JEE Advanced 2025 Paper 2 · Q14 · Galvanic Cells & EMF
An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is $X/F \times 10^3$ volts, where $F$ is the Faraday constant. The value of X is __________.
Use: Standard Gibbs energies of formation at 298 K: $\Delta_f G^o(CO_2) = -394$ kJ/mol; $\Delta_f G^o(H_2O, \text{liquid}) = -237$ kJ/mol; $\Delta_f G^o(\text{butane}) = -18$ kJ/mol.
Reveal answer + step-by-step solution
Correct answer:105.5
Solution
Combustion reaction: $C_4H_{10} + 13/2 O_2 \to 4 CO_2 + 5 H_2O(l)$. $\Delta G_{rxn} = 4 \times (-394) + 5 \times (-237) - (-18) = -1576 - 1185 + 18 = -2743$ kJ = $-2.743 \times 10^6$ J. Number of electrons transferred per mole of butane: each carbon goes from average oxidation state $-2.5$ in butane to $+4$ in $CO_2$; net $4 \times (4 - (-2.5)) = 4 \times 6.5 = 26$ $e^-$ per molecule (this matches H balance: 10 H atoms lose effectively 0 net charge as they end up $+1$ in $H_2O$; the 26-electron count is correct). Cell potential $E = -\Delta G/(n F) = 2.743 \times 10^6$ J $/ (26 \times F) = (2.743/26) \times 10^6 / F$ V $= 1.055 \times 10^5/F$ V $= (105.5/F) \times 10^3$ V. Hence $X = 105.5$.
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