Question

: At 298K standard Gibbs free energy of formation of \[{\text{C}}{{\text{H}}_3}{\text{OH}}({\text{l}}),{{\text{H}}_2}{\text{O}}({\text{l}}){\text{ and C}}{{\text{O}}_2}({\text{g}})\] are \[ - 166.2, - 237.2{\text{ and }} - 394.4{\text{ KJ mo}}{{\text{l}}^{ - 1}}\] respectively. If standard enthalpy of combustion of methanol is \[ - 726{\text{ KJ mo}}{{\text{l}}^{ - 1}}\], efficiency of fuel cell will be:
A.\[87\% \]
B.\[90\% \]
C.\[97\% \]
D.\[80\% \]

Answer 1

Hint: First of all change in Gibbs free energy can be calculated using the given Gibbs free energy values and writing the balanced chemical equation. Then using the formula for the fuel cell and substituting the given data will give us the efficiency of the fuel cell.
Formula used: \[{\text{Efficiency }} = \dfrac{{\Delta {\text{G}}}}{{\Delta {\text{H}}}} \times 100\]
Here \[\Delta {\text{G}}\] is the Gibbs free energy change of reaction and \[\Delta {\text{H}}\] is the enthalpy of reaction.

Complete step by step solution:
The reaction of combustion of methanol produces carbon dioxide and water. The reaction occurs as follow:
\[{\text{C}}{{\text{H}}_3}{\text{OH}}({\text{l}}) + \dfrac{3}{2}{{\text{O}}_2} \to 2{{\text{H}}_2}{\text{O}}({\text{l}}){\text{ }} + {\text{ C}}{{\text{O}}_2}({\text{g}})\]
We need to calculate the difference in Gibbs free energy of the products and reactants in order to put into the formula. Since the given values of Gibbs free energy are in respect to 1 mole of product or reactant so we have to multiply with the number of moles used in the balanced chemical equation: the value of Gibbs free energy change can be calculated as:
\[\Delta {\text{G}} = \Delta {{\text{G}}_{{\text{product}}}} - \Delta {{\text{G}}_{{\text{reactants}}}}\]
Putting the values for the Gibbs free energy of product and reactant multiplying by their respective moles given to us, we will get:
\[\Delta {\text{G}} = \left[ {\left( {2 \times - 237.2 - 394.4} \right) - \left( { - 166.2} \right)} \right]\]
\[ \Rightarrow \Delta {\text{G}} = - 474 - 394.4 + 166.2 = - 702.6{\text{ KJ mo}}{{\text{l}}^{ - 1}}\]
The value of enthalpy change is given to us. Using the formula for efficiency of the fuel cell we will get:
\[{\text{Efficiency }} = \dfrac{{\Delta {\text{G}}}}{{\Delta {\text{H}}}} \times 100\]
Substituting the values:
\[{\text{Efficiency }} = \dfrac{{ - 702.6}}{{ - 726}} \times 100 = 96.77\% \]
\[ \Rightarrow {\text{Efficiency }} \approx 97\% \]

Hence the correct option is C.

Note:
A fuel cell is a cell specifically an electrochemical cell, which converts chemical energy that is present in a fuel into electricity. A basic redox reaction occurs in the fuel cell one substance oxidizes another substance reduces. Combustion is known as the burning of substance in the presence of oxygen.