E'° of the NAD⁺/NADH half reaction is -0.32 V. The E'° of the oxaloacetate/malate half reaction is -0.175 V. When the concentrations of NAD⁺, NADH, oxaloacetate, and malate are all 10^-⁵ M, the "spontaneous" reaction is: A) malate + NAD⁺ $\rightarrow$ oxaloacetate + NADH + H⁺. B) malate + NADH + H⁺ $\rightarrow$ oxaloacetate + NAD⁺. C) NAD⁺ + NADH + H⁺$\rightarrow$ malate + oxaloacetate. D) NAD⁺ + oxaloacetate $\rightarrow$ NADH + H⁺+ malate. E) oxaloacetate + NADH + H⁺ $\rightarrow$ malate + NAD⁺.

Question

Quizplus · Accepted Answer

The reaction with the higher reduction potential (less negative) will proceed as a reduction, and the one with the lower reduction potential will proceed as an oxidation. Since the E'° for oxaloacetate/malate is -0.175 V and for NAD+/NADH is -0.32 V, oxaloacetate will be reduced to malate, and NADH will be oxidized to NAD+. Thus, the spontaneous reaction is oxaloacetate + NADH + H+ → malate + NAD+.

E'° of the NAD+/NADH Half Reaction Is -0 →\rightarrow→ Oxaloacetate + NADH + H+

E'° of the NAD⁺/NADH Half Reaction Is -0 $\rightarrow$ Oxaloacetate + NADH + H⁺