NADH is oxidized by the mitochondrial electron-transfer chain with oxygen as terminal acceptor. Half-reactions: $NAD^++H^++2e^-\to NADH$, $E^{0\prime}=-0.32$ V; and $\tfrac12O_2+2H^++2e^-\to H_2O$, $E^{0\prime}=0.816$ V. With F = 96500 C/mol, find free-energy change for transfer from NADH to oxygen in kJ/mol, rounded to 2 decimals.
The acceptor reduction potential minus donor reduction potential is $\Delta E^{0\prime}=0.816-(-0.32)=1.136$ V. Two electrons are transferred per NADH. Therefore $\Delta G^{0\prime}=-nF\Delta E^{0\prime}=-2(96500)(1.136)=-219248$ J/mol, or −219.25 kJ/mol. The negative sign indicates a favourable standard redox transfer.
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