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Chapter 18: Problem 121

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)$$ The maximum work that 1 mole of alcohol can do is $1.32 \times 10^{3} \mathrm{kJ}$ . What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The theoretical maximum voltage this fuel cell can achieve at 25°C is \(0.907 \text{ V}\).

Step by step solution

01

Write the balanced half-reactions

To find the moles of electrons transferred, we need to write the balanced half-reactions for the oxidation and reduction processes. Oxidation half-reaction (losing electrons by alcohol): $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) \longrightarrow 2 \mathrm{CO}_{2}(g) + 12 \mathrm{H}^{+}(aq) + 12 e^{-} $$ Reduction half-reaction (gaining electrons by oxygen): $$ 3 \mathrm{O}_{2}(g) + 12 \mathrm{H}^{+}(aq) + 12 e^{-} \longrightarrow 6 \mathrm{H}_{2} \mathrm{O}(l) $$
02

Calculate the moles of electrons transferred

From the balanced half-reactions, we can see that 12 moles of electrons are transferred. n = 12 moles
03

Write the values in the appropriate units

Faraday's constant (F) is equal to 96,485 C/mol. The maximum work done (1.32 x 10^3 kJ) should be converted into J/mol. F = 96,485 C/mol Work = (1.32 x 10^3 kJ/mol) x (1,000 J/1 kJ) = 1.32 x 10^6 J/mol
04

Calculate the theoretical maximum voltage

Use the equation Work = -nFE_Voltage to solve for the maximum voltage (E_Voltage). $$E_\text{Voltage} = -\frac{\text{Work}}{nF}$$ $$E_\text{Voltage} = -\frac{1.32 \times 10^6 \text{ J/mol}}{12 \text{ moles} \times 96,485 \text{ C/mol}}$$ $$E_\text{Voltage} = 0.907 \text{ V}$$ The theoretical maximum voltage this fuel cell can achieve at 25°C is 0.907 V.

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Most popular questions from this chapter

Consider the following reduction potentials: $$\begin{array}{ll}{\mathrm{Co}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Co}} & {\mathscr{E}^{\circ}=1.26 \mathrm{V}} \\ {\mathrm{Co}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Co}} & {\mathscr{E}^{\circ}=-0.28 \mathrm{V}}\end{array}$$ a. When cobalt metal dissolves in 1.0\(M\) nitric acid, will \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}^{2+}\) be the primary product (assuming standard conditions)? b. Is it possible to change the concentration of \(\mathrm{HNO}_{3}\) to get a different result in part a? Concentrated \(\mathrm{HNO}_{3}\) is about 16 \(M\) .

Consider a galvanic cell based on the following half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\ \hline {\mathrm{La}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{La}} & {-2.37} \\\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {-0.44}\end{array}$$ a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Fe}^{2+}\right]=2.00 \times 10^{-4} M\) and \(\left[\mathrm{La}^{3+}\right]=3.00 \times 10^{-3} M,\) what is the expected cell potential?

What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at \(25^{\circ} \mathrm{C}\) if the copper electrode is placed in a solution in which \(\left[\mathrm{Cu}^{2+}\right]=\) \(2.5 \times 10^{-4} \mathrm{M} ?\) b. The copper electrode is placed in a solution of unknown \(\left[\mathrm{Cu}^{2+}\right] .\) The measured potential at $25^{\circ} \mathrm{C}\( is 0.195 \)\mathrm{V}\( . What is \)\left[\mathrm{Cu}^{2+}\right] ?$ (Assume \(\mathrm{Cu}^{2+}\) is reduced.)

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?
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