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Chapter 18: Problem 112

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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Most popular questions from this chapter

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm} .\) \(\begin{array}{ll}\text { a. } \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} & \mathscr{6}^{\circ}=1.78 \mathrm{~V} \\ \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} & \mathscr{6}^{\circ}=0.68 \mathrm{~V}\end{array}\) b. \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}\) \(\mathscr{6}^{\circ}=-1.18 \mathrm{~V}\) \(\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{6}^{\circ}=-0.036 \mathrm{~V}\)

Consider the following half-reactions: $$\begin{aligned}\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & & \mathscr{E}^{\circ}=1.188 \mathrm{~V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & & \mathscr{C}^{\circ}=0.755 \mathrm{~V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & & \mathscr{C}^{\circ}=0.96 \mathrm{~V}\end{aligned}$$ Explain why platinum metal will dissolve in aqua regia (a mixture of hydrochloric and nitric acids) but not in either concentrated nitric or concentrated hydrochloric acid individually.

The solubility product for \(\operatorname{Cul}(s)\) is \(1.1 \times 10^{-12}\). Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{CuI}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}+\mathrm{I}^{-}$$

You are told that metal \(\mathrm{A}\) is a better reducing agent than metal \(\mathrm{B}\). What, if anything, can be said about \(\mathrm{A}^{+}\) and \(\mathrm{B}^{+}\) ? Explain.

Consider the cell described below: $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1.00 M) \| \mathrm{Cu}^{2+}(1.00 M)\right| \mathrm{Cu}$$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Zn}^{2+}\right]\) to have changed by \(0.20 \mathrm{~mol} / \mathrm{L}\). (Assume \(\left.T=25^{\circ} \mathrm{C} .\right)\)
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