∆E°' calculation for FADH2 oxidation
∆E°' for Overall Reaction of Oxidative Phosphorylation - starting from FADH2:
Combine these two Half reactions and Calculate ∆E°'. Remember that ∆E°' is in VOLTS = Joules/Coulomb so it is energy PER electron! You do NOT multiple by the number of electrons at this point (we do that next, when we calculate ∆G°').
Find the Reduction potentials from the Table above for the half reaction. Turn one of them around so you get a POSTIVE (= favorable) ∆E°':
FAD + 2H+ + 2e- --> FADH2 E°' = 0
O2 + 4H+ +4e- --> 2H2O E°' = +0.82
Give ∆E°' in Volts (just enter a number - no units!)
Reflection: does the sign of ∆E° tell you this is a favorable or unfavorable process?