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 $$\mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-}$$ $$\mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-}$$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3}\) . Oxide ions can move through this solid at high temperatures (about \(800^{\circ} \mathrm{C} ) . \Delta G\) 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 the fuel cell under the given temperature and concentration conditions is 0.986 V.

Step-by-step solution

Find the number of moles of electrons transferred in the reaction

We are given the two half-cell reactions: 1) CO + O^2- → CO₂ + 2 e^- 2) O₂ + 4 e^- → 2 O^2- To balance the overall reaction, we can add the half-cell reactions and cancel the electrons from both sides: (1) × 2: 2 CO + 2 O^2- → 2 CO₂ + 4 e^- (2): O₂ + 4 e^- → 2 O^2- Adding (1) and (2), we get the overall reaction: 2 CO(g) + O₂(g) → 2 CO₂(g) From the above balanced equation, it is clear that 4 moles of electrons are transferred in the reaction. Hence, n = 4.

Calculate the cell potential using Gibbs free energy

Now that we have the number of moles of electrons transferred (n = 4) and the Gibbs free energy change (\(\Delta G\ = -380\) kJ), we can use the formula \(E_{cell} = -\frac{\Delta G}{nF}\) to calculate the cell potential. To do this, we first convert the Gibbs free energy from kJ to J: \(\Delta G = -380\:kJ \times \frac{1000\:J}{1\:kJ} = -380,\!000\:J\) Next, we substitute the given values into the formula: \(E_{cell} = -\frac{-380,\!000\:J}{(4\:mol)(96,\!485\:C/mol)}\) Now, divide the numbers to get the cell potential: \(E_{cell} = \frac{380,\!000\:J}{(4\:mol)(96,\!485\:C/mol)}\) \(E_{cell} = 0.986\:V\) The cell potential for the fuel cell under the given temperature and concentration conditions is 0.986 V.

Key concepts

Fuel Cells

Fuel cells are devices that convert chemical energy directly into electrical energy through electrochemical reactions. Unlike traditional engines that burn fuel, fuel cells efficiently produce electricity with less waste.
In this context, a fuel cell uses carbon monoxide (CO) as fuel, reacting with oxygen (O₂) to produce carbon dioxide (CO₂). This reaction occurs in two distinct half-cell reactions, generating electrical energy.

• **Fuel Source:** Carbon monoxide serves as the fuel, entering a chemical reaction with oxygen.
• **Efficiency:** These cells are generally more efficient than other power sources because they directly convert chemical energy into electricity.
• **Environmentally Friendly:** They produce fewer emissions than conventional combustion processes.

Fuel cells find applications in various fields such as transportation and backup power systems.

Gibbs Free Energy

Gibbs free energy (\(\Delta G\)) represents the maximum usable energy obtained from a chemical reaction at constant temperature and pressure. It helps determine spontaneity and thermodynamic feasibility of reactions.
In our case, the overall reaction involved in the fuel cell at \(800^{\circ}\) C has a Gibbs free energy (\(\Delta G\)) of \(-380\: \text{kJ}\), indicating that the reaction releases energy.

• If \(\Delta G\) is negative, the process is spontaneous and can proceed without any external work.
• If it's positive, the process is non-spontaneous and requires energy to occur.

This negative \(\Delta G\) tells us that the fuel cell reaction is naturally favorable, making it effective for energy production.

Cell Potential

Cell potential (\(E_{cell}\)) is the measure of the voltage produced by a fuel cell. It's linked with Gibbs free energy and can be calculated using the formula:\[ E_{cell} = -\frac{\Delta G}{nF} \]where:

• \(\Delta G\) is the Gibbs free energy change in joules.
• \(n\) is the number of moles of electrons transferred.
• \(F\) is Faraday's constant (\(96,485\: \text{C/mol}\)).

For this CO fuel cell reaction, \(\Delta G\) is \(-380,000\: \text{J}\), and \(n\) is 4, leading to a cell potential of \(0.986\: \text{V}\).
This potential indicates the ability of the fuel cell to perform electrical work under specific conditions.

Half-Cell Reactions

Half-cell reactions represent distinct processes occurring separately within a fuel cell. Each involves either a reduction or an oxidation, and together they form the overall reaction.
In this fuel cell:

• **First Half-Cell Reaction:** \(\text{CO} + \text{O}^{2-} \rightarrow \text{CO}_2 + 2 \text{e}^-\)—Here, carbon monoxide is oxidized.
• **Second Half-Cell Reaction:** \(\text{O}_2 + 4 \text{e}^- \rightarrow 2 \text{O}^{2-}\)—Here, oxygen is reduced.

Adding these half-reactions gives the complete reaction where electrons are transferred, enabling electricity production.
Each half-cell operates in separate compartments, connected by an electrolyte where oxide ions are transferred, ensuring the movement of charges.