Answer
414.9k+ views
Hint:
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 $1.0968\;{\rm{V}}$.
Note:
The combustion reaction mainly occurs in the presence of atmospheric oxygen. The product of that combustion reaction is always carbon dioxide and water vapour.
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 $1.0968\;{\rm{V}}$.
Note:
The combustion reaction mainly occurs in the presence of atmospheric oxygen. The product of that combustion reaction is always carbon dioxide and water vapour.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)