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Standard free energies of formation $\left( {{\rm{in}}\;{\rm{kJ/mol}}} \right)$ at ${\rm{298}}\;{\rm{K}}$ are $ - 237.2, - 394.4$ and $ - 8.2$ for ${{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right){\rm{,C}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right)$ and pentane (g) respectively. The value of ${\rm{E}}_{{\rm{cell}}}^{\rm{0}}$ for the pentane oxygen fuel cell is:
A. $1.0968\;{\rm{V}}$
B. $0.0968\;{\rm{V}}$
C. $1.968\;{\rm{V}}$
D. $2.0968\;{\rm{V}}$

Last updated date: 13th Jun 2024
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We know that Gibbs free energy is that quantity which can generally be utilized to measure the extreme amount of work done in the thermodynamic system and the temperature and pressure kept constant during the whole process.

Complete step by step solution
The reaction of pentane oxygen fuel cell, which is also the combustion reaction of pentane is shown below.
${{\rm{C}}_{\rm{5}}}{{\rm{H}}_{{\rm{12}}}} + {\rm{8}}{{\rm{O}}_{\rm{2}}} \to {\rm{5C}}{{\rm{O}}_{\rm{2}}} + {\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$
The formula for standard free energies formation is shown below.
\[{\rm{\Delta G}} = {\rm{n}} \times {\rm{G}}\left( {{\rm{Products}}} \right) - {\rm{n}} \times {\rm{G}}\left( {{\rm{Reactants}}} \right)\]
Where, ${\rm{\Delta G}}$ is the standard free energy, ${\rm{G}}\left( {{\rm{Products}}} \right)$ is the free energy of products, and ${\rm{G}}\left( {{\rm{Reactants}}} \right)$ is the free energy of reactants.

Substitute the respective values in the above equation.
{\rm{\Delta G}} = 5 \times \left( { - 394.4} \right) + 6 \times \left( { - 237.2} \right) - 1 \times \left( { - 8.2} \right) + 8 \times 0\\
 = - 1972 - 1423.2 + 8.2\;{\rm{kJ/mol}}\\
 = - 3387\;{\rm{kJ/mol}}

The value of ${\rm{E}}_{{\rm{cell}}}^{\rm{0}}$ can be calculated by using the formula shown below.
${\rm{\Delta G}} = - {\rm{nFE}}_{{\rm{cell}}}^{\rm{0}}$

Where, ${\rm{E}}_{{\rm{cell}}}^{\rm{0}}$ is the standard cell potential, n is the number of moles of electrons involved, F is the faraday constant, and ${\rm{\Delta G}}$ is the standard free energy.
As oxygen is moving from oxidation state $0\;{\rm{to}}\; - 2$.
The number of moles of oxygen involved Is $32$.
The value of faraday in coulomb is ${\rm{1}}\;{\rm{F}} = 96500\;{\rm{coulombs}}$
Substitute the respective values in the above equation.
\Rightarrow - 3387\;{\rm{kJ/mol}} = - {\rm{32}} \times {\rm{96500}} \times {\rm{E}}_{{\rm{cell}}}^{\rm{0}}\\
\Rightarrow - 3387 \times {10^3}\;{\rm{J/mol}} = - {\rm{32}} \times {\rm{96500}} \times {\rm{E}}_{{\rm{cell}}}^{\rm{0}}\\
\Rightarrow {\rm{E}}_{{\rm{cell}}}^{\rm{0}} = \dfrac{{ - 3387 \times {{10}^3}\;{\rm{J/mol}}}}{{ - {\rm{32}} \times {\rm{96500}}}}\\
\Rightarrow {\rm{E}}_{{\rm{cell}}}^{\rm{0}} = 1.0968\;{\rm{V}}

Hence, the correct choice for this question is A that is $1.0968\;{\rm{V}}$.


The combustion reaction mainly occurs in the presence of atmospheric oxygen. The product of that combustion reaction is always carbon dioxide and water vapour.