Skip to content
dronacharya.ai
← JEE Main 2026 · 2 Apr · Shift 2
ChemistryInteger+4 / -1Question 73 of 75

JEE Main 2026 · 2 Apr · Shift 2, Chemistry question 73

Consider the following two half-cell reactions along with the standard reduction potential given :
COX2+6 HX++6 eX−→CHX3OH+HX2OEred∘=0.02 V\ce{CO2 + 6H+ + 6e- -> CH3OH + H2O}\qquad E^\circ_\text{red}=0.02\ \mathrm V
12 OX2+2 HX++2 eX−→HX2OEred∘=1.23 V\ce{1/2O2 + 2H+ + 2e- -> H2O}\qquad E^\circ_\text{red}=1.23\ \mathrm V
A 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 CHX3OH\ce{CH3OH} is used to compress an ideal gas isothermally against a constant pressure of 1 kPa, then the change in the volume of the gas, ΔV\Delta V = ________ m³. (nearest integer)
Given : F=96500F=96500 C mol−1^{-1}

Answer and worked solution

Official answer: 560

Cell. CHX3OH+32 OX2→COX2+2 HX2O\ce{CH3OH + 3/2O2 -> CO2 + 2H2O} with n=6n=6 and E∘=1.23−0.02=1.21E^\circ=1.23-0.02=1.21 V.

Work available. ∣ΔG∘∣=6×96500×1.21=7.006×105|\Delta G^\circ|=6\times96500\times1.21=7.006\times10^5 J; at 80% efficiency, 5.605×1055.605\times10^5 J.

Volume change. W=PΔVW=P\Delta V gives ΔV=5.605×105103=560\Delta V=\dfrac{5.605\times10^5}{10^3}=560 m³.

Answer: 560