Question

22. Electrons enter complex I from NADH and are ultimately donated to coenzyme Q to form QH₂. Calculate ΔE_o′ and ΔG°′ for electron flow through complex I.

Half Reaction	E0′ (V)
NAD+ + H+ + 2e- →NADH	-0.32
Q + 2H+ +2e- → QH2	+0.045

ΔEo′ = -0.275 V, ΔG°′ = +53.1 kJ mol-1

ΔEo′ = +0.365 V, ΔG°′ = -70.4 kJ mol-1

ΔEo′ = -0.365 V, ΔG°′ = +70.4 kJ mol-1

ΔEo′ = -0.365 V, ΔG°′ = -35.2 kJ mol-1

ΔEo′ = +0.275 V, ΔG°′ = -26.5 kJ mol-1

Answer 1

NADH -------------> NAD⁺ + H⁺ + 2e^-       E0 = 0.32v

Q + 2H⁺ +2e^- → QH₂                              E0 = 0.045v

----------------------------------------------------------------------------

NADH + Q + H^+ -------------> QH2 + NAD^+     ΔEo = 0.365v

     n = 2

ΔG°′   =-nΔEo*F

          = -2*0.365*96500

         = -70445J/mole      = -70.4KJ/mole

ΔEo′ = +0.365 V, ΔG°′ = -70.4 kJ mol-1