JEE Advance - Chemistry (2025 - Paper 2 Online - No. 14)
An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is $\frac{\boldsymbol{X}}{F} \times 10^3$ volts, where $F$ is the Faraday constant. The value of $\boldsymbol{X}$ is _____________.
Use: Standard Gibbs energies of formation at 298 K are: $\Delta_f G_{\mathrm{CO}_2}^o=-394 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {water }}^o=$ $-237 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {butane }}^o=-18 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation
Balanced combustion of butane (liquid water):
$$\mathrm{C_4H_{10}} + \tfrac{13}{2}\,\mathrm{O_2}\;\longrightarrow\;4\,\mathrm{CO_2} + 5\,\mathrm{H_2O}$$
Standard Gibbs energy change:
$$\Delta G^\circ =\bigl[4\,\Delta_fG^\circ(\mathrm{CO_2})+5\,\Delta_fG^\circ(\mathrm{H_2O})\bigr]\,-\bigl[\Delta_fG^\circ(\mathrm{C_4H_{10}})+\tfrac{13}{2}\,\Delta_fG^\circ(\mathrm{O_2})\bigr]$$
$$= \bigl[4(-394)+5(-237)\bigr]\,-[-18+0]\;\mathrm{kJ/mol} = -1576 -1185 +18 = -2743\;\mathrm{kJ/mol}$$
Electrons transferred, $n$:
Each C goes from –2.5 to +4 ⇒ loses 6.5 e–, so total $n=4\times6.5=26$.
Cell potential:
$\begin{aligned} & \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ & -2743 \times 1000=-26 \times \mathrm{FE}^{\circ} \\ & \mathrm{E}^{\circ}=\frac{105.5}{\mathrm{~F}} \times 10^3=105.50\end{aligned}$Comments (0)
