Q23 / 75 · JEE Main 2026 (02 April Shift 2) · Chemistry
Consider the following two half-cell reactions along with the standard reduction potential given:
$CO_2 + 6H^+ + 6e^- \rightarrow CH_3 OH + H_2 O \quad E°_{red} = 0.02$ V
$\dfrac{1}{2} O_2 + 2H^+ + 2e^- \rightarrow H_2 O \quad E°_{red} = 1.23$ 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 $CH_3 OH$ is used to compress an ideal gas isothermally against a constant pressure of $1$ kPa, then the change in the volume of the gas, $\Delta V = $ _____ m$^3$. (nearest integer)
Given: $F = 96500$ C mol$^{-1}$
$CO_2 + 6H^+ + 6e^- \rightarrow CH_3 OH + H_2 O \quad E°_{red} = 0.02$ V
$\dfrac{1}{2} O_2 + 2H^+ + 2e^- \rightarrow H_2 O \quad E°_{red} = 1.23$ 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 $CH_3 OH$ is used to compress an ideal gas isothermally against a constant pressure of $1$ kPa, then the change in the volume of the gas, $\Delta V = $ _____ m$^3$. (nearest integer)
Given: $F = 96500$ C mol$^{-1}$
Correct answer: -560
Solution
The half-cell reactions for the fuel cell are:
Anode (Oxidation): $CH_3OH + H_2O \rightarrow CO_2 + 6H^+ + 6e^- \quad E^{\circ}_{ox} = -0.02$ V
Cathode (Reduction): $\dfrac{3}{2}O_2 + 6H^+ + 6e^- \rightarrow 3H_2O \quad E^{\circ}_{red} = 1.23$ V
The overall cell reaction is:
$CH_3OH + \dfrac{3}{2}O_2 \rightarrow CO_2 + 2H_2O$
The standard cell potential is:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.23 - 0.02 = 1.21$ V
The number of electrons transferred per mole of $CH_3OH$ is $n = 6$.
The maximum electrical work that can be obtained from the cell is given by the decrease in Gibbs free energy:
$W_{max} = nFE^{\circ}_{cell} = 6 \times 96500 \times 1.21 = 700590$ J
Since the fuel cell operates at $80\%$ efficiency, the actual work derived is:
$W_{actual} = 0.80 \times 700590 = 560472$ J
This work is used to compress an ideal gas isothermally against a constant external pressure of $1$ kPa ($1000$ Pa). The work done on the gas during compression is positive:
$W = -P_{ext} \Delta V$
Substituting the values:
$560472 = -1000 \times \Delta V$
$\Delta V = -560.472$ m$^3$
Rounding to the nearest integer, the change in volume is $-560$ m$^3$.
Answer: $-560$
Anode (Oxidation): $CH_3OH + H_2O \rightarrow CO_2 + 6H^+ + 6e^- \quad E^{\circ}_{ox} = -0.02$ V
Cathode (Reduction): $\dfrac{3}{2}O_2 + 6H^+ + 6e^- \rightarrow 3H_2O \quad E^{\circ}_{red} = 1.23$ V
The overall cell reaction is:
$CH_3OH + \dfrac{3}{2}O_2 \rightarrow CO_2 + 2H_2O$
The standard cell potential is:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.23 - 0.02 = 1.21$ V
The number of electrons transferred per mole of $CH_3OH$ is $n = 6$.
The maximum electrical work that can be obtained from the cell is given by the decrease in Gibbs free energy:
$W_{max} = nFE^{\circ}_{cell} = 6 \times 96500 \times 1.21 = 700590$ J
Since the fuel cell operates at $80\%$ efficiency, the actual work derived is:
$W_{actual} = 0.80 \times 700590 = 560472$ J
This work is used to compress an ideal gas isothermally against a constant external pressure of $1$ kPa ($1000$ Pa). The work done on the gas during compression is positive:
$W = -P_{ext} \Delta V$
Substituting the values:
$560472 = -1000 \times \Delta V$
$\Delta V = -560.472$ m$^3$
Rounding to the nearest integer, the change in volume is $-560$ m$^3$.
Answer: $-560$
Hence, the correct answer is -560.