁰ = Redox potential)

= 2 x 96.5 kJ/V/mol x (0.82-(-0.32)) V = 220.02 kJ/mol.