Chapter 9, Problem 57TQ

Summary Introduction

To review:

The standard free energy produced by the reduction of sulfur to hydrogen sulfide, and oxygen to water, by NADH (reduced nicotinamide adenine dinucleotide), and the free energy produced by reduction of oxygen in comparison to sulfur.

Introduction:

Gibbs free energy is the potential that can be used for the calculation of maximum reversible work that is performed at constant temperature and pressure. The phosphoryl group transfer potential of a compound can be defined as ameasure of the strength of attachment of a group to amolecule. It usually refers to the differences in the standard free energies of the molecule with and without the group.

Explanation of Solution

The reduction of sulfur to hydrogen sulfide by NADH can be depicted as follows:

$\text{NADH}+{\text{H}}^{\text{+}}+\text{S}+{\text{2e}}^{\text{-}}\to {\text{NAD}}^{\text{+}}+{\text{H}}_{\text{2}}\text{S}$

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 overall cell potential.

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

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

The value of standard reduction potentials (Eº’) for the electron acceptor, in this case, is -0.23V and for the electron donor is -0.32V. Therefore,

$\begin{array}{l}\Delta {\text{E}}^{°\text{'}}=-0.23\text{V}-\left(-0.32\right)\text{V}\\ \Delta {\text{E}}^{°\text{'}}=0.09\text{ V}\end{array}$

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

$\begin{array}{l}{\text{ΔG}}^{o\text{'}}=\text{-nΔFE}°\hfill \\ \Delta {\text{G}}^{\text{o’}}=\text{-2}\times \left(\text{96}\text{.15 kJ/V}\text{.mol}\right)\times 0.09\text{V}\hfill \\ \Delta {\text{G}}^{\text{o’}}=\text{-}15.93\text{kJ/mol}\hfill \end{array}$

Thus, the standard free energy of the reaction is -15.93kJ/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, thenumber 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 potentials (Eº’) for the electron acceptor, in this case, is 0.82 V and for the 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 sulfur to hydrogen sulfide is –15.93 kJ/mo, lnd the standard free energy of reduction of oxygen to water is –219.222 kJ/mol.