Redox active species along the electron transport chain have reduction potentials between the values of the NAD+/NADH redox couple (-320 mV) and the oxygen/water redox couple (+818 mV).

Redox active species along the electron transport chain have reduction potentials between the values of the NAD+/NADH redox couple (-320 mV) and the oxygen/water redox couple (+818 mV). Electrons can thereby move down the energy scale toward progressively more positive reduction potentials! Calculate the free energy change ΔGº’ for the final outcome of the NADH re-oxidation via the electron transport chain. Use the following information: Overall reaction: NADH + H+ + ½ O2 → NAD+ + H2O Standard reduction potential for NAD+/NADH redox couple: NAD+ + 2H+ + 2e- → NADH + H+ E0′ = -0.32 V Standard reduction potential for oxygen/water redox couple: ½ O2 + 2H+ + 2e- → H2O E0′ = +0.816 V The picture below provides ΔE0′ values and ΔG0′ values for segments within the electron transport chain. Demonstrate how the ΔE0′ and ΔG0′ values were calculated for complexes I, III, and IV. What ΔE0′ and ΔG0′ value would you get for complex II

The Correct Answer and Explanation is:

To calculate the standard free energy change (ΔG°’) for redox reactions in the electron transport chain, we use the following relationship:ΔG°′=−nFΔE°′\Delta G°’ = -nF\Delta E°’ΔG°′=−nFΔE°′

Where:

• n is the number of electrons transferred (in this case, 2),
• F is Faraday’s constant (96.485 kJ/V·mol),
• ΔE°’ is the difference in standard reduction potentials between the electron acceptor and donor.

Step 1: Overall NADH Oxidation via Electron Transport Chain

Overall redox reaction:NADH+H++12O2→NAD++H2O\text{NADH} + \text{H}^+ + \frac{1}{2} \text{O}_2 \rightarrow \text{NAD}^+ + \text{H}_2\text{O}NADH+H++21​O2​→NAD++H2​O

Standard reduction potentials:

• NAD⁺/NADH: -0.320 V
• O₂/H₂O: +0.816 V

Now calculate ΔE°’:ΔE°′=E°acceptor′−E°donor′=0.816 V−(−0.320 V)=1.136 V\Delta E°’ = E°’_{\text{acceptor}} – E°’_{\text{donor}} = 0.816\,\text{V} – (-0.320\,\text{V}) = 1.136\,\text{V}ΔE°′=E°acceptor′​−E°donor′​=0.816V−(−0.320V)=1.136V

Calculate ΔG°’:ΔG°′=−nFΔE°′=−(2)(96.485 kJ/V\cdotpmol)(1.136 V)≈−219.3 kJ/mol\Delta G°’ = -nF\Delta E°’ = -(2)(96.485\,\text{kJ/V·mol})(1.136\,\text{V}) \approx -219.3\,\text{kJ/mol}ΔG°′=−nFΔE°′=−(2)(96.485kJ/V\cdotpmol)(1.136V)≈−219.3kJ/mol

So the standard free energy change for NADH re-oxidation through the ETC is approximately -219.3 kJ/mol.

Step 2: Calculating ΔE°’ and ΔG°’ for Complexes

You would use the same method as above for each complex:

Example: Complex I

Suppose Complex I transfers electrons from NADH to ubiquinone (Q), with:

• NAD⁺/NADH = -0.320 V
• Q/QH₂ = +0.045 V

ΔE°′=0.045−(−0.320)=0.365 V\Delta E°’ = 0.045 – (-0.320) = 0.365\,\text{V}ΔE°′=0.045−(−0.320)=0.365VΔG°′=−(2)(96.485)(0.365)≈−70.4 kJ/mol\Delta G°’ = -(2)(96.485)(0.365) \approx -70.4\,\text{kJ/mol}ΔG°′=−(2)(96.485)(0.365)≈−70.4kJ/mol

Complex II

Electrons are transferred from FADH₂ (E°’ = -0.03 V) to ubiquinone (Q/QH₂ = +0.045 V):ΔE°′=0.045−(−0.03)=0.075 V\Delta E°’ = 0.045 – (-0.03) = 0.075\,\text{V}ΔE°′=0.045−(−0.03)=0.075VΔG°′=−(2)(96.485)(0.075)≈−14.5 kJ/mol\Delta G°’ = -(2)(96.485)(0.075) \approx -14.5\,\text{kJ/mol}ΔG°′=−(2)(96.485)(0.075)≈−14.5kJ/mol

Summary

• Overall ΔG°’ for NADH → O₂: -219.3 kJ/mol
• Complex I: ΔE°’ = 0.365 V, ΔG°’ ≈ -70.4 kJ/mol
• Complex II: ΔE°’ = 0.075 V, ΔG°’ ≈ -14.5 kJ/mol
• Complex III and IV: Follow the same formula using their respective redox couples.

These calculations show how electrons flow energetically downhill from NADH to O₂, releasing energy used to pump protons and synthesize ATP.