The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{~K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I) \quad K=1.28 \times 10^{83}$$ a. Calculate \(\mathscr{C}^{\circ}\) 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.

Using the Nernst equation, we can calculate the standard cell potential, \(\mathscr{C}^{\circ}\), from the equilibrium constant, \(K\), as follows: \[\mathscr{C}^{\circ} = \frac{RT}{nF} \ln K\] Where: - \(R\) is the gas constant, \(8.314 \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot K^{-1}\) - \(T\) is the temperature in Kelvin, here \(298 \mathrm{K}\) - \(n\) is the number of electrons transferred, which is \(4\) for \(2 \mathrm{H}_{2} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}\) - \(F\) is Faraday's constant, \(96485 \ \mathrm{C/mol}\) - \(K\) is the equilibrium constant, \(1.28 \times 10^{83}\) Plugging in the values, we get: \[\mathscr{C}^{\circ} = \frac{8.314 \times 298}{4 \times 96485} \ln (1.28 \times 10^{83}) \approx 1.229 \ \mathrm{V}\] Now, we can calculate the standard Gibbs free energy change, \(\Delta G^{\circ}\), using the cell potential and Faraday's constant: \[\Delta G^{\circ} = -nF\mathscr{C}^{\circ}\] \[\Delta G^{\circ} = -4 \times 96485 \times 1.229 \approx -474920 \ \mathrm{J/mol}\]

Since the fuel cell reaction generates a positive cell potential and has a large positive equilibrium constant, we can deduce that the reaction is spontaneous and exergonic, meaning the standard enthalpy change, \(\Delta H^{\circ}\), should be negative (exothermic). The spontaneous nature of the reaction also indicates that the standard entropy change, \(\Delta S^{\circ}\), will likely be positive. An increase in entropy usually accompanies the conversion of reactant gases into fewer product molecules.

To determine how temperature affects the maximum work output of the fuel-cell reaction, we need to consider the following relationships: - \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\) - \(W_{max} = -\Delta G\) As noted earlier, \(\Delta H^{\circ} < 0\) and \(\Delta S^{\circ} > 0\). As the temperature increases, the term \(T\Delta S^{\circ}\) increases, making the standard Gibbs free energy change, \(\Delta G^{\circ}\), less negative or closer to zero. As a result, the maximum work that can be obtained from the reaction, \(W_{max}\), decreases. Therefore, as the temperature increases, the maximum amount of work obtained from the fuel cell reaction decreases.