Answer:
1. NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂
2. The reactant that is reduced is Q
3. The charge on iron on the right side is +2, Fe²⁺
Explanation:
NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂
The reaction above is catalysed by NADH:ubiquinone oxidoreductase (complex 1), which transfers a hydride ion from NADH to FMN, from which two electrons pass through a series of of Fe-S centers to the iron-sulfur protein N-2. Electron transfer from N-2 to Ubiquinone forms QH₂
The species in a reaction which gains hydrogen irons is reduced, Therefore, the reactant that is reduced is Q, ubiquinone to form QH₂, ubiquinol.
To determine the oxidation number of iron on the right side of the reaction below,
QH2 + 2cyt c ( Fe3+) ⟶ Q + 2cyt c(Fex) + 2H^+
Sum of charges on the left side = Sum of charges on the right side
Sum of charges on the left side = 2 *+3 = +6
Therefore 2 * x + 2= 6
2x = 6 -2 = 4
x = 4/
x = 2
Therefore the charge on iron on the right side is +2, Fe²⁺
.