Question

How much energy (in kJ/mol) is absorbed or released for the transfer of electrons from α-ketoglutarate to NAD+?

Answer 1

The two reduction reactions which are half reactions are written below.

α-ketoglutarate is the electron donor (more negative) and NAD⁺ is the e^-acceptor (less negative)

α-ketoglutarate + CO₂+ 2 e^-+ 2H ⁺→Isocitrate (Eº = -0.38 V)

NAD⁺+ 2 e^-+ 2 H⁺→NADH + H ⁺(Eº = -0.32 V)

NAD⁺ is the oxidant (electron acceptor) and Isocitrate is the reductant (electron donor)

α-ketoglutarate + CO₂+ 2 e^-+ 2 H⁺→ Isocitrate (Eº' = -0.38 V)

NAD+ 2 e^-+ 2 H⁺→NADH + H⁺ (Eº' = -0.32 V)

For this coupled redox reaction in which NAD⁺ is the oxidant (electron acceptor) and isocitrate is the reductant (electron donor),

we can calculate ∆Eº using the following equation:

∆Eº = (Eºelectron acceptor) - (Eº electron donor)

∆Eº = (EºNAD⁺) - (Eº iso citrate)

∆Eº =(-0.32 V) - (-0.38 V) = +0.06 V

and then convert we can convert this ∆Eº value to ∆Gº using the relationship

∆Gº = -nF∆Eº

∆Gº = -2 • (96.48 kJ/mol•V) • + 0.06 V

∆Gº = -11.6 kJ/mol

This is the amount of energy released by the transfer of electrons from α-ketoglutarate to NAD⁺