Chemistry · Electrochemistry

JEE Advanced 2025 — Paper 2 — Question 43

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 $\qquad$ .

Use : Standard Gibbs energies of formation at 298 K are : $\Delta_{f} G_{\mathrm{CO}_{2}}^{\mathrm{o}}=-394 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ;

$\Delta_{f} G_{\text {water }}^{\mathrm{o}}=-237 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_{f} G_{\text {butane }}^{\mathrm{o}}=-18 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Answer: 105.5

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{~g})+5 \mathrm{H}_{2} \mathrm{O}(l)$ $\Delta_{\mathrm{r}} \mathrm{G}^{\mathrm{o}}=4 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{CO}_{2}}^{\mathrm{o}}+5 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{H}_{2} \mathrm{O}}^{\mathrm{o}}-\Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{C}_{4} \mathrm{H}_{10}}^{\mathrm{o}}$

$=4 \times(-394)+5(-237)+18$

$=-2743 \mathrm{~kJ} / \mathrm{mol}$

$\Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{nFE}{ }^{\circ}$

$-2743 \times 1000=-26 \times \mathrm{FE}^{\circ}$

$\mathrm{E}^{\mathrm{o}}=\frac{105.5}{\mathrm{~F}} \times 10^{3}=105.50$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Chemistry
Chapter: Electrochemistry
Topic: Basics of Galvanic Cell