Q.No. 55 The mitochondrial electron transfer chain oxidizes NADH with oxygen being the
terminal electron acceptor. The redox potentials for the two half-reactions are given below:

NAD⁺ + H⁺ + 2e⁻ → NADH, E°′ = −0.32V

½ O₂ + 2H⁺ + 2e⁻ → H₂O, E°′ = 0.816V

The free energy change associated with the transfer of electrons from NADH to O₂
is _____________ kJ/mol (round off to 2 decimal places).

Given: F = 96500 C/mol.

Calculation Method

The standard free energy change (ΔG°′) relates to the difference in standard reduction potentials (ΔE°′) by ΔG°′ = -nFΔE°′¹, where:

• n = 2 electrons
• F = 96,500 C/mol
• ΔE°′ = E°′(acceptor) – E°′(donor) = 0.816 V – (-0.32 V) = 1.136 V

Substituting yields: ΔG°′ = -2 × 96,500 × 1.136 = -218,188 J/mol = -218.19 kJ/mol (rounded to two decimal places). This negative value confirms the reaction is spontaneous, driving ATP synthesis in mitochondria.¹

Step-by-Step Derivation

1. NADH oxidation reverses its half-reaction, so use reduction potentials directly: donor NADH at -0.32 V, acceptor O₂ at +0.816 V.
2. Compute ΔE°′ = 1.136 V first.
3. Multiply by -nF for energy in joules: -2 × 96,500 × 1.136 = -218,188 J/mol.
4. Convert to kJ/mol: -218.19 kJ/mol.

Common pitfalls include sign errors (using oxidation potentials incorrectly) or forgetting n=2, yielding ~109 kJ/mol instead.²

Introduction to NADH Oxidation Free Energy

In the mitochondrial electron transfer chain, NADH donates electrons to O₂, releasing energy calculated as free energy change using redox potentials. This process powers oxidative phosphorylation, with ΔG°′ = -218.19 kJ/mol driving ~2.5 ATP per NADH. Understanding this calculation is key for biochemistry exams like IIT JAM.³

Redox Potentials Explained

Half-reactions define potentials:

• NAD⁺ + H⁺ + 2e⁻ → NADH, E°′ = -0.32 V (low, electron donor)
• ½O₂ + 2H⁺ + 2e⁻ → H₂O, E°′ = +0.816 V (high, terminal acceptor)

ΔE°′ = 0.816 – (-0.32) = 1.136 V indicates favorable transfer. Electrons flow spontaneously from negative to positive potential, conserving energy via proton gradient.⁴

Detailed Free Energy Calculation

Apply ΔG°′ = -nFΔE°′:

Parameter	Value
n	2 (electrons from NADH)
F	96,500 C/mol
ΔE°′	1.136 V
Result	-218.19 kJ/mol

Verify units: V·C = J, divided by 1,000 for kJ.²

Common Errors

• Incorrect ΔE°′ sign (+1.136 V yields positive ΔG, non-spontaneous).
• Using n=1 halves magnitude.
• Forgetting biochemical standard (pH 7, E°′ not E°).⁵

Biological Significance

This -218 kJ/mol funds ~10 proton translocations, yielding ATP via chemiosmosis. In exams, round precisely to two decimals as instructed.³

Answer: -218.19 kJ/mol