JEE Advanced 2025 Paper 2, Chemistry Section 3 Q6: Electrochemical Cell And Nernst Equation

JEE Advanced2025Paper 2Chemistry Section 3

An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is $\frac{X}{F} \times 1 0^{3}$ volts, where $F$ is the Faraday constant. The value of $X$ is ______.

Use : Standard Gibbs energies of formation at 298 K are : $Δ_{f} G_{CO_{2}}^{\circ} = - 394$ kJ mol $^{- 1}$ ; $Δ_{f} G_{water}^{\circ} = - 237$ kJ mol $^{- 1}$ ; $Δ_{f} G_{butane}^{\circ} = - 18$ kJ mol $^{- 1}$

Your answer

Solution

Combustion reaction:

C $_{4}$ H $_{10}$ (g) + $\frac{13}{2}$ O $_{2}$ (g) $\to$ 4 CO $_{2}$ (g) + 5 H $_{2}$ O(l)

$Δ_{r} G^{\circ} = 4 Δ_{f} G_{CO_{2}}^{\circ} + 5 Δ_{f} G_{H_{2} O}^{\circ} - Δ_{f} G_{C_{4} H_{10}}^{\circ}$

$= 4 (- 394) + 5 (- 237) - (- 18) = - 1576 - 1185 + 18 = - 2743$ kJ mol $^{- 1}$ .

The number of electrons transferred per molecule of butane: each C goes from oxidation state $- 5/2$ (average in C $_{4}$ H $_{10}$ , but cleaner to count atom-by-atom) to $+ 4$ in CO $_{2}$ . Direct calculation by balancing half-reactions gives $n = 26$ electrons (each of 4 C loses 4 to 6 electrons depending on its position; total electrons in combustion of C $_{4}$ H $_{10}$ to 4 CO $_{2}$ + 5 H $_{2}$ O is 26).

$Δ_{r} G^{\circ} = - n F E^{\circ}$

$- 2743 \times 1 0^{3} = - 26 \times F \times E^{\circ}$

$E^{\circ} = \frac{2743 \times 1 0 ^{3}}{26 F} = \frac{105.5 \times 1 0 ^{3}}{F}$ V

Comparing with $E^{\circ} = \frac{X}{F} \times 1 0^{3}$ V gives $X = 105.50$ .

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JEE Advanced 2025 Paper 2, Chemistry Section 3 Q6: Electrochemical Cell And Nernst Equation

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