Question

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)

Short answer

Question: Calculate the electrical efficiency (fuel to electricity) of an electrolyte-supported SOFC, given the following information: the Gibbs free energy change is -188.6 kJ/mol, cell voltage is 0.7 V, cell resistance is 0.1 Ω, cell conductivity is 5 Ω-1·m-1, electrolyte thickness is 100 µm, and cell active area is 2x10-4 m2. Answer: The electrical efficiency (fuel to electricity) of the electrolyte-supported SOFC is 13.6%.

Step-by-step solution

1. Calculate the Inlet Nernst Voltage

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: \(E_{Nernst}=\frac{-188.6 \times 10^3}{2 \times 96485} = -0.977 \mathrm{V}\).

2. Calculate the Rates of Hydrogen and Oxygen Consumption

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: \(r_{H_2} = \frac{J}{2 F} \times 2 = \frac{7}{96485} \times 2 = 1.449 \times 10^{-4} \,\mathrm{mol} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}^{-1}\). \(r_{O_2} = \frac{r_{H_2}}{2}= \frac{1.449 \times 10^{-4}}{2} = 7.247 \times 10^{-5} \,\mathrm{mol} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}^{-1}\).

3. Calculate the Electrical Efficiency (Fuel to Electricity)

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%.

Key concepts

Nernst Equation

The Nernst Equation is a key principle in electrochemistry, helping us calculate the electric potential of an electrochemical cell. It links the measurable voltage between the electrodes to the chemical reaction quotient, providing insights into the behavior of the cell under non-standard conditions. For a Solid Oxide Fuel Cell (SOFC), the Nernst Equation is expressed as follows:\[E_{Nernst} = \frac{\Delta \bar{G_{f}}}{nF} + \frac{RT}{nF}\ln\left(\frac{p_{H_2}p_{O_2}^{1/2}}{p_{H_2O}}\right)\]Where:

• \(E_{Nernst}\) is the Nernst voltage.
• \(\Delta \bar{G_{f}}\) is the change in Gibbs free energy under standard conditions.
• \(n\) is the number of moles of electrons transferred per mole of hydrogen.
• \(F\) is Faraday’s constant, representing the charge of one mole of electrons.
• \(R\) is the universal gas constant.
• \(T\) is the absolute temperature in Kelvin.
• \(p_{H_2}, p_{O_2},\) and \(p_{H_2O}\) are the partial pressures of hydrogen, oxygen, and water, respectively.

The Nernst Equation is crucial for understanding and calculating the potential difference at which an SOFC operates under specific conditions. Understanding this equation helps predict the performance of SOFCs in real-world applications.

Gibbs Free Energy

Gibbs Free Energy (\(\Delta G\)) is a thermodynamic property that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. Within the context of Solid Oxide Fuel Cells, it serves as a measure of the achievable work from electrochemical reactions. This energy change is fundamental for calculating the cell potential because it directly links to the chemical and thermal energies involved.In the context of SOFCs:

• \(\Delta \bar{G_{f}}\) is the Gibbs Free Energy change from reactants to products in their standard states.
• It determines the theoretical maximum voltage (or electromotive force) for every mole of fuel consumed.

Hence, SOFCs aim to maximize the conversion of Gibbs Free Energy into electrical energy, achieving higher efficiency. Mastery of this concept allows better assessment and optimization of fuel cells, crucial for the development of efficient, sustainable energy solutions.

Electrochemical Reactions

Electrochemical Reactions are fundamental to the operation of SOFCs. These reactions involve electron transfer across chemical compounds, converting chemical energy into electrical energy; precisely what SOFCs are designed for. In SOFCs, hydrogen gas (\(\text{H}_2\)) reacts with oxygen gas (\(\text{O}_2\)) to produce water vapor and electric current:The general reactions at the electrodes are:

• Anode (Hydrogen oxidation):\[\text{H}_2 + 2\text{O}^{2-} \rightarrow 2\text{H}_2\text{O} + 4e^-\]
• Cathode (Oxygen reduction):\[\text{O}_2 + 4e^- \rightarrow 2\text{O}^{2-}\]

These reactions must be well-balanced and matched with the electron flow through an external circuit to generate power. Understanding these reactions is integral to appreciate how SOFCs convert pure fuels into electricity and the inherent efficiencies involved.

Electrical Efficiency

Electrical Efficiency in a Solid Oxide Fuel Cell is a measure of how effectively the chemical energy of the fuel is converted into electrical energy. The efficiency is determined by comparing the electrical power output to the fuel’s energy content.Electrical efficiency (\(\eta\)) can be calculated as:\[\eta = \frac{P_{output}}{-P_{input}}\]Where:

• \(P_{output}\) is the electrical power generated by the cell.
• \(P_{input}\) is the total chemical energy input from the fuel consumed.

Understanding the electrical efficiency is crucial for evaluating the performance and viability of an SOFC. Higher efficiencies imply better conversion rates, less fuel wastage, and more sustainable energy production, which are essential for meeting clean energy goals and reducing environmental impact. Therefore, continuously improving this efficiency ensures the competitiveness and adoption of SOFC technologies.