Answer:
ΔG∘′ = -200.68kj/mol
Explanation:
check the attachment for explanations
Reduction reaction of O2 with FADH2
FADH2 + 1/2 O2 → H2O + FAD+
Half cell reaction;
Reduction of half reaction; 2H+ + 1/2 O2 +2e- → 2H2O( at cathode)
oxidation half reaction,
FADH2 → FAD+ 2H+ + 2e- ( at anode)
E°Cell = E°cathode - E°anode
E°Cell = +0.82 -(-0.22)
E°Cell = 1.04v
ΔG∘′ = -nf E°cell
ΔG∘′ = -2 x96.48 kj/mol/ v x 1.04v
ΔG∘′ = -200.68kj/mol