Chapter 14, Problem 1P

Interpretation Introduction

Interpretation:

The value of $\Delta {\text{G}}^{0\text{'}}$ for the oxidation of malate in the presence of malate dehydrogenase should be calculated.

Concept introduction:

The standard free energy change for a reaction is related to the standard cell potential by the following equation,

  ${\text{ΔG}}^{0\text{'}}=-{\text{nFE}}_{\text{0}}^{\text{'}}$

Here, ${\text{ΔG}}^{0\text{'}}$ is the standard change in Gibbs free energy.

n is number of electrons involved

F is Faraday's constant (96485 C/mol)

E_o is the standard cell potential.

Explanation of Solution

The reaction is as follows:

  $\text{Malate (aq)}{\text{+ NAD}}^{+}(\text{aq})\rightleftharpoons \text{Oxaloacetate(aq)}+\text{NADH(aq)}+{\text{H}}^{+}(\text{aq)}$

The standard reduction potentials for the following half reactions are,

  $\begin{array}{l}\text{Oxaloacetate (aq)}\text{+ 2}{\text{H}}^{\text{+}}\text{(aq)}\to \text{Malate}\text{(aq)}{\text{E}}_{\text{0}}^{\text{'}}=-0.17\text{V}\\ {\text{NAD}}^{+}(\text{aq})+{\text{H}}^{\text{+}}(\text{aq})\to \text{NADH}\text{(aq)}{\text{E}}_{\text{0}}^{\text{'}}=-0.32\text{V}\end{array}$

The standard potential for the overall reaction will be,

  $\begin{array}{l}{\text{ΔE}}_{\text{0}}^{\text{'}}{\text{=E}}_{\text{cathode}}^{\text{'}}{\text{-E}}_{\text{anode}}^{\text{'}}\\ =(-0.32\text{V})-(-0.17\text{V})\\ =-0.15\text{V}\end{array}$

Calculating the standard free energy change,

  $\begin{array}{l}{\text{ΔG}}^{\text{0'}}{\text{=-nFE}}_{\text{0}}^{\text{'}}\\ =-(2)(96485\frac{\text{C}}{1\text{mol}{\text{e}}^{-}})(-0.15\text{V)}\\ =2895\frac{\text{C}\text{.V}}{\text{mol}}\text{×}\frac{\text{1}\text{J}}{\text{1}\text{C}\text{.V}}\text{×}\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}\\ =28.95\text{kJ/mol}\end{array}$