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In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l) At 298 K standard Gibbs energies of formation for CH3OH(l), H2O(l) and CO2(g) are -166.2, -237.2 and -394.4 kJ mol⁻¹ respectively. If standard enthalpy of combustion of methanol is -726 kJ mol⁻¹, efficiency of the fuel cell will be:

1. 87%
2. 90%
3. 97%
4. 80%

Correct answer: 97%

Solution

The efficiency of the fuel cell is calculated by comparing the Gibbs free energy change of the reaction to the enthalpy change. Given the high Gibbs energy change relative to the enthalpy of combustion, the fuel cell operates at a high efficiency, which is reflected in the calculated value of 97%.

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