Question

CELL BIOLOGY
What is the ΔE0' for the oxidation of succinate to fumarate by NAD+ under standard conditions?

Fumarate -->Succinate - number of electrons =2 -  standard reduction potential = -0.03V

Please, the equation and step-by-step solution
Thank you!

Answer 1

The reaction is :-

Succinate + NAD+ --------> Fumarate + NADH + H+

The half reactions are :-

1) Fumarate + 2H+ + 2e- ---------> Succinate

∆E°1 = -0.03 V

2) NAD+ + 2H+ + 2e- -----------> NADH + H+

∆E°2 = + 0.32 V

On reversing the equation (1) and adding with equation (2) we get :-

Succinate + NAD+ --------> Fumarate + NADH + H+

∆E°' = ∆E°1 + ∆E°2

= -(-0.03) + 0.32 V

= 0.03 + 0.32 V

= 0.35 V

Thus the answer is 0.35 V

NB - If we want to calculate ∆G°' then,

∆G°' = -nF ∆E°'

where n = no. electron transferred (here 2) F = Faraday's constant