/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 116 Cytochrome, a complicated molecu... [FREE SOLUTION] | 91Ó°ÊÓ

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Cytochrome, a complicated molecule that we will represent as \(\mathrm{CyFe}^{2+}\), reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions (Section 19.7). At \(\mathrm{pH} 7.0\) the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}\) $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{~V} \\\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} & \longrightarrow \mathrm{CyFe}^{2+}(a q) & E_{\mathrm{red}}^{\circ} &=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air? \((\mathbf{b})\) If the synthesis of \(1.00 \mathrm{~mol}\) of ATP from adenosine diphosphate (ADP) requires a \(\Delta G\) of \(37.7 \mathrm{~kJ},\) how many moles of ATP are synthesized per mole of \(\mathrm{O}_{2} ?\)

Short Answer

Expert verified
The oxidation of \(\mathrm{CyFe}^{2+}\) by air has a Gibbs free energy change of \(-231.6\,\text{kJ/mol}\). Approximately \(6.14\,\text{mol}\) of ATP are synthesized per mole of \(\mathrm{O}_{2}\).

Step by step solution

01

Write the balanced redox reaction between \(\mathrm{CyFe}^{2+}\) and \(\mathrm{O}_2\)

First, we need to write the balanced redox reaction by combining the given half-reactions: $$ \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{CyFe}^{2+}(a q) $$ To balance the number of electrons being transferred, we multiply the second half-reaction by 4: $$ 4[\mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{CyFe}^{2+}(a q)] $$ Now, we can add the two half-reactions: $$ \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{CyFe}^{3+}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + 4 \mathrm{CyFe}^{2+}(a q) $$ This is the balanced redox reaction we will use for the calculations.
02

Calculate the overall cell potential, \(E_{cell}\), for the redox reaction

To find the overall cell potential, we subtract the reduction potential of the cathode from the reduction potential of the anode: $$ E_{cell} = E_{red}^{\circ{\text{(cathode)}}} - E_{red}^{\circ{\text{(anode)}}} $$ In this case, the oxygen reduction half-reaction acts as the cathode and the \(\mathrm{CyFe}^{2+}\) oxidation half-reaction acts as the anode: $$ E_{cell} = (+0.82\,\text{V}) - (+0.22\,\text{V}) = +0.60\,\text{V} $$ So the overall cell potential is \(+0.60\,\text{V}\).
03

Calculate the Gibbs free energy change, \(\Delta G\), using the relationship between \(\Delta G\) and \(E_{cell}\)

To find \(\Delta G\), we use the following equation: $$ \Delta G = -nFE_{cell} $$ where \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant \((96,485\,\text{C/mol})\), and \(E_{cell}\) is the overall cell potential. In the balanced redox reaction, \(4\) moles of electrons are being transferred: $$ \Delta G = -(4\,\text{mol})(96,485\,\text{C/mol})(+0.60\,\text{V}) = -231,564\,\text{J/mol} = -231.6\,\text{kJ/mol} $$ So, \(\Delta G = -231.6\,\text{kJ/mol}\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air.
04

Calculate the moles of ATP synthesized per mole of \(\mathrm{O}_{2}\) using the given information

We are given that the synthesis of \(1.00\,\text{mol}\) of ATP requires a \(\Delta G\) of \(37.7\,\text{kJ}\). We can use the stoichiometry of the balanced redox reaction to find the number of moles of ATP synthesized per mole of \(\mathrm{O}_2\): $$ \frac{-\Delta G_\text{redox}}{-\Delta G_\text{ATP}} = \frac{231.6\,\text{kJ/mol}}{37.7\,\text{kJ/mol}} = 6.14\,\text{mol} $$ Thus, approximately \(6.14\,\text{mol}\) of ATP are synthesized per mole of \(\mathrm{O}_{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cytochrome
Cytochromes are proteins that play a crucial role in the electron transport chain, a series of chemical reactions that generate energy in cells. Think of cytochromes as tiny machines transporting electrons. Each cytochrome contains a heme group, which includes an iron atom that can easily gain or lose electrons. This property makes cytochromes effective at shuttling electrons between other molecules in the electron transport chain. This process is essential for cellular respiration, allowing cells to convert nutrients into energy.

In this context, we see cytochrome represented as \(\mathrm{CyFe}^{2+}\), where the iron is in a state ready to accept electrons. When \(\mathrm{CyFe}^{2+}\) reacts with oxygen in a redox reaction, it facilitates the generation of energy, which can be used for ATP synthesis, underscoring its critical function in bioenergetics and energy metabolism.
ATP synthesis
Adenosine triphosphate, or ATP, is often referred to as the "energy currency" of the cell. This molecule stores and provides energy for various cellular processes. ATP synthesis primarily occurs through oxidative phosphorylation, a complex process in cells involving the electron transport chain, where cytochromes play a key role.

During this process, energy released from redox reactions, like the oxidation of \(\mathrm{CyFe}^{2+}\) by oxygen, is harnessed to convert adenosine diphosphate (ADP) into ATP. This conversion is crucial because ATP then powers cellular activities such as muscle contraction, nerve impulse propagation, and molecule synthesis, essential for life.
  • The balanced reaction in the step-by-step solution illustrates how electrons transfer to oxygen, releasing energy.
  • This energy drives the synthesis of ATP from ADP.
The efficiency of this process is highlighted by how many molecules of ATP can be synthesized per molecule of \(\mathrm{O}_2\), with this exercise showing approximately 6.14 moles of ATP formed, emphasizing the effectiveness of ATP synthesis.
Gibbs free energy
Gibbs free energy, denoted as \(\Delta G\), is a thermodynamic quantity that indicates the amount of work a thermodynamic system can perform. It helps determine whether a reaction is spontaneous, with negative \(\Delta G\) values indicating spontaneous reactions.

In the context of the given oxidation of \(\mathrm{CyFe}^{2+}\), the calculated \(\Delta G\) is \(-231.6\,\text{kJ/mol}\), reflecting a favorable or spontaneous reaction. This negative value shows that the reaction releases energy, which can be used by the body to perform work, such as synthesizing ATP. The relationship between \(\Delta G\) and the cell potential \(E_{cell}\) is given by the equation \(\Delta G = -nFE_{cell}\), where \(n\) is the number of moles of electrons transferred and \(F\) is the Faraday constant. This relationship connects the electrochemical potentials with the thermodynamic properties, showing the interplay of chemistry and biology. Understanding \(\Delta G\) is vital for interpreting how efficiently cells manage their energy resources through biochemical reactions.

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Most popular questions from this chapter

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger reducing agent: (a) \(\mathrm{Al}(s)\) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Fe}(s)\) or \(\mathrm{Ni}(s)\) (c) \(\mathrm{H}_{2}(g\), acidic solution) or \(\operatorname{Sn}(s)\) (d) \(\mathrm{I}^{-}(a q)\) or \(\mathrm{Br}^{-}(a q)\)

Disulfides are compounds that have \(S-S\) bonds, like peroxides have \(\mathrm{O}-\mathrm{O}\) bonds. Thiols are organic compounds that have the general formula \(\mathrm{R}-\mathrm{SH}\), where \(\mathrm{R}\) is a generic hydrocarbon. The \(\mathrm{SH}^{-}\) ion is the sulfur counterpart of hydroxide, \(\mathrm{OH}^{-}\). Two thiols can react to make a disulfide, \(\mathrm{R}-\mathrm{S}-\mathrm{S}-\mathrm{R} .\) (a) What is the oxidation state of sulfur in a thiol? (b) What is the oxidation state of sulfur in a disulfide? (c) If you react two thiols to make a disulfide, are you oxidizing or reducing the thiols? (d) If you wanted to convert a disulfide to two thiols, should you add a reducing agent or oxidizing agent to the solution? (e) Suggest what happens to the H's in the thiols when they form disulfides.

Hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) and dinitrogen tetroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) form a self-igniting mixture that has been used as a rocket propellant. The reaction products are \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Write a balanced chemical equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance serves as the reducing agent and which as the oxidizing agent?

For a spontaneous reaction \(\mathrm{A}(a q)+\mathrm{B}(a q) \longrightarrow \mathrm{A}^{-}(a q)+\) \(\mathrm{B}^{+}(a q),\) answer the following questions: (a) If you made a voltaic cell out of this reaction, what halfreaction would be occurring at the cathode, and what half reaction would be occurring at the anode? (b) Which half-reaction from (a) is higher in potential energy? (c) What is the sign of \(E_{\text {cell }}^{\circ}\) ?

A voltaic cell is based on \(\mathrm{Cu}^{2+}(a q) / \mathrm{Cu}(s)\) and \(\mathrm{Br}_{2}(l) /\) \(\mathrm{Br}^{-}(a q)\) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode and which at the anode of the cell? (c) Use \(S^{\circ}\) values in Appendix \(\mathrm{C}\) and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above \(25^{\circ} \mathrm{C}\). (Thestandard entropy of \(\mathrm{Cu}^{2+}(a q)\) is \(\left.S^{\circ}=-99.6 \mathrm{~J} / \mathrm{K}\right)\)

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