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The ultimate electron acceptor in the respiration process is molecular oxygen. Electron transfer through the respiratory chain takes place through a complex series of oxidation-reduction reactions. Some of the electron transport steps use iron-containing proteins called cytochromes. All cytochromes transport electrons by converting the iron in the cytochromes from the \(+3\) to the \(+2\) oxidation state. Consider the following reduction potentials for three different cytochromes used in the transfer process of electrons to oxygen (the potentials have been corrected for \(\mathrm{pH}\) and for temperature): \(\begin{aligned} \text { cytochrome } \mathrm{a}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{a}\left(\mathrm{Fe}^{2+}\right) & \\ \mathscr{B} &=0.385 \mathrm{~V} \\ \text { cytochrome } \mathrm{b}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{b}\left(\mathrm{Fe}^{2+}\right) & \\ \mathscr{E} &=0.030 \mathrm{~V} \\ \text { cytochrome } \mathrm{c}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \text { cytochrome } \mathrm{c}\left(\mathrm{Fe}^{2+}\right) & \\ \mathscr{Z} &=0.254 \mathrm{~V} \end{aligned}\) In the electron transfer series, electrons are transferred from one cytochrome to another. Using this information, determine the cytochrome order necessary for spontaneous transport of electrons from one cytochrome to another, which eventually will lead to electron transfer to \(\mathrm{O}_{2}\).

Short Answer

Expert verified
The correct order for spontaneous electron transfer through the cytochromes is from cytochrome b to cytochrome c, then to cytochrome a (b → c → a). This is determined by comparing the reduction potentials of the cytochromes and looking for a negative difference in reduction potentials, indicating a spontaneous transfer. In this case, the only negative ΔE value occurs between cytochrome b and c (-0.224 V).

Step by step solution

01

Understand the concept of electron transfer

The electron transfer process involves electrons being passed from one molecule to another and is a fundamental aspect of cellular respiration. In the given problem, the electron transfer takes place between three cytochromes - a, b, and c - each with its own reduction potential. A negative difference in reduction potentials would indicate a spontaneous transfer of electrons.
02

Compare the reduction potentials of the cytochromes

In order for the electron transfer process to be spontaneous, we need the reducing agent (electron donor) to have a higher reduction potential and the oxidizing agent (electron acceptor) to have a lower reduction potential. Let's compare the reduction potentials for each cytochrome to find the order necessary for spontaneous transfer by subtracting the value of the reduction potential of the acceptor from the value of the reduction potential of the donor (ΔE = E_don - E_acc).
03

Comparison 1: Cytochrome a and b

ΔE_ab = E_a - E_b = 0.385 - 0.030 = 0.355 V
04

Comparison 2: Cytochrome a and c

ΔE_ac = E_a - E_c = 0.385 - 0.254 = 0.131 V
05

Comparison 3: Cytochrome b and c

ΔE_bc = E_b - E_c = 0.030 - 0.254 = -0.224 V
06

Determine the order

From the comparisons made in step 2, we can see that only the cytochrome b to c transfer has a negative ΔE value (-0.224 V), which is essential for spontaneous electron transfer. Therefore, the correct order for spontaneous electron transfer through the cytochromes is from cytochrome b to cytochrome c, then to cytochrome a: b → c → a

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Key Concepts

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

Cellular Respiration
Cellular respiration is a vital process that takes place in the cells of organisms. It is how cells extract energy from nutrients to fuel various cellular processes. The essence of cellular respiration is the transfer of electrons from an electron donor to an electron acceptor. This transfer is carried out through a chain of oxidation-reduction reactions, often involving multiple intermediates and structures.

A primary feature of cellular respiration is the electron transport chain (ETC), located in the inner mitochondrial membrane. The ETC is composed of protein complexes that transfer electrons derived from nutrients to oxygen, producing water as a byproduct. In this process, energy is captured and used to produce ATP, the energy currency of the cell. Efficient electron transfer is crucial because it allows for maximum ATP production, making cellular respiration essential for energy metabolism. Understanding the role of the electron transport chain highlights why electron transfer is integral to cellular respiration.
Cytochromes
Cytochromes are a group of heme proteins critically involved in electron transport within the electron transport chain. They play a pivotal role in cellular respiration by facilitating the transfer of electrons between complexes.

Cytochromes contain an iron atom within a porphyrin ring, and their ability to transfer electrons is due to cycling between two oxidation states: ferrous ( ext{Fe}^{2+}) and ferric ( ext{Fe}^{3+}). As electrons are transferred, the cytochrome protein undergoes a reversible change between these states.
  • Cytochrome a: Functions in transferring electrons to oxygen. It has a higher reduction potential compared to cytochromes b and c. This higher reduction potential makes it suitable for the final steps in the ETC.
  • Cytochrome b and c: Serve as intermediary electron carriers in the ETC, with cytochrome c eventually transferring electrons to cytochrome a.
This cycling of iron between oxidation states is fundamental for electron transfer, enabling the sequential transport of electrons through the electron transport chain.
Reduction Potential
Reduction potential, or redox potential, is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. It is often expressed in volts (V) and can be used to predict the direction of electron flow.

In cellular respiration, reduction potentials help determine the sequence of electron transfer between different molecules, such as cytochromes. Electrons will spontaneously transfer from a substance with a lower reduction potential to one with a higher reduction potential. This ensures that energy is favorably captured during the electron transfer process.
  • Higher reduction potential: Indicates a stronger tendency to gain electrons. Thus, substances with high reduction potential act effectively as electron acceptors.
  • Lower reduction potential: Typically indicates a better ability to donate electrons, making such substances effective electron donors.
Understanding reduction potentials is crucial for deciphering how and why electrons flow through the electron transport chain in a specific order.
Oxidation-Reduction Reactions
Oxidation-reduction reactions, also known as redox reactions, involve the transfer of electrons between molecules. These reactions are foundational to cellular respiration and broader biochemical processes.

In these reactions, the substance that donates electrons is oxidized, while the substance that gains electrons is reduced. Oxidizing and reducing agents work together, where one is oxidized and the other is reduced. This electron transfer is crucial for the creation of energy-carrying molecules like ATP.
  • Oxidation: Loss of electrons from a molecule, increasing its oxidation state.
  • Reduction: Gain of electrons by a molecule, decreasing its oxidation state.
Redox reactions not only function within the electron transport chain but are also essential for regulating metabolic pathways. The intricate balance of oxidation and reduction ensures that energy is carefully managed and sustainably used in cellular processes.

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Most popular questions from this chapter

The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\begin{array}{ll}\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V}\end{array}$$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q)\) ?

What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M \mathrm{NiBr}_{2}\) solution b. \(1.0 \mathrm{M} \mathrm{AlF}_{3}\) solution c. \(1.0 \mathrm{M} \mathrm{MnI}_{2}\) solution

The measurement of pH using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$\mathscr{C}_{\text {meas }}=\mathscr{E}_{\text {ref }}+0.05916 \mathrm{pH}$$ where \(\mathscr{E}_{\text {ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\text {ref }}=0.250 \mathrm{~V}\) and that \(\mathscr{C}_{\text {tme\pi }}=0.480 \mathrm{~V}\) a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the uncertainty in the measured potential is \(\pm 1 \mathrm{mV}(\pm 0.001 \mathrm{~V})\) ? b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

In making a specific galvanic cell, explain how one decides on the electrodes and the solutions to use in the cell.

Calculate \(K_{\mathrm{sp}}\) for iron(II) sulfide given the following data: $$\begin{aligned}\mathrm{FeS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)+\mathrm{S}^{2-}(a q) & & \mathscr{b}^{\circ} &=-1.01 \mathrm{~V} \\\\\mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & & \mathscr{b}^{\circ} &=-0.44 \mathrm{~V} \end{aligned}$$

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