JEE Main 2026 — Chemistry Physical Chemistry

The half-cell reactions for the fuel cell are:

Anode (Oxidation): $CH_3OH + H_2O \rightarrow CO_2 + 6H^+ + 6e^- \quad E^{\circ}_{ox} = -0.02$CH3​OH+H2​O→CO2​+6H++6e−Eox∘​=−0.02 V

Cathode (Reduction): $\dfrac{3}{2}O_2 + 6H^+ + 6e^- \rightarrow 3H_2O \quad E^{\circ}_{red} = 1.23$23​O2​+6H++6e−→3H2​OEred∘​=1.23 V

The overall cell reaction is:

$$CH_3OH + \dfrac{3}{2}O_2 \rightarrow CO_2 + 2H_2O$$CH3​OH+23​O2​→CO2​+2H2​O

The standard cell potential is:

$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.23 - 0.02 = 1.21$Ecell∘​=Ecathode∘​−Eanode∘​=1.23−0.02=1.21 V

The number of electrons transferred per mole of $CH_3OH$CH3​OH is $n = 6$n=6.

The maximum electrical work that can be obtained from the cell is given by the decrease in Gibbs free energy:

$W_{max} = nFE^{\circ}_{cell} = 6 \times 96500 \times 1.21 = 700590$Wmax​=nFEcell∘​=6×96500×1.21=700590 J

Since the fuel cell operates at $80\%$80% efficiency, the actual work derived is:

$W_{actual} = 0.80 \times 700590 = 560472$Wactual​=0.80×700590=560472 J

This work is used to compress an ideal gas isothermally against a constant external pressure of $1$1 kPa ($1000$1000 Pa). The work done on the gas during compression is positive:

$$W = -P_{ext} \Delta V$$W=−Pext​ΔV

Substituting the values:

$$560472 = -1000 \times \Delta V$$560472=−1000×ΔV

$\Delta V = -560.472$ΔV=−560.472 m$^3$3

Rounding to the nearest integer, the change in volume is $-560$−560 m$^3$3.

Answer: $-560$−560