Chapter 9, Problem 59TQ

Summary Introduction

To review:

The standard free energy change for the reaction ${\text{FADH}}_{\text{2}}+{\text{2H}}^{\text{+}}+\frac{1}{2}{\text{O}}_{\text{2}}\to \text{ FAD}+{\text{H}}_{\text{2}}\text{O}$. Also, compare the values with that of NADH(nicotinamide adenine dinucleotide).

Introduction:

The phosphoryl group transfer potential of a compound can be defined as the measure of the strength of attachment of a group to the molecule. It refers to the differences in the standard free energy of the molecule with and without the group. NADH has a higher phosphoryl group transfer potential than FADH₂ (flavin adenine dinucleotide).

Explanation of Solution

The reduction of oxygen to water by FADH₂can be depicted as follows:

${\text{FADH}}_{\text{2}}+{\text{2H}}^{\text{+}}+\frac{1}{2}{\text{O}}_{\text{2}}\to \text{ FAD}+{\text{H}}_{\text{2}}\text{O}$

For calculating the standard free energy, thenumber of electrons transferred needs to be balanced. For a reaction, the standard free energy can be calculated by using the Nernst equation, which is as follows:

$\Delta \text{G}=-\text{nF}\Delta \text{E}°\text{'}$ .

Where,

n is the number of electrons transferred,

F is Faraday’s constant, which is 96.15 kJ/V.mol (kilojoule per Volt. mole), and

∆E°’ is the overall cell potential.

∆E°’ is the overall cell potential.

∆E^°’ can be calculated using the following formula:

$\begin{array}{l}\Delta {\text{E}}^{°\text{'}}={\text{E}}^{°\text{'}}\left(\text{electron acceptor}\right)-{\text{E}}^{°\text{'}}\left(\text{electron donor}\right)\text{V}\\ \Delta {\text{E}}^{°\text{'}}={\text{E}}^{°\text{'}}\left(\frac{{\text{2H}}^{\text{+}}+\frac{1}{2}{\text{O}}_{\text{2}}}{{\text{H}}_{\text{2}}\text{O}}\right)\text{V}-{\text{E}}^{°\text{'}}\left(\frac{{\text{FADH}}_{\text{2}}}{\text{FAD}+{\text{H}}^{\text{+}}}\right)\text{V}\end{array}$

In the given case, E°’ of electron acceptor is +0.82 V and that of electron donor is -0.22 V.

$\begin{array}{l}\Delta {\text{E}}^{\text{°’}}=\text{0}\text{.82V}-\left(-\text{0}\text{.22V}\right)\hfill \\ {\text{ΔE}}^{\text{°’}}=\text{1}\text{.04 V}\hfill \end{array}$

Putting the values of n, F, and ∆E^°’ in the Nernst equation:

$\begin{array}{l}{\text{ΔG}}^{o\text{'}}=-\text{nΔFE}°\hfill \\ \Delta {\text{G}}^{\text{o’}}=-\text{2}\times \left(\text{96}\text{.15 kJ/V}\text{.mol}\right)\times \text{1}\text{.04V}\hfill \\ \Delta {\text{G}}^{\text{o’}}=-\text{199}\text{.992 kJ/mol}\hfill \end{array}$

Thus, the standard free energy of the reaction is –199.992 kJ/mol.

The reduction of oxygen to water by NADH can be depicted as follows:

$\text{NADH}+{\text{H}}^{\text{+}}+\frac{1}{2}{\text{O}}_{\text{2}}\to {\text{NAD}}^{\text{+}}+{\text{H}}_{\text{2}}\text{O}$

For calculating the standard free energy, the number of electrons transferred needs to be balanced. For a reaction, the standard free energy can be calculated by using the Nernst equation.

Thus,

$\begin{array}{l}\Delta {\text{E}}^{°\text{'}}={\text{E}}^{°\text{'}}\left(\text{electron acceptor}\right)V-{\text{E}}^{°\text{'}}\left(\text{electron donor}\right)\text{V}\\ \Delta {\text{E}}^{°\text{'}}={\text{E}}^{°\text{'}}\left(\frac{{\text{2H}}^{\text{+}}+\frac{1}{2}{\text{O}}_{\text{2}}}{{\text{H}}_{\text{2}}\text{O}}\right)V-{\text{E}}^{\text{°'}}\left(\frac{\text{NADH}}{{\text{NAD}}^{\text{+}}+{\text{H}}^{\text{+}}}\right)\text{V}\end{array}$

The value of standard reduction potential (Eº’) for electron acceptor in this case is 0.82 V and for electron donor is –0.32 V. Therefore,

$\begin{array}{l}\Delta {\text{E}}^{°\text{'}}=0.82\text{V}-(-0.32\text{V)}\\ \Delta {\text{E}}^{°\text{'}}=1.14\text{ V}\end{array}$

Putting the values of n, F, and ∆E^°’ in the Nernst equation:

$\begin{array}{l}{\text{ΔG}}^{o\text{'}}=\text{-nΔFE}°\hfill \\ \Delta {\text{G}}^{\text{o’}}=\text{-2}\times \left(\text{96}\text{.15 kJ/V}\text{.mol}\right)\times \text{1}\text{.14V}\hfill \\ \Delta {\text{G}}^{\text{o’}}=\text{-219}\text{.222kJ/mol}\hfill \end{array}$

Thus, the standard free energy of the reaction is -219.222 kJ/mol.

Conclusion

Thus, it can be concluded that the standard free energy of reduction of oxygen to water by FADH₂ is -199.992 kJ/mo, lnd the standard free energy of reduction of oxygen to water by NADH is -219.222 kJ/mol.