JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

Consider the following two half-cell reactions along with the standard reduction potential given :

CO $_{2} + 6$ H $^{+} + 6$ e $^{-} \to$ CH $_{3}$ OH $+$ H $_{2}$ O, $E_{red}^{\circ} = 0.02$ V

$\frac{1}{2}$ O $_{2} + 2$ H $^{+} + 2$ e $^{-} \to$ H $_{2}$ O, $E_{red}^{\circ} = 1.23$ V

The fuel cell was set up using the above two reactions such that the cell operates under the standard condition of $1$ bar pressure and $298$ K temperature. The fuel cell works with $80%$ efficiency. If the work derived from the cell using $1$ mol of CH $_{3}$ OH is used to compress an ideal gas isothermally against a constant pressure of $1$ kPa, then the change in the volume of the gas, $Δ V =$ __________ m $^{3}$ . (Nearest integer)

Given : F = $96500$ C mol $^{- 1}$

Your answer

Solution

The methanol fuel cell uses CH $_{3}$ OH as the fuel, so CH $_{3}$ OH is oxidised (reverse of the first half-reaction) and O $_{2}$ is reduced.

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 1.23 - 0.02 = 1.21$ V.

The oxidation of one CH $_{3}$ OH transfers $n = 6$ electrons. For $1$ mol of methanol:

$∣Δ G^{\circ} ∣ = n F E_{cell}^{\circ} = 6 \cdot 96500 \cdot 1.21$ J.

Usable work at $80%$ efficiency: $W = 0.80 \cdot 6 \cdot 96500 \cdot 1.21 \approx 560472$ J.

For isothermal compression against constant external pressure $P_{ext} = 1$ kPa $= 1 0^{3}$ Pa: $W = P_{ext} Δ V$ , so $Δ V = \frac{W}{P _{ext}} = \frac{560472}{1 0 ^{3}} \approx 560$ m $^{3}$ .

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