Chapter 13, Problem 26P

(a)

Summary Introduction

To determine: The redox pair out of NAD⁺/NADH and pyruvate/lactate that has the greater tendency to lose electrons.

Introduction: The coenzyme being used in the “oxidation-reduction” reactions in the cell associated with the cellular respiration is known as Nicotinamide adenine dinucleotide (NAD+). It acts as an electron carrier in the electron transport chain (ETC).

Explanation of Solution

Explanation:

The capability of gaining electrons by any compound is termed as standard reduction potential. The capability increases with the greater value of $E^{\prime }°$.

The value of $E^{\prime }°$ for NAD⁺/NADH is -0.32V and the value of $E^{\prime }°$ for Pyruvate/ Lactate is -0.19V. This implies that the NAD⁺/NADH has more capability to lose electrons.

(b)

Summary Introduction

To determine: The redox pair out of NAD+/NADH and pyruvate/lactate that is the stronger oxidizing agent.

Introduction:

The coenzyme being used in the “oxidation-reduction” reactions in the cell associated with the cellular respiration is known as Nicotinamide adenine dinucleotide (NAD⁺). It acts as an electron carrier in the electron transport chain (ETC).

Explanation of Solution

Explanation:

The “oxidizing agent” is the term for the compound, which can easily acquire electrons. The NAD⁺/NADH have more capability to lose electrons. So, the NAD⁺/NADH are the strong reducing agent. Therefore, the pyruvate/lactate is stronger oxidizing agent.

(c)

Summary Introduction

To determine: The direction, in which the reaction proceeds if the reactants have a concentration of 1M and product at pH 7and at 25°C.

Introduction:

The coenzyme being used in the “oxidation-reduction” reactions in the cell associated with the cellular respiration is known as Nicotinamide adenine dinucleotide (NAD⁺). It acts as an electron carrier in the electron transport chain (ETC).

Explanation of Solution

Explanation: All the given conditions are favorable for the proceeding of reaction in further direction. This will result in formation of lactate from pyruvate. The equation is given below:

$\text{Pyruvate}\text{+}\text{NADH}\text{+}{\text{H}}^{\text{+}}\rightleftharpoons \text{lactate}\text{+}{\text{NAD}}^{\text{+}}$

Conclusion

Conclusion:

The reaction will proceed in right direction with given conditions.

(d)

Summary Introduction

To determine: The standard free energy for conversion of pyruvate to lactate.

Introduction:

The biochemical process, in which conversion of a sugar molecule (glucose) into lactic acid molecule takes place and cellular energy is released is called lactic acid fermentation. Lactic acid fermentation occurs in the animal cells, such as muscle cells and bacteria. The pyruvate molecule is utilized in the reaction and two molecules of lactic acid are formed. The products, such as cheese and yogurt are formed by the use of lactic acid fermentation. One enzyme which is involved in catalyzing the reaction is lactate dehydrogenase.

Explanation of Solution

Explanation:

The $E^{\prime }°$ for the formation of acetate from NADH is -0.32V, but the sign changes when the reaction is reversed. The $E^{\prime }°$ for formation of lactate is -0.185V.

The summation of the reactions is as follows:

$\begin{array}{c}\begin{array}{cc}\text{NADH}\to {\text{NAD}}^{\text{+}}+{\text{2H}}^{+}+{\text{2e}}^{-}& (E^{\prime }°)=+0.320\text{V}\\ \text{Pyruvate}+2{\text{H}}^{+}+{\text{2e}}^{-}\to \text{Lactate}& (E^{\prime }°)=-0.185\text{V}\end{array}\\ \underset{\_}{}\\ \\ \begin{array}{cc}\text{Pyruvate}+\text{NADH}\to \text{Lactate}+{\text{NAD}}^{+}& (E^{\prime }°)=+0.135\text{V}\end{array}\end{array}$

The “standard free-energy change” for an “oxidation-reduction reaction” is directly proportional to the “difference in standard reduction potentials” of “two half-cells”. The formula is $\Delta G^{\prime }°=-\eta F\Delta E^{\prime }°$ and $\Delta G^{\prime }°=-RT\ln \Delta {K^{\prime }}_{\text{eq}}$.

$F$ is the Faraday constant (96,480 JV mol) and the value of $E^{\prime }°$ is $\text{0}\text{.135}\text{V}$. The value of $\eta$ is $2$.

$\Delta G^{\prime }°=-\eta F\Delta E^{\prime }°$

$\begin{array}{c}\Delta G^{\prime }°=-(2)(96.5\text{kJ}/\text{mol})(\text{0}\text{.135}\text{V})\\ =-2(13.02)\text{kJ}/\text{mol}\\ =-26\text{kJ}/\text{mol}\end{array}$

Conclusion

Conclusion:

The standard free energy for conversion of pyruvate to lactate is -26kj/mol.

(e)

Summary Introduction

To determine: The equilibrium constant for the conversion of pyruvate to lactate.

Introduction:

The biochemical process, in which conversion of a sugar molecule (glucose) into lactic acid molecule takes place and cellular energy is released is called lactic acid fermentation. Lactic acid fermentation occurs in the animal cells, such as muscle cells and bacteria. The pyruvate molecule is utilized in the reaction and two molecules of lactic acid are formed. The products, such as cheese and yogurt are formed by the use of lactic acid fermentation. One enzyme which is involved in catalyzing the reaction is lactate dehydrogenase.

Explanation of Solution

Explanation:

The value of gas constant $R$ is 8.315J/mole K, temperature is 298K, and $\Delta G^{\prime }°$ is -26kJ mol^-1.

The equation $\Delta G^{\prime }°=-RT\ln \Delta {K^{\prime }}_{\text{eq}}$ can also be written as follows:

${K^{\prime }}_{\text{eq}}=e^{-\Delta G^{\prime }°/RT}$

The value of ${K^{\prime }}_{\text{eq}}$ is calculated as follows:

$\begin{array}{c}{K^{\prime }}_{\text{eq}}=e^{-(-26000\text{J}\cdot {\text{mol}}^{-1})/(8.315\text{J}/\text{mole}\cdot \text{K})(\text{298K})}\\ =e^{(10.49)}\\ =3.63\times {10}^4\end{array}$

Conclusion

Conclusion:

The equilibrium constant for the conversion of pyruvate to lactate is $\underset{\_}{3.63\times 10^4}$.