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JEE Advanced 2025 · Paper 2, Chemistry question 14

An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is XF×103\dfrac{\boldsymbol{X}}{F}\times10^3 volts, where FF is the Faraday constant. The value of X\boldsymbol{X} is _____.

Use: Standard Gibbs energies of formation at 298 K are: ΔfGCOX2∘=−394 kJ mol−1\Delta_fG^\circ_{\ce{CO2}} = -394\ \mathrm{kJ\,mol^{-1}}; ΔfGwater∘=−237 kJ mol−1\Delta_fG^\circ_{\text{water}} = -237\ \mathrm{kJ\,mol^{-1}}; ΔfGbutane∘=−18 kJ mol−1\Delta_fG^\circ_{\text{butane}} = -18\ \mathrm{kJ\,mol^{-1}}

Answer and worked solution

Official answer: 105.4

Reaction. CX4HX10+132 OX2→4 COX2+5 HX2O\ce{C4H10 + 13/2 O2 -> 4CO2 + 5H2O}.

Free-energy change.
ΔG∘=4(−394)+5(−237)−(−18)=−2743 kJ.\Delta G^\circ=4(-394)+5(-237)-(-18)=-2743\ \text{kJ}.

Electrons transferred. Carbon goes from −2.5-2.5 to +4+4, so n=4×6.5=26n=4\times6.5=26.

Cell potential.
E∘=−ΔG∘nF=2743×10326F=105.5F×103 V.E^\circ=-\frac{\Delta G^\circ}{nF}=\frac{2743\times10^3}{26F}=\frac{105.5}{F}\times10^3\ \text{V}.
So X≈105.5X\approx105.5.

Topics:Gibbs energy and EMFBatteries and corrosionOxidation number
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