Question

At pH 7, the reaction for the oxidation of NADH at 298 K and a pressure of 1 atm is given below:

Hint:

Use $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ directlyMake sure to identify correct cathode and anode potentials

Answer 1

Correct Answer: -219.09

Step 1: Cell potential.
\[ E^\circ_{\text{cell}} = 0.816 - (-0.320) = 1.136 \text{ V} \]

Step 2: Use the Gibbs relation.
With two electrons, \[ \Delta G^\circ = -nFE^\circ = -2 \times 96500 \times 1.136 \]

Step 3: Evaluate.
\[ \Delta G^\circ \approx -219.09 \text{ kJ} \]

Step 4: Answer.
\[ \boxed{-219.09 \text{ kJ}} \]