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JEE-Advanced 2025
PAPER-2 2025
Question
An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K . Its cell potential is $\frac{\mathbf{X}}{F} \times 10^3$ volts, where $F$ is the Faraday constant. The value of $\mathbf{X}$ is $\_\_\_\_$ . Use: Standard Gibbs energies of formation at 298 K are: $\Delta_f G_{\mathrm{CO}_2}^o=-394 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {water }}^o=-237 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {butane }}^o=-18 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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Question Tags
JEE Advance
Chemistry
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