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: Identify oxidation and reduction half-reactions.
NADH is oxidized to NAD$^+$ and O$_2$ is reduced to H$_2$O

Step 2: Determine standard cell potential.
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = 0.816 - (-0.320) = 1.136 \text{ V} \]

Step 3: Determine number of electrons transferred.
For NADH $\rightarrow$ NAD$^+$, number of electrons transferred = 2
\[ n = 2 \]

Step 4: Use relation between $\Delta G^\circ$ and $E^\circ$.
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]

Step 5: Substitute values.
\[ \Delta G^\circ = -2 \times 96500 \times 1.136 \]
\[ \Delta G^\circ = -219,088 \text{ J} \]

Step 6: Convert into kJ.
\[ \Delta G^\circ = -219.09 \text{ kJ} \]

Step 7: Conclusion.
\[ \boxed{-219.09 \text{ kJ}} \]