Question

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)$$ The maximum work that 1 mole of alcohol can do is \(1.32 \times 10^{3} \mathrm{kJ}\) . What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C}\) ?

Short answer

The theoretical maximum voltage this fuel cell can achieve at 25°C is \(0.907 \text{ V}\).

Step-by-step solution

Write the balanced half-reactions

To find the moles of electrons transferred, we need to write the balanced half-reactions for the oxidation and reduction processes. Oxidation half-reaction (losing electrons by alcohol): $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) \longrightarrow 2 \mathrm{CO}_{2}(g) + 12 \mathrm{H}^{+}(aq) + 12 e^{-} $$ Reduction half-reaction (gaining electrons by oxygen): $$ 3 \mathrm{O}_{2}(g) + 12 \mathrm{H}^{+}(aq) + 12 e^{-} \longrightarrow 6 \mathrm{H}_{2} \mathrm{O}(l) $$

Calculate the moles of electrons transferred

From the balanced half-reactions, we can see that 12 moles of electrons are transferred. n = 12 moles

Write the values in the appropriate units

Faraday's constant (F) is equal to 96,485 C/mol. The maximum work done (1.32 x 10^3 kJ) should be converted into J/mol. F = 96,485 C/mol Work = (1.32 x 10^3 kJ/mol) x (1,000 J/1 kJ) = 1.32 x 10^6 J/mol

Calculate the theoretical maximum voltage

Use the equation Work = -nFE_Voltage to solve for the maximum voltage (E_Voltage). $$E_\text{Voltage} = -\frac{\text{Work}}{nF}$$ $$E_\text{Voltage} = -\frac{1.32 \times 10^6 \text{ J/mol}}{12 \text{ moles} \times 96,485 \text{ C/mol}}$$ $$E_\text{Voltage} = 0.907 \text{ V}$$ The theoretical maximum voltage this fuel cell can achieve at 25°C is 0.907 V.

Key concepts

Electrochemistry

Electrochemistry is the study of chemical processes that cause electrons to move. This movement of electrons is what we call electricity. The core idea behind electrochemistry is that certain chemical reactions can either produce electric currents or be driven by them. Fuel cells, like the one in our exercise, are a great example of electrochemical systems.

Fuel cells make use of electrochemical reactions to convert chemical energy directly into electrical energy. This is different from combustion engines which first convert chemical energy into heat and then into mechanical work. In a fuel cell, substances are oxidized and reduced at separate electrodes, causing electrons to flow through an external circuit. This flow of electrons is what we harness as electricity.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, are chemical processes where electrons are transferred between substances. One substance loses electrons, known as oxidation, and another gains those electrons, known as reduction. In the provided problem, the grain alcohol oxidizes, losing electrons, while oxygen is reduced, gaining electrons.

Understanding these reactions is crucial in the operation of fuel cells. For fuel cells to operate, they rely on these half-reactions: oxidation occurs at the anode, providing electrons, and reduction occurs at the cathode, consuming electrons. The net flow of electrons through an external circuit is what creates electrical energy in a fuel cell. Grasping the half-reactions helps comprehend how a fuel cell transforms chemical energy into electrical power.

Theoretical Maximum Voltage

The theoretical maximum voltage of a fuel cell is the highest electromotive force (emf) that can be achieved under ideal conditions. It is calculated from the energy change of the reaction and the number of electrons involved, using the formula: \[ E_\text{Voltage} = -\frac{\text{Work}}{nF} \]where "Work" is the maximum chemical energy converted to electrical energy, "n" represents the number of moles of electrons exchanged during the complete reaction, and "F" is Faraday's constant, representing the charge of one mole of electrons.

In the solution, we calculate this by setting the work equal to the Gibbs free energy change, which is converted into joules per mole. Dividing this energy by the product of electron moles and Faraday's constant gives us the voltage. This theoretical value tells us how efficient the cell might be under perfect conditions, helping us understand its potential performance.

Work-Energy Principle

The work-energy principle in electrochemistry relates the chemical work done by a system to the electrical energy it can produce. The work done in a chemical reaction, especially within a fuel cell, comes from breaking and forming chemical bonds—transforming chemical potential energy into electricity.

Fuel cells convert fuel into electricity through controlled reactions. Using the equation\[ \text{Work} = -nFE_\text{Voltage} \]we can deduce the energy conversion efficiency of a fuel cell. "Work" here represents the maximum work done by the fuel cell when driving the electric current through. A high work value means more energy can be extracted, and hence, a higher theoretical voltage.

Understanding this principle helps in optimizing the design and materials used in fuel cells to maximize their energy conversion efficiency.