Given that the standard reduction potential of oxaloacetate is -0.166 V and the standard reduction potential of NAD⁺ is -0.315 V.What is the $\Delta$ $\varepsilon$ °' for the oxidation of malate by NAD⁺:
Malate + NAD⁺ $\to$ Oxaloacetate + NADH + H⁺

Question

Given that the standard reduction potential of oxaloacetate is -0.166 V and the standard reduction potential of NAD⁺ is -0.315 V.What is the

\Delta

\varepsilon

°' for the oxidation of malate by NAD⁺:
Malate + NAD⁺

\to

Oxaloacetate + NADH + H⁺

Quizplus · Accepted Answer

The standard potential of the overall reaction can be calculated by adding the reduction potentials of the half-reactions for the reduction of NAD+ to NADH and the oxidation of malate to oxaloacetate:
NAD+ + 2e- + H+ → NADH E° = -0.320 V
Malate → Oxaloacetathe + 2e- E° = +0.166 V
Overall Reaction: Malate + NAD+ + H+ → Oxaloacetate + NADH + H+ E° = -0.154 V
However, since the question asks for the F1F1F1S1F1F1F10F1F1F1S1F1F1F10°', we need to use the equation:
ΔG° = -nFE°
ΔG° = -RTlnKeq from which we can calculate the value of Keq
F1F1F1S1ΔF1F1F10F1F1F1S1εF1F1F10°' = -nFE°/ln(Keq)
For this reaction:
n = 2 (since 2 electrons are transferred in the half-reactions)
F1F1F1S1ΔF1F1F10F1F1F1S1εF1F1F10°' = (2)(96,485 J/mol V)(-0.154 V)/ln(Keq)
F1F1F1S1ΔF1F1F10F1F1F1S1εF1F1F10°' = -9,364 J/mol
F1F1F1S1ΔF1F1F10F1F1F1S1εF1F1F10°' = -0.0094 kJ/mol
F1F1F1S1ΔF1F1F10F1F1F1S1εF1F1F10°' = -0.149 eV/mol
Therefore, the answer is C.

Given That the Standard Reduction Potential of Oxaloacetate Is -0 Δ\DeltaΔ

Given That the Standard Reduction Potential of Oxaloacetate Is -0 $\Delta$