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Chapter 17: 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 $$ \begin{array}{c} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \\\ \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ 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 \(800^{\circ} \mathrm{C}\) ). \(\Delta G\) for the overall reaction at \(800^{\circ} \mathrm{C}\) under certain concentration conditions is -380 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 at 800°C and under certain concentration conditions is approximately 0.983 V.

Step by step solution

01

Convert ∆G to Joules

We are given that the Gibbs free energy change ∆G = -380 kJ. To work with the equation mentioned in the analysis, we need to convert this value to Joules. 1 kJ = 1000 J So, ∆G = -380 * 1000 J = -380,000 J
02

Determine the number of transferred electrons (n)

To find the number of transferred electrons (n), we need to look at the two half-cell reactions: - CO + O²⁻ → CO₂ + 2 e⁻ - O₂ + 4 e⁻ → 2 O²⁻ By multiplying the first half-cell reaction by 2, we will achieve the same number of transferred electrons as in the second half-cell reaction (4): - 2(CO + O²⁻ → CO₂ + 2 e⁻) This gives us: - 2 CO + 2 O²⁻ → 2 CO₂ + 4 e⁻ Now, add the second half-cell reaction: - 2 CO + 2 O²⁻ → 2 CO₂ + 4 e⁻ - O₂ + 4 e⁻ → 2 O²⁻ We get the overall reaction: - 2 CO + O₂ → 2 CO₂ Therefore, the number of transferred electrons, n = 4.
03

Calculate the cell potential (E)

Now that we have the values of ∆G (-380,000 J) and n (4), we can calculate the cell potential (E) using the relationship between ∆G, n, F (Faraday's constant), and E: $$ \Delta G = -nFE $$ Rearrange the equation to isolate E: $$ E = -\frac{\Delta G}{nF} $$ Plugging in the values, ∆G = -380,000 J, n = 4, and F = 96,485 C/mol (Faraday's constant), we get: $$ E = -\frac{-380,000 \text{ J}}{4 \times 96,485 \text{ C/mol}} $$ Calculating this expression, we find: $$ E = \frac{380,000}{4 \times 96,485} = 0.983 \text{ V} $$ So, the cell potential for this fuel cell at the same temperature and concentration conditions is approximately 0.983 V.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \\ \mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ}=-1.20 \mathrm{V} \end{array}$$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 \mathrm{M},\) and the vanadium compartment contains a vanadium electrode and \(V^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{L}\) of solution) was titrated with \(0.0800 M \space \mathrm{H}_{2} \mathrm{EDTA}^{2-},\) resulting in the reaction $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \space \mathrm{K=?}$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{mL} \space \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{V}\). The solution was buffered at a pH of \(10.00 .\) a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K,\) for the titration reaction. c. Calculate \(\mathscr{E}_{\text {cell }}\) at the halfway point in the titration.

A galvanic cell is based on the following half-reactions: $$\begin{array}{cl} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \end{array}$$ In this cell, the silver compartment contains a silver electrode and excess \(\mathrm{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right),\) and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 \mathrm{M}\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. Assuming \(1.0 \mathrm{L}\) of \(2.0 \mathrm{M}\space \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3}\) ). $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) & \rightleftharpoons \ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$

It took 2.30 min using a current of 2.00 A to plate out all the silver from 0.250 L of a solution containing Ag \(^{+} .\) What was the original concentration of \(\mathrm{Ag}^{+}\) in the solution?

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$ \mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q) $$ (Hint: Reference the \(K_{\mathrm{sp}}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+} .\) )

In the electrolysis of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what reactions occur at the anode and the cathode (assuming standard conditions)? $$\begin{array}{lr} \mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-} & 80^{\circ} \\ \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow_{2 \mathrm{H}_{2} \mathrm{O}} & 2.01 \mathrm{V} \\ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-} & -0.83 \mathrm{V} \\ \mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na} & -2.71 \mathrm{V} \end{array}$$
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