Question

Lactate dehydrogenase catalyzes the reversible reaction: Pyruvate + NADH + H+ —> Lactate + NAD+. Given that: NAD+ + H+ + 2e– —> NADH, E°’ = –0.32 V and pyruvate + 2H+ + 2e– —-> lactate, E°’ = –0.19 V. Calculate deltaG'° for this reaction.

Answer 1

The reaction is : Pyruvate + NADH + H⁺ <----> Lactate + NAD⁺

Oxidation : NADH ----> NAD⁺ + H⁺ + 2e^- So, E^{​​​​​​0} = +0.32V Reduction : Pyruvate + 2H⁺ + 2e^- ----> lactate So, E^{​​​​​​0} = -0.19V We know, deltaG'^o  = -nFE⁰

Therefore, E^{​​​​0} = E_{​​​reduced} + E^​​​​_{​​​​​oxidized}   =(-0.19)V + 0.32V = 0.13 V = 0.13 J/C

F = 96485 C/mol and n = 2 mol

So, deltaG'^o = - (2mol)(96485C/mol)(0.13J/C) = - 25086.1 J = - 25kJ

Therefore, as delta G is negative (- 25kJ) the reaction is spontaneous.