Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 125

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)$$ The two half-cell reactions are $$\mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-}$$ $$\mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-}$$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3}\) . Oxide ions can move through this solid at high temperatures (about \(800^{\circ} \mathrm{C} ) . \Delta G\) for the overall reaction at $800^{\circ} \mathrm{C}\( under certain concentration conditions is \)-380 \mathrm{kJ}$ . Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

Short Answer

Expert verified
The cell potential for the fuel cell under the given temperature and concentration conditions is 0.986 V.

Step by step solution

01

Find the number of moles of electrons transferred in the reaction

We are given the two half-cell reactions: 1) CO + O^2- → CO₂ + 2 e^- 2) O₂ + 4 e^- → 2 O^2- To balance the overall reaction, we can add the half-cell reactions and cancel the electrons from both sides: (1) × 2: 2 CO + 2 O^2- → 2 CO₂ + 4 e^- (2): O₂ + 4 e^- → 2 O^2- Adding (1) and (2), we get the overall reaction: 2 CO(g) + O₂(g) → 2 CO₂(g) From the above balanced equation, it is clear that 4 moles of electrons are transferred in the reaction. Hence, n = 4.
02

Calculate the cell potential using Gibbs free energy

Now that we have the number of moles of electrons transferred (n = 4) and the Gibbs free energy change (\(\Delta G\ = -380\) kJ), we can use the formula \(E_{cell} = -\frac{\Delta G}{nF}\) to calculate the cell potential. To do this, we first convert the Gibbs free energy from kJ to J: \(\Delta G = -380\:kJ \times \frac{1000\:J}{1\:kJ} = -380,\!000\:J\) Next, we substitute the given values into the formula: \(E_{cell} = -\frac{-380,\!000\:J}{(4\:mol)(96,\!485\:C/mol)}\) Now, divide the numbers to get the cell potential: \(E_{cell} = \frac{380,\!000\:J}{(4\:mol)(96,\!485\:C/mol)}\) \(E_{cell} = 0.986\:V\) The cell potential for the fuel cell under the given temperature and concentration conditions is 0.986 V.

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.

Start your free trial

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

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

The black silver sulfide discoloration of silverware can be removed by heating the silver article in a sodium carbonate solution in an aluminum pan. The reaction is $$3 \mathrm{Ag}_{2} \mathrm{S}(s)+2 \mathrm{Al}(s) \rightleftharpoons 6 \mathrm{Ag}(s)+3 \mathrm{S}^{2-}(a q)+2 \mathrm{Al}^{3+}(a q)$$ a. Using data in Appendix \(4,\) calculate \(\Delta G^{\circ}, K,\) and \(\mathscr{E}^{\circ}\) for the above reaction at $25^{\circ} \mathrm{C} .\left[\text { For } \mathrm{Al}^{3+}(a q), \Delta G_{\mathrm{f}}^{\circ}=-480 . \mathrm{kJ} / \mathrm{mol.}\right]$ b. Calculate the value of the standard reduction potential for the following half-reaction: $$2 \mathrm{e}^{-}+\mathrm{Ag}_{2} \mathrm{S}(s) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)$$

Estimate \(\mathscr{E}^{\circ}\) for the half-reaction $$2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}$$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ} :\) $$\quad\quad\quad \mathrm{H}_{2} \mathrm{O}(l)=-237 \mathrm{kJ} / \mathrm{mol}$$ $$\mathrm{H}_{2}(g)=0.0$$ $$\quad\quad\quad \mathrm{OH}^{-}(a q)=-157 \mathrm{kJ} / \mathrm{mol}$$ $$\quad \mathrm{e}^{-}=0.0$$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 18.1

What reactions take place at the cathode and the anode when each of the following is electrolyzed? a. molten \(\mathrm{NiBr}_{2} \quad\) b. molten \(\mathrm{AlF}_{3} \quad\) c. molten \(\mathrm{MgI}_{2}\)

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. $\mathrm{SO}_{3}^{2-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Mn}^{2+}(a q)$

It takes 15 kWh (kilowatt-hours) of electrical energy to produce 1.0 kg aluminum metal from aluminum oxide by the Hall-Heroult process. Compare this to the amount of energy necessary to melt 1.0 kg aluminum metal. Why is it economically feasible to recycle aluminum cans? [The enthalpy of fusion for aluminum metal is 10.7 $\mathrm{kJ} / \mathrm{mol}(1 \text { watt }=1 \mathrm{J} / \mathrm{s}) . ]$
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation
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