A simple equation relates the standard free‑energy change, ΔG∘′, to the change in reduction potential. ΔE0′. ΔG∘′ = −nFΔE0′ The n represents the number of transferred electrons, and F is the Faraday constant with a value of 96.48 kJ⋅mol^(−1)⋅V^(−1). Use the standard reduction potentials provided to determine the standard free energy released by reducing O2 with FADH2. FADH2 + 1/2O2 → FAD + H2O
given that the standard reduction potential for the reduction of oxygen to water is +0.82 V and for the reduction of FAD to FADH2 is +0.03 V.

Answer :  The value of standard free energy is, -152.4 kJ/mol

Explanation :

The given balanced cell reaction is:

The half reaction will be:

Reaction at anode (oxidation) :    

Reaction at cathode (reduction) :    

First we have to calculate the standard electrode potential of the cell.

Relationship between standard Gibbs free energy and standard electrode potential follows:

where,

= standard free energy = ?

n = number of electrons transferred = 2

F = Faraday constant =

 = standard electrode potential of the cell = 0.79 V

Now put all the given values in the above formula, we get:

Therefore, the value of standard free energy is, -152.4 kJ/mol

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