Question

Calculate the change in free energy of NADH oxidation by molecular oxygen

Answer 1

Ans. Step 1:

Reduction-half reaction:

            0.5 O₂ + 2 H⁺ + 2e -----------> H₂O          ; E⁰ = + 0.82 V

Oxidation-half reaction:

            NADH ----------------> NAD⁺ + H⁺ + 2e             ; E⁰ = + 0.320V

# Step 2: Net Reaction for reduction of NO₃^-     -

            0.5 O₂ + 2 H⁺ + 2e -----------> H₂O                                  ; E⁰ = + 0.82 V

     (+) NADH ------------------------> NAD⁺ + H⁺ + 2e            ; E⁰ = + 0.320V

            -------------------------------------------------------------------------------

            0.5 O₂ + NADH --------------> H₂O + NAD⁺                ; E⁰net = ?

Net E⁰ for overall reaction = (E⁰, Reduction-half) – (E⁰, Oxidation-half)

            Or, E⁰_net = 0.82 V – 0.320 V = 0.50 V

#Step 3: Now, dG⁰ for oxidation of NADH by O₂ is given by-

                        dG⁰ = - n F E⁰                ; [n = number of electrons transferred]

                      dG⁰ = - 2 x (96.48 kJ V^-1 mol) x (+ 0.50 V)

Hence, dG⁰= - 96.48 kJ mol^-1