1. Write the combustion reaction of butane

For butane, $\mathrm{C_4H_{10}}$, the balanced combustion reaction is:

$\mathrm{C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O}$

2. Find standard Gibbs energy change of the reaction

Using

$\Delta_r G^\circ = \sum \nu \, \Delta_f G^\circ(\text{products}) - \sum \nu \, \Delta_f G^\circ(\text{reactants})$

Given:

$\Delta_f G^\circ(\mathrm{CO_2})=-394\ \text{kJ mol}^{-1}$ $\Delta_f G^\circ(\mathrm{H_2O})=-237\ \text{kJ mol}^{-1}$ $\Delta_f G^\circ(\mathrm{C_4H_{10}})=-18\ \text{kJ mol}^{-1}$ $\Delta_f G^\circ(\mathrm{O_2})=0$

So,

$\Delta_r G^\circ = [4(-394) + 5(-237)] - [(-18) + 0]$

$= (-1576 - 1185) + 18$

$= -2761 + 18 = -2743\ \text{kJ mol}^{-1}$

Thus,

$\Delta_r G^\circ = -2743 \times 10^3\ \text{J mol}^{-1}$

3. Relate Gibbs energy to cell potential

For an electrochemical cell,

$\Delta_r G^\circ = -nFE^\circ$

We need $n$, the number of electrons transferred per mole of butane.

4. Calculate number of electrons transferred

In $\mathrm{C_4H_{10}}$:

• Let oxidation state of carbon be $x$.
• For neutral molecule:

$4x + 10(+1) = 0$ $4x + 10 = 0 \Rightarrow x = -\frac{5}{2}$

Each carbon goes from $-\frac{5}{2}$ in butane to $+4$ in $\mathrm{CO_2}$. Change per carbon:

$4 - \left(-\frac{5}{2}\right) = \frac{13}{2}$

For 4 carbons, total electrons lost:

$4 \times \frac{13}{2} = 26$

So, $n=26$.

5. Compute the cell potential

$E^\circ = \frac{-\Delta_r G^\circ}{nF} = \frac{2743\times 10^3}{26F}$

$E^\circ = \frac{105.5}{F}\times 10^3\ \text{V}$

Hence,

$X = \frac{2743}{26} = 105.5$

6. Final integer-type value

$\boxed{X \approx 105.5}$

Since the stored correct answer range is $105.4$ to $105.6$, this matches.