Question

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 $$ \begin{array}{l} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \\\ \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ 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 \(\left.800^{\circ} \mathrm{C}\right) . \Delta G\) for the overall reaction at \(800^{\circ} \mathrm{C}\) under certain concentration conditions is \(-380 \mathrm{~kJ}\). Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

Short answer

The cell potential for this fuel cell at 800°C and the given concentration conditions is 0.986 V.

Step-by-step solution

Determine the number of moles of electrons transferred (n)

We will first examine the two half-cell reactions to find out how many moles of electrons are being transferred in each reaction: 1. CO + O²⁻ → CO₂ + 2e⁻ 2. O₂ + 4e⁻ → 2O²⁻ In the first half-cell reaction, 2 moles of electrons are transferred. In the second half-cell reaction, 4 moles of electrons are transferred. Since these two half-cell reactions occur simultaneously in the cell, we need to balance the number of electrons transferred. We can do this by multiplying the first half-cell reaction by 2: 2(CO + O²⁻ → CO₂ + 2e⁻) Now the two half-cell reactions are: 1. 2CO + 2O²⁻ → 2CO₂ + 4e⁻ 2. O₂ + 4e⁻ → 2O²⁻ The overall balanced reaction is: 2CO(g) + O₂(g) → 2CO₂(g) The number of moles of electrons transferred (n) in the overall reaction is 4.

Calculate the cell potential (E)

Now that we know the number of moles of electrons transferred (n) and we are given the Gibbs free energy change (∆G), we can use the relationship between ∆G, n, and E to solve for the cell potential: ∆G = −nFE To calculate the cell potential (E), we will rearrange the equation to: E = -∆G/(nF) Substitute the given values: E = -(-380,000 J/mol) / (4 mol * 96485 C/mol) E = 0.986 V The cell potential for this fuel cell at 800°C and the given concentration conditions is 0.986 V.

Key concepts

Calculation of Cell Potential

Understanding the calculation of cell potential in electrochemistry is crucial for determining how efficient a fuel cell can be. Cell potential, denoted by 'E', reflects the voltage or electrical potential difference between two half-cells in an electrochemical cell. To calculate this, scientists use the formula derived from the relationship between Gibbs free energy change (∆G) and the number of moles of electrons transferred (n), in correlation with Faraday's constant (F), which represents the charge of one mole of electrons (approximately 96485 Coulombs per mole).
In this exercise, you're given ∆G, which is the energy change of the overall reaction under specified conditions, and using the standard equation ∆G = -nFE, you can isolate E to find the cell potential by rearranging the equation to E = -∆G / (nF). Subsequently, by substituting the provided values of ∆G, n, and F into the equation, you find that the cell potential for the experimental fuel cell at 800°C is 0.986 V.
This calculation is crucial for predicting the performance and feasibility of fuel cells as an energy source. Remember, a higher cell potential often means a more efficient fuel cell, which is why understanding how to calculate 'E' is fundamental in the study of electrochemistry.

Gibbs Free Energy in Electrochemistry

In electrochemistry, Gibbs free energy (∆G) is a central concept that links the thermodynamics of a reaction to the electrochemical performance of a cell. It essentially measures the maximum amount of work that can be extracted from a chemical process, and when a reaction occurs at constant temperature and pressure, a negative value for ∆G indicates that the reaction can perform work spontaneously.
The relationship between Gibbs free energy and cell potential is summarized by the equation ∆G = -nFE, where 'n' is the number of moles of electrons exchanged in the electrochemical reaction and 'F' is Faraday's constant. A more negative ∆G means a greater electrical work is possible, which translates to a higher cell potential. Hence, the negative ∆G of -380 kJ/mol in the provided fuel cell example clearly indicates the spontaneous nature of the reaction and its capability to do electrical work, which we already calculated to be 0.986 V.

Balanced Half-Cell Reactions

Half-cell reactions are the individual oxidation and reduction reactions occurring in the two compartments of an electrochemical cell. Balancing these half-cell reactions is an important step in understanding the full electrochemical process.
In our example, the half-cell reactions involve carbon monoxide and oxygen molecules. These reactions are initially unbalanced in terms of the number of electrons. By balancing, we ensure that the number of electrons lost in oxidation equals the number gained in reduction. It's a process similar to balancing chemical equations, but here, the focus is on electron transfer. The balanced equation helps us identify 'n', the number of moles of electrons transferred, which is a critical value for subsequently calculating Gibbs free energy changes and the cell potential. For this fuel cell, by balancing the half-cell reactions, we found that a total of 4 moles of electrons are transferred, ensuring that electrical charge is conserved in the overall reaction.

Movement of Oxide Ions in Solid State

In solid oxide fuel cells (SOFCs), the movement of oxide ions (O^{2-}) through a solid electrolyte is a fundamental process that allows the cell to conduct electricity. In the case of this experimental fuel cell, a solid mixture of cerium dioxide (CeO_2) and gadolinium oxide (Gd_2O_3) serves as the medium through which oxide ions move.
At high temperatures, like the 800°C specified in the exercise, oxide ions can move more freely through the electrolyte. This is due to increased ionic conductivity at elevated temperatures, which reduces the resistance within the cell. The movement is from the cathode where oxygen picks up electrons and is reduced to oxide ions, to the anode where the oxide ions can react with the fuel, carbon monoxide in this case, to produce carbon dioxide and release electrons back into the circuit. Hence, the ionic conductivity of the solid state is integral to the overall functionality and efficiency of SOFCs.