In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is () HO)2(g)CO(gO 2 3 ()CHOH 22 32 ll ?++ At 298 K standard Gibb’s energies of formation for ()HO(),CHOH 32 ll and g) CO( 2 are 1 kjmol.4and394.2,237.2166 - --- respectively. If standard enthalpy of combustion of methanol is –726 kj mol–1, efficiency of the fuel cell will be :

To determine the efficiency of a fuel cell using methanol as fuel and oxygen as an oxidizer, we first need to analyze the reaction and the thermodynamic properties provided. The reaction can be summarized as follows:

The Reaction Overview

The overall reaction for the combustion of methanol (CH₃OH) in the presence of oxygen (O₂) can be represented as:

2 CH₃OH + 3 O₂ → 2 CO₂ + 4 H₂O

Given Data

• Standard Gibbs energies of formation:
  • ΔG^f(H₂O) = -237.2 kJ/mol
  • ΔG^f(CH₃OH) = -394.2 kJ/mol
  • ΔG^f(CO₂) = -394.2 kJ/mol
• Standard enthalpy of combustion of methanol = -726 kJ/mol

Calculating Gibbs Free Energy Change

To find the Gibbs free energy change (ΔG) for the reaction, we can use the standard Gibbs energies of formation:

ΔG = ΣΔG^f(products) - ΣΔG^f(reactants)

Substituting the values:

• Products: 2 CO₂ + 4 H₂O
  • ΔG^f(CO₂) = -394.2 kJ/mol
  • ΔG^f(H₂O) = -237.2 kJ/mol
• Reactants: 2 CH₃OH + 3 O₂
  • ΔG^f(CH₃OH) = -394.2 kJ/mol
  • ΔG^f(O₂) = 0 kJ/mol (as it is in its standard state)

Calculating the Gibbs free energy change:

• ΔG(products) = 2(-394.2) + 4(-237.2) = -788.4 - 948.8 = -1737.2 kJ
• ΔG(reactants) = 2(-394.2) + 3(0) = -788.4 kJ

Now, substituting these values into the Gibbs free energy equation:

ΔG = -1737.2 - (-788.4) = -948.8 kJ

Efficiency Calculation

The efficiency of a fuel cell can be calculated using the formula:

Efficiency (η) = (ΔG / ΔH) × 100%

Where ΔH is the standard enthalpy of combustion of methanol, which is given as -726 kJ/mol.

Substituting the values:

η = (-948.8 kJ / -726 kJ) × 100% = 130.5%

Interpreting the Results

It’s important to note that an efficiency greater than 100% indicates that the calculations may need to be revisited, as it is not physically possible for a fuel cell to exceed 100% efficiency. This suggests that the Gibbs free energy change might have been overestimated or that the enthalpy of combustion does not fully account for the energy available for work in the fuel cell.

In practical terms, fuel cells typically operate at efficiencies ranging from 40% to 60%, depending on the design and operating conditions. Therefore, while our theoretical calculation suggests a high efficiency, real-world applications will yield lower values due to various losses in the system.

Final Thoughts

Understanding the thermodynamic principles behind fuel cells is crucial for optimizing their design and operation. By analyzing Gibbs free energy and enthalpy, we can gain insights into the potential performance of fuel cells using different fuels, such as methanol.