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Chapter 18: Problem 124

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{~K}\) is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83} $$ a. Calculate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

Short Answer

Expert verified
In summary: a. For the fuel cell reaction at 298 K, the standard cell potential \(\mathscr{E}^{\circ} = 1.229 V\), and the standard Gibbs free energy \(\Delta G^{\circ} = -4.740 \times 10^5 J/mol\). b. The signs of the standard enthalpy and standard entropy for the fuel cell reaction are predicted to be negative (exothermic) and positive (increase in entropy), respectively. c. As the temperature increases, the maximum amount of work obtained from the fuel cell reaction will decrease.

Step by step solution

01

Determine n, R, T and F

The balanced chemical equation is given by \(2H_2 (g) + O_2 (g) \longrightarrow 2H_2O (l)\). We can determine the number of moles of electrons (n) involved in the reaction by looking at the oxidation numbers of hydrogen and oxygen. Hydrogen is reduced from oxidation number 0 to +1, and oxygen is reduced from oxidation number 0 to -2. Therefore, n = 4 (each \(H_2\) has 2 electrons; there are 2 moles of \(H_2\)). We are given T = 298 K. The gas constant R is \(8.314 J/mol \cdot K\). The Faraday constant F is \(9.648 \times 10^4 C/mol\).
02

Calculate E°

We can now calculate E° using the Nernst equation: \(\mathscr{E}^{\circ}=\frac{RT}{nF} \ln K = \frac{8.314 \cdot 298}{4 \cdot 9.648 \times 10^4} \ln (1.28 \times 10^{83}) \) After solving for E°, we get: \(\mathscr{E}^{\circ} = 1.229 V\)
03

Calculate ΔG°

Next, we can calculate the standard Gibbs free energy ΔG° using the Gibbs free energy equation: \(\Delta G^{\circ}=-nFE^{\circ} = -4 \cdot (9.648 \times 10^4 C/mol) \cdot 1.229 V\) After solving for ΔG°, we get: \(\Delta G^{\circ} = -4.740 \times 10^5 J/mol\) #b. Predict the signs of ΔH° and ΔS°#
04

Predict the signs of ΔH° and ΔS°

We know that ΔG° = ΔH° - TΔS°. Since ΔG° is negative and T is positive, we can predict that ΔH° will be negative (exothermic reaction) and ΔS° will be positive (increase in entropy). This makes sense because the reaction results in the formation of water, which releases heat. #c. Maximum work obtained from the fuel cell reaction as temperature increases#
05

Determine the effect of temperature on maximum work

We can analyze the effect of temperature on the maximum amount of work obtained from the fuel cell reaction by examining the relationship between ΔG°, ΔH°, and ΔS°: \(\Delta G^{\circ}=\Delta H^{\circ} - T\Delta S^{\circ}\) Since we have determined that ΔH° is negative and ΔS° is positive, as the temperature (T) increases, the TΔS° term will become larger, causing ΔG° to become less negative. Since ΔG° represents the maximum work that can be obtained from the reaction, as ΔG° becomes less negative, the maximum work decreases. Therefore, as temperature increases, the maximum amount of work obtained from the fuel cell reaction will decrease.

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Most popular questions from this chapter

Under standard conditions, what reaction occurs, if any, when each of the following operations is performed? a. Crystals of \(\mathrm{I}_{2}\) are added to a solution of \(\mathrm{NaCl}\). b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of NaI. c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\). d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. For the reactions that occur, write a balanced equation and calculate \(\mathscr{8}^{\circ}, \Delta G^{\circ}\), and \(K\) at \(25^{\circ} \mathrm{C}\).

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