The following half-cell biochemical standard reduction electrode potentials appl

Question

The following half-cell biochemical standard reduction electrode potentials apply at 298 K: E degree + (V) at pH 7 NAD^+/NADH NAD^+ + H^+ + 2e- rightarrow NADH -0.320 V NO_3^-/NO_2^-/Pt NO_3^- + 2H^+ + 2e^- rightarrow NO_2^- + H_2 O +0.420 V Many microorganisms use NO_3^- as an oxidizing agent under anaerobic conditions. Calculate the cell EMF, E degree and the corresponding Delta G degree for the reaction NO_3^- + H^+ + NADH rightarrow NO_2^- + NAD^+ + H_2 O under biochemical standard state conditions. Given that the Gibbs free energy Delta G degree' made available by the above reaction under biochemical standard state conditions is used in making ATP with an efficiency of 50 %. calculate the number of moles of ATP made from ADP and P_1 per mole of NADH oxidized at 298 K, also under standard state conditions: (Delta G degree' = 31.0 kJ mol^-1 for ATP synthesis) Calculate E and Delta G at 298 K for the cell reaction at the following stated concentrations: NO_3^- (1 mM) + H^+ (10^-8 M) + NADH (1 mM) rightarrow NO_2^- (100 mM) + NAD^+ (20 mM) + H_2 O(l)

Explanation / Answer

(A)

Ecell = Ecathode – Eanode

Here NADH half cell reaction takes place at the anode.

So Ecell = 0.420-(-0.320) = 0.740 V

dG = -n*Ecell*F = -2*0.740*96500 = -142.82 kJ/mol

(B)

1 mole of NADH undergoes oxidation and produces 142.82 kJ free energy, since energy usage is only 50%, so useful energy produced = 142.82/2 = 71.41 kJ

Energy required for producing 1 mole of ATP = 31 kJ

So number of moles of ATP produced = 71.41/31 = 2.30 moles

(C)

E = E0 – 0.0591/n * log( ([NO2-]*[NAD+]) / ([NO3-]*[H+]*[NADH]) ) = 0.740 – 0.0591/2 * log( (100*20) / (1*10^-5*1) ) = 0.494 V

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