What is the E^$\circ$' for the spontaneous reaction that would result if the following reactions were coupled? NAD⁺ + H⁺ + 2e^- $\rarr$ NADH E^$\circ$' = -0.32 V
Pyruvate + 2H⁺ + 2e^- $\rarr$ 0 lactate E^$\circ$' = -0.19 V

Question

What is the E^$\circ$' for the spontaneous reaction that would result if the following reactions were coupled? NAD⁺ + H⁺ + 2e^-

\rarr

NADH E^$\circ$' = -0.32 V
Pyruvate + 2H⁺ + 2e^-

\rarr

0 lactate E^$\circ$' = -0.19 V

Quizplus · Accepted Answer

To determine the overall cell potential of the coupled reactions, we need to add the individual potentials of the two half-reactions. However, we also need to make sure that the individual half-reactions are properly balanced.

First, we need to flip the lactate half-reaction, so that it undergoes reduction instead of oxidation:

lactate + 2H+ + 2e- -> pyruvate             E°' = +0.19 V

Now, the electrons cancel out, and we can add the two half-reactions:

NAD+ + pyruvate + 2H+ -> NADH + lactate          E°' = -0.32 V + 0.19 V = -0.13 V

Therefore, the best answer is C: +0.13 V (the positive value indicates that the reaction is spontaneous).

What Is the E ∘\circ∘ ' for the Spontaneous Reaction That Would Result If

What Is the E^$\circ$' for the Spontaneous Reaction That Would Result If