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

Use: Standard Gibbs energies of formation at 298 K are: $$\Delta_f G^o_{CO_2} = -394$$ kJ mol$$^{-1}$$; $$\Delta_f G^o_{water} = -237$$ kJ mol$$^{-1}$$; $$\Delta_f G^o_{butane} = -18$$ kJ mol$$^{-1}$$

The balanced combustion of butane under standard conditions is
$$C_4H_{10}+ \frac{13}{2}\,O_2 \rightarrow 4\,CO_2 + 5\,H_2O$$

1. Standard Gibbs energy change, $$\Delta_rG^\circ$$
Using $$\Delta_rG^\circ = \sum \Delta_fG^\circ(\text{products})-\sum \Delta_fG^\circ(\text{reactants})$$,

$$\Delta_rG^\circ = \bigl[4(-394) + 5(-237)\bigr] - \bigl[(-18) + 0\bigr] \text{ kJ}$$
$$= (-1576 -1185) + 18 \text{ kJ}$$
$$= -2743 \text{ kJ mol}^{-1}$$

2. Electrons transferred, $$n$$
Each $$O_2$$ molecule is reduced from oxidation state 0 to -2, accepting $$4$$ electrons. Number of $$O_2$$ molecules $$= \dfrac{13}{2}=6.5$$, hence

$$n = 6.5 \times 4 = 26 \text{ mol e}^-$$

3. Cell potential and definition of $$X$$
Maximum electrical work $$= -\Delta_rG^\circ = nF E^\circ$$, so

$$E^\circ = \frac{-\Delta_rG^\circ}{nF}$$

The problem states $$E^\circ = \dfrac{X}{F}\times 10^{3}\ \text{V}$$, that is
$$E^\circ = \frac{1000\,X}{F}$$

Equating the two expressions for $$E^\circ$$ gives

$$\frac{-\Delta_rG^\circ}{nF} = \frac{1000\,X}{F} \Longrightarrow X = \frac{-\Delta_rG^\circ}{1000\,n}$$

Substituting $$\Delta_rG^\circ = -2743 \text{ kJ mol}^{-1}$$ and $$n = 26$$,

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

The required value therefore lies in the range $$105.4 - 105.6$$.