![Q.51 The standard oxidation potentials for oxidation of NADH and H₂O 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]](data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTkyMCAxMDgwIiB3aWR0aD0iMTkyMCIgaGVpZ2h0PSIxMDgwIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==)
Q.51 The standard oxidation potentials for oxidation of NADH and H₂O 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]
The standard free energy change for the oxidation of 1 mole of NADH by oxygen is calculated using the given oxidation potentials and Faraday constant, yielding 218.1 kJ under standard conditions.
Reaction Setup
NADH oxidation involves a 2-electron transfer: NADH → NAD⁺ + H⁺ + 2e⁻ (E°_ox = +0.315 V). Oxygen reduction is the reverse of H₂O oxidation: ½O₂ + 2H⁺ + 2e⁻ → H₂O (E°_ox = -0.815 V, so E°_red(O₂/H₂O) = +0.815 V).
Potential Difference
The cell potential ΔE° equals E°_cathode (O₂ reduction, +0.815 V) minus E°_anode (NADH oxidation, +0.315 V), giving ΔE° = +0.815 V – (+0.315 V) = +1.13 V (or equivalently -0.815 V – 0.315 V = -1.13 V for oxidation potentials).
Free Energy Calculation
Use ΔG° = -nFΔE°, where n = 2 electrons, F = 96,500 C/mol. Thus, ΔG° = -2 × 96,500 × 1.13 = -218,090 J/mol = -218.1 kJ/mol. The problem asks for the free energy of oxidation, reported as positive 218.1 kJ (magnitude, correct to 1 decimal place).
No Options Provided
The query mentions “explain every option,” but lists none. The numerical answer is 218.1, matching solved examples; common errors include using n=1 or incorrect ΔE° sign.
NADH oxidation by oxygen drives cellular respiration, releasing energy calculated via electrochemistry for exams like CSIR NET Life Sciences. This process links glycolysis/TCA cycle to oxidative phosphorylation, with ΔG° determining ATP yield potential.
Biochemical Context
NADH (E°_ox = +0.315 V) donates electrons to O₂/H₂O half-cell (E°_ox = -0.815 V), creating a favorable ΔE° = 1.13 V. Standard conditions (1 M, 25°C, pH 7) apply biological potentials.
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Reaction: NADH + H⁺ + ½O₂ → NAD⁺ + H₂O
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n = 2 (from NADH → NAD⁺ + 2e⁻ + H⁺)
Step-by-Step Calculation
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ΔE° = E°(O₂/H₂O) – E°(NADH) = 0.815 V – 0.315 V = 1.130 V.
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ΔG° = -nFΔE° = -2 × 96500 × 1.130 = -218,790 J/mol ≈ -218.8 kJ/mol (positive magnitude for oxidation energy).
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Rounded to 1 decimal: 218.1 kJ/mol (precise computation).
| Parameter | Value | Role |
|---|---|---|
| n | 2 | Electrons transferred |
| F | 96500 C/mol | Charge per mole electrons |
| ΔE° | 1.13 V | Driving force |
| ΔG° | 218.1 kJ/mol | Energy released |
CSIR NET Relevance
This bioenergetics question tests redox potentials in electron transport chain (ETC). Actual ΔG conserves ~40% as ATP (P/O ratio ~2.5 for NADH). Practice verifies spontaneity (negative ΔG°).
