The ultimate electron acceptor in the respiration process is molecular oxygen. Electron transfer through the respiratory chain takes place through a complex series of oxidationreduction reactions. Some of the electron transport steps use iron-containing proteins called \(c y\) tochromes. 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): cytochrome $\mathrm{a}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow\( cytochrome \)\mathrm{a}\left(\mathrm{Fe}^{2+}\right)$ $$ \begin{array}{c}\mathscr{C}=0.385 \mathrm{~V}\end{array} $$ cytochrome $\mathrm{b}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow\( cytochrome \)\mathrm{b}\left(\mathrm{Fe}^{2+}\right)$ $$ \begin{array}{l}\mathscr{E}=0.030 \mathrm{~V}\end{array} $$ cytochrome $\mathrm{c}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \mathrm{cytochrome} \mathrm{c}\left(\mathrm{Fe}^{2+}\right)$ $$ \begin{array}{c}\mathscr{C}=0.254 \mathrm{~V}\end{array} $$ 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 in the cytochromes is: Cytochrome b \(\rightarrow\) Cytochrome c \(\rightarrow\) Cytochrome a. This order ensures that the transfer process occurs spontaneously and ultimately leads to electron transfer to molecular oxygen.

Step by step solution

01

List down the reduction potentials

We are given three reduction potentials for cytochromes: Cytochrome a: \(E_{a}=0.385 \mathrm{~V}\) Cytochrome b: \(E_{b}=0.030 \mathrm{~V}\) Cytochrome c: \(E_{c}=0.254 \mathrm{~V}\)
02

Arrange the cytochromes in order of increasing reduction potential

First, we need to arrange the three given cytochromes in order of increasing reduction potential: Cytochrome b: \(E_{b}=0.030 \mathrm{~V}\) < Cytochrome c: \(E_{c}=0.254 \mathrm{~V}\) < Cytochrome a: \(E_{a}=0.385 \mathrm{~V}\)
03

Determine the correct order for spontaneous electron transfer

For the electron transfer process to occur spontaneously, the final acceptor must have a more positive reduction potential than the initial donor. Therefore, the correct order for spontaneous electron transfer is Cytochrome b \(\rightarrow\) Cytochrome c \(\rightarrow\) Cytochrome a. This means that electrons are first transferred from Cytochrome b (with the lowest reduction potential) to Cytochrome c, and then from Cytochrome c to Cytochrome a (with the highest reduction potential). This order ensures that the electron transfer process occurs spontaneously and ultimately leads to the electron transfer to molecular oxygen.

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Most popular questions from this chapter

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains 1.0\(M \mathrm{M}^{2+} .\) Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s)$$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C} .\) Assume that the process $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}$$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C} .\)

The measurement of \(\mathrm{pH}\) 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 $$\mathscr{E}_{\text { meas }}=\mathscr{E}_{\text { ref }}+0.05916 \mathrm{pH}$$ 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 \(\mathscr{E}_{\mathrm{ref}}=0.250 \mathrm{V}\) and that $\mathscr{E}_{\text { meas }}=0.480 \mathrm{V}$ a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the nncertainty in the measured potential is \(+1 \mathrm{mV}\) \(( \pm 0.001 \mathrm{V}) ?\) b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

A galvanic cell consists of a standard hydrogen electrode and a copper electrode immersed in a Cu(NO \(_{3} )_{2}(a q)\) solution. If you wish to construct a calibration curve to show how the cell potential varies with \(\left[\mathrm{Cu}^{2+}\right],\) what should you plot to obtain a straight line? What will be the slope of this line?

The overall reaction in the lead storage battery is $\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow$ $$\quad\quad\quad\quad\quad\quad\quad 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ a. For the cell reaction \(\Delta H^{\circ}=-315.9 \mathrm{kJ}\) and $\Delta S^{\circ}=\( 263.5 \)\mathrm{J} / \mathrm{K}\( . Calculate \)\mathscr{E}^{\circ} \mathrm{at}-20 .^{\circ} \mathrm{C}\( . Assume \)\Delta H^{\circ}\( and \)\Delta S^{\circ}$ do not depend on temperature. b. Calculate \(\mathscr{E}\) at \(-20 .^{\circ} \mathrm{C}\) when \(\left[\mathrm{HSO}_{4}^{-}\right]=\left[\mathrm{H}^{+}\right]=4.5 \mathrm{M}\) . c. Consider your answer to Exercise 69. Why does it seem that batteries fail more often on cold days than on warm days?

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. $\mathrm{Cr}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{NO}(g)$ b. $\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Ce}^{4+}(a q) \rightarrow \mathrm{CO}_{2}(a q)+\mathrm{Ce}^{3+}(a q)$ c. $\mathrm{SO}_{3}^{2-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Mn}^{2+}(a q)$

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