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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Chemistry
Chapter
Electrochemistry

JEE Advanced 2025 · Paper 2 · Chemistry · Electrochemistry

An electrochemical cell is fueled by the combustion of butane at 1...

An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is XF×103\frac{\boldsymbol{X}}{\boldsymbol{F}} \times 10^3 volts, where F\boldsymbol{F} is the Faraday constant. The value of X\boldsymbol{X} is \qquad .
Use : Standard Gibbs energies of formation at 298 K are : ΔfGCO2∘=−394 kJ mol−1\Delta_f G_{\mathrm{CO}_2}^{\circ}=-394 \mathrm{~kJ} \mathrm{~mol}^{-1}; ΔfGwater ∘=−237 kJ mol−1;ΔfGbutane ∘=−18 kJ mol−1\Delta_f G_{\text{water }}^{\circ}=-237 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text{butane }}^{\circ}=-18 \mathrm{~kJ} \mathrm{~mol}^{-1}

Correct answer

0

Step-by-step solution

C4H10( g)+132O2( g)→4CO2( g)+5H2O(l)ΔrG∘=4ΔfGCO2∘+5ΔfGH2O∘−ΔfGC4H10∘=4×(−394)+5(−237)+18=−2743 kJ/molΔrG∘=−nFE∘−2743×1000=−26×FE∘E∘=105.5 F×103=105.50\begin{aligned} & \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}^{\circ}=4 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{CO}_2}^{\circ}+5 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{H}_2 \mathrm{O}}^{\circ}-\Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{C}_4 \mathrm{H}_{10}}^{\circ} \\ & =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}^{\circ}=\frac{105.5}{\mathrm{~F}} \times 10^3=105.50\end{aligned}
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