/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 127 An experimental fuel cell has be... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 127

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 $$ \begin{array}{l} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \\\ \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3} . \mathrm{Ox}-\) ide ions can move through this solid at high temperatures (about \(\left.800^{\circ} \mathrm{C}\right) . \Delta G\) 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 this fuel cell under the given temperature and concentration conditions is 0.986 V.

Step by step solution

01

Calculate the number of moles of electrons transferred (n)

In the overall reaction, $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ From the half-cell reactions, we can see that 4 moles of electrons are transferred in the process: $$ \begin{array}{l} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \end{array} $$ and $$ \begin{array}{l} \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ Now we know n = 4 moles.
02

Use the relationship between Gibbs free energy change and cell potential

We can calculate the cell potential (E) by using the formula: $$ \Delta G = -nFE $$ Where 鈭咷 is the Gibbs free energy change (-380 kJ), n is the number of moles of electrons transferred (4 moles), F is the Faraday constant (96485 C/mol), and E is the cell potential which we want to find. First, we need to convert the 鈭咷 from kJ to J by multiplying by 1000: 鈭咷 = -380 kJ 脳 1000 J/kJ = -380000 J Now, we can rearrange the formula to solve for E: $$ E = -\frac{\Delta G}{nF} $$ Substitute the values, $$ E = -\frac{-380000 \mathrm{J}}{4 \times 96485 \mathrm{C/mol}} $$
03

Calculate the cell potential (E)

Finally, calculate the cell potential E: $$ E = \frac{380000 \mathrm{J}}{4 \times 96485 \mathrm{C/mol}} = 0.986 \mathrm{V} $$ Hence, the cell potential for this fuel cell under the given temperature and concentration conditions is 0.986 V.

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Most popular questions from this chapter

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times\) \(10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{~V}\). Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume \(\mathrm{Al}\) is oxidized.)

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). The \(\mathscr{E}^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{~V}\) relative to the standard hydrogen electrode: $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{c}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}{ }^{2-} $$ a. Calculate \(\mathscr{E}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \mathrm{~mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \times 10^{-5} M\), what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{~V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\). What is \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1\), calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M \mathrm{KF}\) solution b. \(1.0 \mathrm{M} \mathrm{CuCl}_{2}\) solution c. \(1.0 \mathrm{MgI}_{2}\) solution

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of \(4.00 \mathrm{~A}\). How long will it take to plate out \(99 \%\) of the platinum in \(0.50 \mathrm{~L}\) of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+}\) ?

Consider the following half-reactions: $$ \begin{aligned} \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & \mathscr{E}^{\circ} &=1.188 \mathrm{~V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ} &=0.755 \mathrm{~V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & \mathscr{E}^{\circ} &=0.96 \mathrm{~V} \end{aligned} $$ 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.
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