Chapter 20, Problem 8RE

MATHEMATICAL Using the information in Table 20.2, calculate $\Delta G°\text{'}$ for the following reaction:

$\begin{array}{l}\text{2 Cyt }a{a}_3\left[\text{oxidized; Fe}\left(\text{III}\right)\right]+\text{2 Cyt }b\left[\text{reduced; Fe}\left(\text{II}\right)\right]\to \\ \text{2 Cyt }a{a}_3\left[\text{reduced; Fe}\left(\text{II}\right)\right]+\text{2 Cyt }b\left[\text{oxidized; Fe}\left(\text{III}\right)\right]\end{array}$

Interpretation Introduction

Interpretation:

The standard free energy change for the given biological reaction is to be determined.

Concept introduction:

Oxidation is the addition of an electronegative element or the removal of an electropositive element in a chemical reaction.

Reduction is the addition of an electropositive element or the removal of an electronegative element in a chemical reaction

The chemical reaction in which oxidation process and reduction process takes place and reduction take place simultaneously is called a redox reaction.

The relation between standard free energy change and standard reduction potential for the biological process is shown as follows:

ΔG∘' = − n F E∘'

Here, ΔG∘'is standard Gibbs free energy change for a biological process, E∘'is standard reduction potential, Fis the Faraday’s constant(1F = 96500 J/V . mol e−)and n is the transfer of electrons in the half reactions.

Answer to Problem 8RE

Solution: 60.0 kJ mol−1

Explanation of Solution

The half reactions that are paired during the redox reaction are shown below, along with their standard free energy change:

Reaction−1: 2Cyt b[Fe(II)] + 2 Cyt c1[Fe(III)] →2Cyt c1[Fe(II)]+2Cyt b [Fe(III)] ΔG∘'= −34.7 kJ mol−1

Reaction−2: 2 Cyt c1[Fe(II)] + 2 Cyt c[Fe(III)] →2Cyt c[Fe(II)]+2Cyt  c1 [Fe(III)] ΔG∘'= −5.8 kJ mol−1

Reaction−3: 2 Cyt c[Fe(II)] + 2 Cyt (aa3)[Fe(III)] →2Cyt (aa3)[Fe(II)]+2Cyt  c [Fe(III)] ΔG∘'= −7.7 kJ mol−1

Reaction−4:  2 Cyt (aa3)[Fe(II)]+12O2+2 H+ →2Cyt (aa3)[Fe(III)]+ H2O ΔG∘'= −102.3 kJ mol−1

The aim reaction for which the standard free energy is to be calculated is shown as follows:

2 Cyt aa3 [oxidized;Fe(III)] + 2 Cyt b [reduced;Fe(II)]→ 2 Cyt aa3 [reduced;Fe(II)]+2 Cyt b [oxidized;Fe(III)]

As, the aim reaction can be calculated as follows:

Aim reaction =(Reaction−1)+ (Reaction−2)+ (Reaction−3)− (Reaction−4)

Therefore the standard free energy is calculated as follows:

ΔG∘'= −34.7 kJ mol−1+(−5.8 kJ mol−1)+(−7.7 kJ mol−1)−(−102.3 kJ mol−1)ΔG∘'= −34.7 kJ mol−1− 5.8 kJ mol−1−7.7 kJ mol−1+102.3 kJ mol−1ΔG∘'= + 59.9 kJ mol−1ΔG∘'≅ +60.0 kJ mol−1

Hence, the standard free energy of the given biological reaction will be calculated.

Conclusion

The standard free energy of the given biological reaction is60.0 kJ mol−1.