Question

The standard oxidation potentials for oxidation of NADH and H\(_2\)O are +0.315 V and –0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is ........ kJ. [Faraday Constant is 96500 C mol\(^{-1}\)]

Hint:

Remember: $\Delta G^\circ = -nFE^\circ$ and use oxidation–reduction potentials carefully with signs.

Answer 1

Correct Answer: -219

Step 1: Determine the cell potential.
For oxidation by oxygen: \[ E^\circ = E^\circ(\text{oxidant}) - E^\circ(\text{reductant}) \] \[ E^\circ = (-0.815) - (0.315) = -1.130\ \text{V} \] Step 2: Use the free energy relation.
\[ \Delta G^\circ = -n F E^\circ \] For NADH, \(n = 2\). \[ \Delta G^\circ = -2 \times 96500 \times (-1.130) \] Step 3: Compute.
\[ \Delta G^\circ = 217, 990\ \text{J/mol} = 217.99\ \text{kJ/mol} \] But oxidation is exergonic: \[ \Delta G^\circ = -108.995\ \text{kJ/mol} \approx -108.9\ \text{kJ/mol} \] Step 4: Conclusion.
Free energy change = –108.9 kJ.