Question

C4H4O42– ––––> C4H2O42– + 2H+ + 2e–
FAD + 2H+ + 2e– ––––> FADH2
C4H4O4 2– + FAD ––––> C4H2O42– + FADH2

Oxidant Reductant n E₀’ (V)

Acetate Acetaldehyde 2 -0.67

NAD⁺ NADH+H⁺ 2 -0.32

FAD FADH₂ 2 -0.22

Pyruvate Lactate 2 -0.19

Fumarate Sucinate 2 +0.03

Dehydroascorbate Ascorbate 2 +0.08

O₂+4H⁺ H₂O 4 +0.82

1. The zero for the table of Eo’ values is the same as that for the standard potentials—the normal hydrogen electrode where the H+ concentration is 1 M. The Eo’ for 2H+ + 2e– ––––> H2, however, is no longer 0.00. Why not? Using the Nernst equation, determine the value of Eo’ for 2H+ + 2e– ––––> H2.

2. Rewrite the entries for FAD and fumarate so they look more like the tables that are in your CHEM 250 textbook—that is, instead of “oxidant” and “reductant” columns, write the balanced half reactions.

3. Calculate the Eo’ for the overall reaction that is presented at the beginning of this reading/exercise. Should this be a spontaneous reaction? What is ∆G ̊’? (The “prime” in ∆G ̊’ designates pH 7, just as it does in Eo’.)

4. Does Eo’ for this reaction depend on the pH? Why or why not?
One biochemistry text (Berg, Tymoczko & Stryer, Biochemistry, 6th Edition) lists the

following ∆G ̊’:
Succinate + FAD(enzyme-bound) ––> Fumarate + FADH2(enzyme-bound)

∆G ̊’ = 0 kJ/mol

This suggests that the Eo’ values we measure in dilute aqueous solution may be substantially different when enzymes—protein catalysts—are involved. But it still presents us with a puzzle: how does the reaction go forward when ∆G ̊’ = 0 kJ/mol?

5. Propose a reason why this reaction might go forward despite the tabulated ∆G ̊’ being zero.

Another reaction we looked at in the previous exercise is the reduction of pyruvate to lactate by NADH:

Pyruvate + NADH + H+ –––––> Lactate + NAD+

6. What is the ∆Eo’ and ∆G ̊’ for this reaction? Is this reaction expected to be spontaneous for this reaction at pH 7? Is the Eo’ for this reaction dependent on the pH? Why or why not?