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

i think it is d. e because the next one is igneous.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.