$\text{An electrochemical cell is fueled by the combustion of butane at}\mathrm{ 1}\mathrm{ bar}\text{and}\mathrm{ 298}\mathrm{ K}.\text{Its cell potential is }\frac{X}{F}\times {10}^3\text{volts,}$

$\text{where}\mathrm{ F }\text{is the Faraday constant.}\text{The value of }X\text{ is}\text{\_\_\_\_\_}$
$\text{Use: Standard Gibbs energies of formation at}\mathrm{ 298}K\text{ are :}$
${\Delta }_f{G}^{\ominus }{\mathrm{CO}}_2=-394\mathrm{kJ}{\mathrm{mol}}^{-1};{\Delta }_f{G}^{\ominus }{\mathrm{water}}_{\mathrm{}}=-237\mathrm{kJ}{\mathrm{mol}}^{-1};{\Delta }_f{G}^{\ominus }{\mathrm{butane}}_{\mathrm{}}=-18\mathrm{kJ}{\mathrm{mol}}^{-1}$

The correct answer is 105.50.

To find the value of $X$, we need to determine the standard cell potential for the combustion of butane.

First, we write the balanced chemical equation for the combustion of butane ($C_4{H}_{10}$):
$C_4{H}_{10}\left(g\right)+\frac{13}{2}{O}_2\left(g\right)\to 4C{O}_2\left(g\right)+5H_{2}O\left(l\right)$

The standard Gibbs energy change of the reaction (${\Delta }_r{G}^{\ominus }$) is calculated as:
${\Delta }_r{G}^{\ominus }=\left[4{\Delta }_f{G}^{\ominus }\right(C{O}_2)+5{\Delta }_f{G}^{\ominus }(\text{water}\left)\right]-\left[{\Delta }_f{G}^{\ominus }\right(\text{butane})+\frac{13}{2}{\Delta }_f{G}^{\ominus }(O_2\left)\right]$

Substituting the given standard Gibbs energies of formation:
${\Delta }_f{G}^{\ominus }\left(C{O}_2\right)=-394\text{kJ/mol}$
${\Delta }_f{G}^{\ominus }\left(\text{water}\right)=-237\text{kJ/mol}$
${\Delta }_f{G}^{\ominus }\left(\text{butane}\right)=-18\text{kJ/mol}$
${\Delta }_f{G}^{\ominus }\left(O_2\right)=0\text{kJ/mol}$

Calculating the value:
${\Delta }_r{G}^{\ominus }=\left[4\right(-394)+5(-237\left)\right]-[-18+0]$
${\Delta }_r{G}^{\ominus }=[-1576-1185]+18$
${\Delta }_r{G}^{\ominus }=-2761+18=-2743\text{kJ/mol}=-2743\times {10}^3\text{J/mol}$

Next, we determine the number of electrons transferred ($n$) in this reaction. We look at the reduction of oxygen:
$\frac{13}{2}{O}_2+26e^{-}\to 13O^{2-}$
Therefore, $n=26$ electrons are transferred.

The relation between the standard Gibbs energy change and the cell potential is:
${\Delta }_r{G}^{\ominus }=-nF{E}^{\ominus }$
Substitute the values into the equation:
$-2743\times {10}^3=-26\times F\times E^{\ominus }$
$E^{\ominus }=\frac{2743\times {10}^3}{26F}=\frac{2743/26}{F}\times {10}^3$
$E^{\ominus }=\frac{105.50}{F}\times {10}^3\text{V}$

Comparing this with the given potential form $\frac{X}{F}\times {10}^3\text{V}$:
$X=105.50$