An electrolyte-supported SOFC is operated at atmospheric pressure and \(800^{\circ} \mathrm{C}\) with the following mole fractions of the reactant and product species: \(x_{\mathrm{H}_{2}}=\) \(0.95\) and \(x_{\mathrm{H}_{2} \mathrm{O}}=0.05\) (anode) and \(x_{\mathrm{O}_{2}}=0.21\) (cathode). At \(800^{\circ} \mathrm{C}\), the fuel cell has \(\Delta \bar{g}_{\mathrm{f}}=-188.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) and \(\Delta h_{\mathrm{r}}=-248.3 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) of \(\mathrm{H}_{2}\), and the conductivity of the cell is \(5 \Omega^{-1} \cdot \mathrm{m}^{-1}\). The cell active area is \(2 \times 10^{-4} \mathrm{~m}^{2}\), and the electrolyte thickness is \(100 \mu \mathrm{m}\). If the cell is operated at \(0.7 \mathrm{~V}\), then determine the following: a. The inlet Nernst voltage b. The rates at which hydrogen and oxygen are consumed c. The electrical efficiency (fuel to electricity)

In this step, we will use the Nernst equation to determine the theoretical voltage of the SOFC. The Nernst equation is given as follows: \(E_{Nernst}=\frac{\Delta \bar{G_{f}}}{nF}\), where \(E_{Nernst}\) is the Nernst voltage, \(\Delta \bar{G_{f}}\) is the Gibbs free energy change, \(n\) is the number of electrons transferred per mole of hydrogen, and \(F\) is Faraday's constant. We know that the Gibbs free energy change \(\Delta \bar{G_{f}}\) is \(-188.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\), \(n=2\) and Faraday's constant \(F=96485 \mathrm{~C} \cdot \mathrm{mol}^{-1}\). The inlet Nernst voltage can be calculated as follows: \(E_{Nernst}=\frac{-188.6 \times 10^3}{2 \times 96485} = -0.977 \mathrm{V}\).

In this step, we will calculate the current density and use it to determine the rates of hydrogen and oxygen consumption. First, let's calculate the current density (\(J\)). The formula for current density is as follows: \(J = \frac{V_{cell}}{R_{total}}\), where \(V_{cell}\) is the cell voltage and \(R_{total}\) is the SOFC's resistance. In this problem, we have a cell voltage of \(0.7 \mathrm{V}\), and we can calculate the resistance, given cell's conductivity (\(5 \Omega^{-1} \cdot \mathrm{m}^{-1}\)), the electrolyte thickness (\(100 \mu m\)), and the cell active area (\(2 \times 10^{-4} \mathrm{m}^{2}\)). \(R_{total} = \frac{100 \times 10^{-6}}{2 \times 10^{-4} \times 5} = 0.1 \Omega\). Now, let's calculate the current density: \(J = \frac{0.7}{0.1} = 7 \mathrm{A} \cdot \mathrm{m}^{-2}\). To find the rates of hydrogen and oxygen consumption, we will use the stoichiometry and Faraday's constant: \(r_{H_2} = \frac{J}{2 F} \times 2 = \frac{7}{96485} \times 2 = 1.449 \times 10^{-4} \,\mathrm{mol} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}^{-1}\). \(r_{O_2} = \frac{r_{H_2}}{2}= \frac{1.449 \times 10^{-4}}{2} = 7.247 \times 10^{-5} \,\mathrm{mol} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}^{-1}\).

In this step, we will calculate the fuel to electricity efficiency of the SOFC. First, let's calculate the power output (\(P_{output}\)) of the cell: \(P_{output} = J \times V_{cell} = 7 \times 0.7 = 4.9 \mathrm{W} \cdot \mathrm{m}^{-2}\). Next, we need to calculate the fuel input power (\(P_{input}\)): \(P_{input} = r_{H_2} \times \Delta h_{r} = 1.449 \times 10^{-4} \times (-248.3 \times 10^3) = -35.99 \mathrm{W} \cdot \mathrm{m}^{-2}\). Finally, let's find the electrical efficiency: \(\eta = \frac{P_{output}}{-P_{input}} = \frac{4.9}{35.99} = 0.136 = 13.6\%\). Thus, the electrical efficiency (fuel to electricity) is 13.6%.