JEE Advanced · Chemistry · 17. Electrochemistry
An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is \(\frac{\boldsymbol{X}}{\boldsymbol{F}} \times 10^3\) volts, where \(\boldsymbol{F}\) is the Faraday constant. The value of \(\boldsymbol{X}\) is \(\qquad\) .
Use : Standard Gibbs energies of formation at 298 K are : \(\Delta_f G_{\mathrm{CO}_2}^{\circ}=-394 \mathrm{~kJ} \mathrm{~mol}^{-1}\);
\(\Delta_f G_{\text {water }}^{\circ}=-237 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {butane }}^{\circ}\) \(=-18 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- A 105.5
- B 152.46
- C 452.56
- D 10.25
Answer & Solution
Correct Answer
(A) 105.5
Step-by-step Solution
Detailed explanation
\(\mathrm{C}_4 \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+5 \mathrm{H}_2 \mathrm{O}(l) \)
\( \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=4 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{CO}_2}^{\circ}+5 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{H}_2 \mathrm{O}}^{\circ}-\Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{C}_4 \mathrm{H}_{10}}^{\circ} \)
\( =4 \times(-394)+5(-237)+18 \)
\( =-2743 \mathrm{~kJ} / \mathrm{mol} \)
\( \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \)
\( -2743 \times 1000=-26 \times \mathrm{FE}^{\circ} \)
\( \mathrm{E}^{\circ}=\frac{105.5}{\mathrm{~F}} \times 10^3=105.50\)
\( \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=4 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{CO}_2}^{\circ}+5 \Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{H}_2 \mathrm{O}}^{\circ}-\Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{C}_4 \mathrm{H}_{10}}^{\circ} \)
\( =4 \times(-394)+5(-237)+18 \)
\( =-2743 \mathrm{~kJ} / \mathrm{mol} \)
\( \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \)
\( -2743 \times 1000=-26 \times \mathrm{FE}^{\circ} \)
\( \mathrm{E}^{\circ}=\frac{105.5}{\mathrm{~F}} \times 10^3=105.50\)
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