Standard free energy change (Biochemistry)

Calculate the standard free energy change associated with reduction of oxygen to water by NADH. Write down the different steps of your calculation.

$$\begin{gathered} 1/2 O_2 +2H^{+} +2e^{-}\to H_{2}O E_0^{\text{'}} = +0.82 V \left(A\right) \\ NA{D}^{+} + H^{+} + 2e^{-} \to NADH E_0^{\text{'}} = -0.32 V \left(B\right) \\ Combination of the two half-reactions yeilds \\ 1/2 O_2 + NADH + H^{+} \to H_{2}O + NA{D}^{+} \left(C\right) \\ S\tan dard free energi \\ ∆G^{°\text{'}}=(-2 \times 96.48 kJ mo{l}^{-1} V^{-1} \times +0.82 V) + (-2 \times 96.48 kJ mo{l}^{-1} V^{-1} \times 0.32 V) \\ = 158.2 kJ mo{l}^{-1} + (-61.9) kJ mo{l}^{-1} \\ = -220.1 kJ mo{l}^{-1} (-52.6 kcal mo{l}^{-1}) \end{gathered}$$

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