JEE Main 2026 — 2 April, Evening Shift — Question 65

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

$$\begin{array}{ll} \mathrm{CO}_{2}+6 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}_{2} \mathrm{O} & \mathrm{E}_{\text {red }}^{\mathrm{o}}=0.02 \mathrm{~V} \frac{1}{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O} & \mathrm{E}_{\text {red }}^{\mathrm{o}}=1.23 \mathrm{~V} \end{array}$$

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 $\mathrm{CH}_{3} \mathrm{OH}$ is used to compress an ideal gas isothermally against a constant pressure of 1 kPa , then the change in the volume of the gas, $\Delta \mathrm{V}=$ $\_\_\_\_$ $\mathrm{m}^{3}$. (Nearest integer) Given : $\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}$