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The drop in  Go of a pair of electrons is -52.6 kcal/mol, and the  Go of ATP formation is +7.3 kcal/mol. If three ATPs are formed for each pair of electrons removed from substrate, can you conclude that oxidative phosphorylation is only 21.9/52.6 or 42 percent efficient? Why or why not?

This question is confusing........

Just like any other spontaneous reaction, the oxidation -reduction reactions are also accompanies by a loss in free energy. The standard free-energy change during a reaction as follows:

A(ox)  + B(red)   <--> A(red) + B(ox)

Can be calculated for the standard redox potentials of the two couples involved in the reaction according to the equation:

delta Gnaught' = -nFdeltaEnaught

Where n is the number of electrons transferred in the reaction, F is the Faraday constant (23.063 kcal/V.mol), and delta Enaught is the difference in volts between the standard redox potentials of the two couples. The greater the difference in standard redoc potential between the two couples, the farther the reaction proceeds under standard conditions to the formation of products before an equilibrium state is reached. Similar, the reaction in which NADH, a strong reducing agent, is oxidized by molecular oxygen, a strong oxidizing agent.

NADH + 1/2 O2 + H+ --> H2O + NAD+ ;

The standard redox potentials of the two couples can be written as:

1/2 O2 + H+ + 2e- --> H2O    E0 = +.82 V
NAD+ + H+ + 2e- --> NADH + H+     E0 = -0.32 V

The voltage change for the total reaction is equal to the difference between two E0 values (delta E0):

+0.82 V -(-0.32 V) = +1.14 = delta E0; which is a measure of the free energy released when NADH is oxidized by molecular oxygen under standard conditions. Substituting the values in above equations:

delta Gnaught' = (-2)(23.063 kcal/V.mol)(1.14 V) = -52.6 kcal/mol of NADH oxidized

So, the standard free energy difference is -52.6 kcal/mol. The actual delta G values depend on the relative concentrations of reactants and products (oxidized and reduced versions of the compounds) present in the cell at a given instant. Regardless, it would apprear that the drop in free energy of a pair of electrons as they pass from NADH to molecular oxygen (delta Gnaught' = -52.6 kcal/mol) should be sufficient to drive the formation of several molecules of ATP (delta Gnaught'=+7.3 kcal/mol) even under conditions of the cell, where ATP/ADP rations are much higher than those at standard conditions. The transfer of this energy from NADH to ATP within the mitochondria occurs in a series of small, energy-releasing steps.