Question

In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidizer.
The reaction is: $C{H}_{3}OH(l)+\frac{3}{2}{O}_2(g)\to C{O}_2(g)+H_{2}O(l)$
At 298 K, standard Gibbs energy of formation for $C{H}_{3}OH(l),H_{2}O(l),andC{O}_2(g)$ are -166.2, -237.2 and 394.4 kJ / mol respectively.
If the standard enthalpy of combustion of methanol is -726 kJ / mol, the efficiency of the fuel cell will be:
(a)- 80%
(b)- 87%
(c)- 90%
(d)- 97%

Answer 1

Hint: The efficiency of the fuel cell is calculated by dividing the standard free energy with the standard enthalpy of combustion. The standard free energy of a reaction is calculated by taking the difference of the sum of standard free energy of the product and the sum of standard free energy of the reactant.

Complete answer:
Standard free energy change of a reaction is defined as the change in free energy which takes place when the reactants in the standard state are converted into the products in their standard state.
It is calculated by the formula:
$\mathrm{\Delta }{G^{\circ }}_r=\sum \mathrm{\Delta }{G^{\circ }}_f(Products)-\sum \mathrm{\Delta }{G^{\circ }}_f(Reactants)$
Standard free energy of formation of a compound is defined as the free energy change which takes place when 1 mole of the compound is formed from its elements taken in their standard states.
So, according to the reaction:
$C{H}_{3}OH(l)+{\displaystyle \frac{3}{2}}{O}_2(g)\to C{O}_2(g)+H_{2}O(l)$
The reaction’s standard free energy change will be:
$\mathrm{\Delta }{G^{\circ }}_{reaction}=[\mathrm{\Delta }{G^{\circ }}_{f}C{O}_2+2\mathrm{\Delta }{G^{\circ }}_f{H}_{2}O]-[\mathrm{\Delta }{G^{\circ }}_{f}C{H}_{3}OH+\frac{3}{2}\mathrm{\Delta }{G^{\circ }}_f{O}_2]$
The standard free energy of carbon dioxide, water, methanol, and oxygen are -394.4, -237.2, 166.2, and 0 respectively. So,
$\mathrm{\Delta }{G^{\circ }}_{reaction}=[-394.4+2(-237.2)]-[-166.2+0]$
$\mathrm{\Delta }{G^{\circ }}_{reaction}=-702.3kJ/mol$
It is found that in a fuel cell the heat evolved during the process is not completely converted into energy, i.e., useful work. The ratio of$\mathrm{\Delta }G/\mathrm{\Delta }H$ is called the efficiency of the fuel cell.
$Efficiency={\displaystyle \frac{\mathrm{\Delta }G}{\mathrm{\Delta }H}}$
The free energy is -702.3kJ / mol and standard enthalpy are -726kJ / mol.
$Efficiency={\displaystyle \frac{\mathrm{\Delta }G}{\mathrm{\Delta }H}}={\displaystyle \frac{-702.3}{-726}}=0.97$
For percentage, 0.97 x 100 = 97%
The efficiency of the fuel cell is 97%
So, the correct answer is “Option D”.

Note: While calculating the standard free energy change of the reaction the number of atoms must be multiplied before each compound according to the reaction. The reaction must be balanced.