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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 pH 7.0 the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}:\) $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow & 2 \mathrm{H}_{2} \mathrm{O}(l) & & E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{~V} \\\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{CyFe}^{2+}(a q) & & E_{\mathrm{red}}^{\mathrm{o}}=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air? (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 CyFe虏鈦 by air has a 鈭咷 of -230.36 kJ/mol. Approximately 6.11 moles of ATP are synthesized per mole of O鈧 in this reaction.

Step by step solution

01

Half-cell reactions

We are given two half-cell reactions: 1. \(\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}(l)\) with \(E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{V}\) 2. \(\mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{CyFe}^{2+}(a q)\) with \(E_{\mathrm{red}}^{\circ}=+0.22 \mathrm{V}\) The first reaction is a reduction reaction, and the second one is an oxidation reaction. **Step 2: Write the overall balanced redox reaction**
02

Overall balanced redox reaction

Combine the half-cell reactions to obtain the overall balanced redox reaction: \(\mathrm{O}_{2}(g) + 4 \mathrm{H}^{+}(a q) + 4 \mathrm{CyFe}^{2+}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + 4 \mathrm{CyFe}^{3+}(a q)\) There are four electrons exchanged in this reaction, and the stoichiometry between O鈧 and CyFe虏鈦 is one-to-four. **Step 3: Calculate E掳 of the reaction**
03

Calculate E掳 of the reaction

The standard cell potential E掳 for this reaction can be computed using the Nernst equation: \(E掳_\text{cell} = E掳_\text{red} - E掳_\text{ox}\) Substitute the given values: \(E掳_\text{cell} = (0.82 \, \text{V}) - (0.22 \, \text{V}) = 0.60 \, \text{V}\) **Step 4: Calculate 鈭咷 for the reaction**
04

Calculate 鈭咷 for the reaction

Use the Nernst equation to find 鈭咷: \(\Delta G = -nFE掳_\text{cell}\) where - n = the number of electrons exchanged in the balanced equation, - F = Faraday's constant (96,485 C/mol), and - E掳 = standard cell potential. Substitute the values: \(\Delta G = - (4\,\text{mol})(96,485 \frac{\text{C}}{\text{mol}})(0.60\text{V}) \Longrightarrow \Delta G = -230,364\text{J}\) So, \(\Delta G = -230.36\text{kJ/mol}\) **Step 5: Calculate the number of moles of ATP synthesized per mole of O鈧**
05

Number of moles of ATP synthesized per mole of O鈧

The 鈭咷 for the synthesis of 1 mol of ATP is given as 37.7 kJ. To find the number of moles of ATP synthesized per mole of O鈧, we'll divide the 鈭咷 for the oxidation of CyFe虏鈦 (Step 4) by the 鈭咷 for the synthesis of 1 mol of ATP: \(\text{Moles of ATP synthesized per mole of O鈧倉 = -\frac{\Delta G_\text{oxygen}}{\Delta G_\text{ATP}} \Longrightarrow -\frac{-230.36 \,\text{kJ/mol}}{37.7 \,\text{kJ/mol}} \) \(\text{Moles of ATP synthesized per mole of O鈧倉 \approx 6.11\) So, approximately 6.11 moles of ATP are synthesized per mole of O鈧 in this reaction.

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Key Concepts

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

Cytochrome
Cytochrome plays a crucial role in the electron transport chain within our cells. This complex protein contains a heme group, which is essential for its function in redox reactions. In simple terms, redox reactions involve the transfer of electrons between molecules. Cytochromes act as electron carriers that facilitate this transfer.
These proteins are pivotal in cellular respiration, a process that extracts energy from nutrients we consume. Within mitochondria, cytochromes transfer electrons from donors like
  • NADH
  • 贵础顿贬鈧
Ultimately passing them to oxygen, which combines with protons to form water. This process is not only key to generating energy but also ensures the efficient use of oxygen.
Interestingly, during these electron transfers, a proton gradient is created across the mitochondrial membrane. This gradient is later used to produce ATP, the primary energy currency of cells. Hence, cytochromes are vital in both energy generation and maintaining cellular homeostasis.
Adenosine Triphosphate (ATP)
Adenosine Triphosphate, commonly known as ATP, is often referred to as the energy currency of the cell. This molecule stores and provides energy for various cellular functions.
The structure of ATP includes an adenosine molecule bonded to three phosphate groups. These phosphate bonds are highly energetic. When one of these bonds breaks, ATP is converted to Adenosine Diphosphate (ADP), releasing energy in the process.
Cells utilize this released energy for:
  • Muscle contraction
  • Nerve impulse propagation
  • Protein synthesis
Additionally, in the context of the problem presented, ATP is synthesized from ADP through the energy supplied by redox reactions. The synthesis requires a specific amount of energy, exttt{37.7 kJ/mol}, making ATP production efficient yet regulated. In essence, ATP serves as the bridge between energy-yielding reactions and energy-consuming processes in cells.
Standard Cell Potential
Standard Cell Potential exttt{(E^{ ext{掳}})} is a measure of the potential difference between two electrodes in a cell under standard conditions exttt{(1M concentration, 1 atm pressure, and 25掳C temperature)}. It reveals the cell's ability to push electrons through a circuit.
This potential is determined by the half-cell reactions occurring in the electrochemical cell. Each half-cell has a specific standard reduction potential, denoted as exttt{E^{ ext{掳}}_{ ext{red}}}. For example, in the exercise, the reduction potential of oxygen is higher exttt{(+0.82 V)} compared to cytochrome exttt{(+0.22 V)}. Having a positive cell potential indicates a spontaneous redox reaction.
To find the standard cell potential for the overall reaction, subtract the lower reduction potential exttt{(E^{ ext{掳}}_{ ext{ox}})} from the higher one:\[E^{ ext{掳}}_{ ext{cell}} = E^{ ext{掳}}_{ ext{red}} - E^{ ext{掳}}_{ ext{ox}}\]This calculation gives exttt{0.60 V for the reaction}, indicating the intrinsic ability of the system to generate electrical energy, which can be utilized in various bioenergetic processes.

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Most popular questions from this chapter

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s)\) (acidic solution) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution) (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)\) (acidic solution) (d) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (acidic solution) (e) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution) (f) \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (basic solution) (g) \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) (basic solution)

The \(K_{s p}\) value for \(\mathrm{PbS}(s)\) is \(8.0 \times 10^{-28} .\) By using this value together with an electrode potential from Appendix \(\mathrm{E}\), determine the value of the standard reduction potential for the reaction $$\mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q)$$

A voltaic cell utilizes the following reaction: $$2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q)$$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}\), \(P_{\mathrm{H}_{2}}=0.95 \mathrm{~atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M},\) and the \(\mathrm{pH}\) in both half-cells is \(4.00 ?\)

Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution: $$ \begin{array}{lr} \hline \text { Reduction Half-Reaction } & \multicolumn{1}{c} {E^{\circ}(\mathbf{V})} \\ \hline \mathrm{A}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{A}(s) & 1.33 \\\ \mathrm{~B}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{B}(s) & 0.87 \\\ \mathrm{C}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{C}^{2+}(a q) & -0.12 \\ \mathrm{D}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{D}(s) & -1.59 \\\ \hline \end{array} $$ (a) Which substance is the strongest oxidizing agent? Which is weakest? (b) Which substance is the strongest reducing agent? Which is weakest? (c) Which substance(s) can oxidize \(\mathrm{C}^{2+}\) ? [Sections 20.4 and 20.5\(]\)

Metallic gold is collected from below the anode when crude copper metal is refined by electrolysis. Explain this behavior.

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