The theoretical \(E_{\text {cell }}^{\circ}\) for the methane-oxygen fuel cell is \(1.06 \mathrm{V} .\) What is \(E^{\circ}\) for the reduction half-reaction \(\mathrm{CO}_{2}(\mathrm{g})+8 \mathrm{H}^{+}(\mathrm{aq})+8 \mathrm{e}^{-} \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1) ?\)

Short Answer

Expert verified
The \(E^{\circ}\) for the given reduction half-reaction is \(0.53 \mathrm{V}\)

Step by step solution

01

Confirmed half reaction

For the fuel cell reaction given, the reduction half-reaction is \[\mathrm{CO}_{2}(\mathrm{g})+8 \mathrm{H}^{+}(\mathrm{aq})+8 \mathrm{e}^{-} \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(l)\]
02

Determine the entire reaction

Before obtaining the E° for this half-reaction, you must first work out the full reaction for the methane-oxygen fuel cell. The full reaction is \[\mathrm{CH}_{4}(\mathrm{g}) + 2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O}(l)\]
03

Determine Oxidation half-reaction

Next is to determine the oxidation half-reaction which is obtained by reversing the reduction half-reaction. The oxidation half-reaction is \[\mathrm{CH}_{4}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) + 8 \mathrm{H}^{+}(\mathrm{aq}) + 8 \mathrm{e}^{-}\]
04

Calculate E° for the reduction half-reaction

Finally, knowing that \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{oxidation}} + E^{\circ}_{\text{reduction}}\), and that for the oxidation half-reaction, \(E^{\circ}_{\text{oxidation}} = -E^{\circ}_{\text{reduction}}\) (since you're reversing the reaction), you can deduce that \(E^{\circ}_{\text{reduction}} = \frac{E^{\circ}_{\text{cell}}}{2}\), which equals to 1.06 V / 2

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Standard Electrode Potential
The standard electrode potential, often represented as \(E^\circ\), is a measurement that indicates how readily a chemical species gains electrons in a half-reaction, which is a significant concept in electrochemistry. It describes the tendency of an electrode to be reduced when it's compared to the standard hydrogen electrode (SHE), which is set at 0 volts. By convention, the more positive the \(E^\circ\), the greater the species' affinity for electrons and the stronger its oxidizing ability.

For instance, in a methane-oxygen fuel cell, these potentials are used to calculate the voltage output of the cell by combining the potential of the oxidation half-reaction and the reduction half-reaction. Knowing the standard potentials also allows us to predict whether a reaction will occur spontaneously under standard conditions.
Reduction Half-Reaction
A reduction half-reaction involves the gain of electrons by a substance. During this process, the oxidation state of the substance is decreased as electrons (e-) are added. In electrochemical cells, the cathode is where reduction occurs, and this is typically represented in a half-cell notation. For example, in the exercise provided, \(\text{CO}_2\) is reduced to \(\text{CH}_4\) by gaining electrons.

In the calculation of the standard electrode potential for the methane-oxygen fuel cell, the reduction half-reaction is central. By identifying which species are being reduced and calculating the associated potential, you can determine the performance characteristics of the fuel cell and predict which reactions can occur spontaneously.
Oxidation Half-Reaction
The oxidation half-reaction is the counterpart to the reduction half-reaction and involves the loss of electrons. It raises the oxidation state of a substance as it releases electrons. For instance, in our exercise, methane (\(\text{CH}_4\)) is oxidized to carbon dioxide (\(\text{CO}_2\)) and hydrogen ions (\(\text{H}^+\)) with the release of electrons. The oxidation half-reaction takes place at the anode of an electrochemical cell.

Understanding oxidation processes is vital when considering overall cell reactions in fuel cells and determining the standard electrode potential for these reactions. Being able to write and balance oxidation half-reactions is as crucial as understanding reduction, since both processes occur simultaneously in electrochemical cells.
Methane-Oxygen Fuel Cell
A methane-oxygen fuel cell is a type of electrochemical cell that generates electricity through the reaction of methane with oxygen. It has two half-reactions, as with any fuel cell: an oxidation half-reaction at the anode and a reduction half-reaction at the cathode. In the particular problem discussed, methane (\(\text{CH}_4\)) reacts with oxygen (\(\text{O}_2\)) to form carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)).

The overall efficiency and potential of the fuel cell rely on the careful balance of these reactions and understanding the standard electrode potentials involved. Components like the catalysts used at the electrodes and the membrane that separates them also play a significant role in the performance of these cells. Methane-oxygen fuel cells are considered for their high energy efficiency and lower environmental impact compared to traditional fossil fuels.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following electrochemical cell: $$ \operatorname{Pt}(\mathrm{s})\left|\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm})\right| \mathrm{H}^{+}(1 \mathrm{M}) \| \mathrm{Ag}^{+}(x \mathrm{M}) | \mathrm{Ag}(\mathrm{s}) $$ (a) What is \(E_{\text {cell }}^{\circ}-\) that is, the cell potential when \(\left[\mathrm{Ag}^{+}\right]=1 \mathrm{M} ?\) (b) Use the Nernst equation to write an equation for \(E_{\text {cell }}\) when \(\left[\mathrm{Ag}^{+}\right]=x\) (c) Now imagine titrating \(50.0 \mathrm{mL}\) of \(0.0100 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) in the cathode half-cell compartment with 0.0100 M KI. The titration reaction is $$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \longrightarrow \mathrm{AgI}(\mathrm{s})$$ Calculate \(\left[\mathrm{Ag}^{+}\right]\) and then \(E_{\text {cell }}\) after addition of the following volumes of \(0.0100 \mathrm{M} \mathrm{KI}:(\mathrm{i}) 0.0 \mathrm{mL} ;(\mathrm{ii}) 20.0 \mathrm{mL}\) (iii) \(49.0 \mathrm{mL} ;(\text { iv }) 50.0 \mathrm{mL} ;(\mathrm{v}) 51.0 \mathrm{mL} ;(\mathrm{vi}) 60.0 \mathrm{mL}\) (d) Use the results of part (c) to sketch the titration curve of \(E_{\text {cell }}\) versus volume of titrant.

Ultimately, \(\Delta G_{\mathrm{f}}^{\mathrm{Q}}\) values must be based on experimental results; in many cases, these experimental results are themselves obtained from \(E^{\circ}\) values. Early in the twentieth century, G. N. Lewis conceived of an experimental approach for obtaining standard potentials of the alkali metals. This approach involved using a solvent with which the alkali metals do not react. Ethylamine was the solvent chosen. In the following cell diagram, \(\mathrm{Na}(\text { amalg, } 0.206 \%)\) represents a solution of \(0.206 \%\) Na in liquid mercury. 1\. \(\mathrm{Na}(\mathrm{s}) | \mathrm{Na}^{+}(\text {in ethylamine }) | \mathrm{Na}(\text { amalg }, 0.206 \%)\) \(E_{\text {cell }}=0.8453 \mathrm{V}\) Although Na(s) reacts violently with water to produce \(\mathrm{H}_{2}(\mathrm{g}),\) at least for a short time, a sodium amalgam electrode does not react with water. This makes it possible to determine \(E_{\text {cell }}\) for the following voltaic cell. 2\. \(\mathrm{Na}(\text { amalg }, 0.206 \%)\left|\mathrm{Na}^{+}(1 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\right|\) $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) \quad E_{\mathrm{cell}}=1.8673 \mathrm{V}$$ (a) Write equations for the cell reactions that occur in the voltaic cells (1) and (2) (b) Use equation (20.14) to establish \(\Delta G\) for the cell reactions written in part (a). (c) Write the overall equation obtained by combining the equations of part (a), and establish \(\Delta G^{\circ}\) for this overall reaction. (d) Use the \(\Delta G^{\circ}\) value from part (c) to obtain \(E_{\text {cell }}^{\circ}\) for the overall reaction. From this result, obtain \(E_{\mathrm{Na}^{+}}^{\circ} / \mathrm{Na}\) Compare your result with the value listed in Appendix D.

Of the following statements concerning electrochemical cells, the correct ones are: (a) The cathode is the negative electrode in both voltaic and electrolytic cells. (b) The function of a salt bridge is to permit the migration of electrons between the half-cell compartments of an electrochemical cell. (c) The anode is the negative electrode in a voltaic cell. (d) Electrons leave the cell from either the cathode or the anode, depending on what electrodes are used. (e) Reduction occurs at the cathode in both voltaic and electrolytic cells. (f) If electric current is drawn from a voltaic cell long enough, the cell becomes an electrolytic cell. (g) The cell reaction is an oxidationreduction reaction.

Given that \(E_{\text {cell }}^{\circ}=3.20 \mathrm{V}\) for the reaction $$2 \mathrm{Na}(\mathrm{in} \mathrm{Hg})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{Na}^{+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})$$ What is \(E^{\circ}\) for the reduction \(2 \mathrm{Na}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow\) \(2 \mathrm{Na}(\text { in } \mathrm{Hg}) ?\)

Show that for a combination of half-cell reactions that produce a standard reduction potential for a half-cell that is not directly observable, the standard reduction potential is $$E^{\circ}=\frac{\sum n_{i} E_{i}^{\circ}}{\sum n_{i}}$$ where \(n_{i}\) is the number of electrons in each half-reaction of potential \(E_{i}^{\circ} .\) Use the following half-reactions: $$ \begin{array}{c} \mathrm{H}_{5} \mathrm{IO}_{6}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{IO}_{3}^{-}(\mathrm{aq})+ \\ 3 \mathrm{H}_{2} \mathrm{O}(1) \quad E^{\circ}=1.60 \mathrm{V} \\ \mathrm{IO}_{3}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq})+5 \mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{I}_{2}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(1) \\ E^{\circ}=1.19 \mathrm{V} \\ 2 \mathrm{HIO}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1) \\ E^{\circ}=1.45 \mathrm{V} \\ \mathrm{I}_{2}(\mathrm{s})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(\mathrm{aq}) \quad \quad E^{\circ}=0.535 \mathrm{V} \end{array} $$ Calculate the standard reduction potential for $$ \mathrm{H}_{6} \mathrm{IO}_{6}+5 \mathrm{H}^{+}+2 \mathrm{I}^{-}+3 \mathrm{e}^{-} \longrightarrow $$ $$ \frac{1}{2} \mathrm{I}_{2}+4 \mathrm{H}_{2} \mathrm{O}=2 \mathrm{HIO} $$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free