Question

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 1

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

Explanation :

The given balanced cell reaction is:

$FADH_2+\frac{1}{2}O_2\rightarrow FAD+H_2O$

The half reaction will be:

Reaction at anode (oxidation) : $FADH_2\rightarrow FAD+2H^++2e^-$     $E^0_{Anode}=+0.03V$

Reaction at cathode (reduction) : $\frac{1}{2}O_2+2H^++2e^-\rightarrow H_2O$     $E^0_{Cathode}=+0.82V$

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

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=(+0.82V)-(+0.03V)=+0.79V$

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

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard free energy = ?

n = number of electrons transferred = 2

F = Faraday constant = $96.48kJ.mol^{-1}V^{-1}$

$E^o_{cell}$  = standard electrode potential of the cell = 0.79 V

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

$\Delta G^o=-(2)\times (96.48kJ.mol^{-1}V^{-1})\times (0.79V)$

$\Delta G^o=-152.4kJ/mol$

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