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Calculate Volume Change of Ideal Gas Compressed by Fuel Cell Work

Consider the following two half-cell reactions along with the standard reduction potential given : CO2+6H++6e−⟶CH3OH+H2OEred0=0.02 V\text{CO}_2 + 6\text{H}^+ + 6e^- \longrightarrow \text{CH}_3\text{OH} + \text{H}_2\text{O} \quad \text{E}^0_{\text{red}} = 0.02\text{ V} 12O2+2H++2e−⟶H2OEred0=1.23 V\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \longrightarrow \text{H}_2\text{O} \quad \text{E}^0_{\text{red}} = 1.23\text{ V} A fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar1\text{ bar} pressure and 298 K298\text{ K} temperature. The fuel cell works with 80%80\% efficiency. If the work derived from the cell using 1 mol1\text{ mol} of CH3OH\text{CH}_3\text{OH} is used to compress an ideal gas isothermally against a constant pressure of 1 kPa1\text{ kPa}, then the change in the volume of the gas, ΔV=\Delta V = ______ m3\text{m}^3. (nearest integer) Given : F=96500 C mol−1\text{F} = 96500\text{ C mol}^{-1}

Official Numerical Answer560

Topics & Concepts

ElectrochemistryElectrochemical CellNernst EquationFirst Law of Thermodynamics

Step-by-Step Solution

To find the change in volume of the gas, we first determine the standard cell potential (Ecell∘E^\circ_{\text{cell}}) for the fuel cell operating with methanol (CH3OH\text{CH}_3\text{OH}) as the fuel.

The two half-cell reduction reactions are given as:

  1. CO2+6H++6e−⟶CH3OH+H2OE1∘=0.02 V\text{CO}_2 + 6\text{H}^+ + 6e^- \longrightarrow \text{CH}_3\text{OH} + \text{H}_2\text{O} \quad E^\circ_1 = 0.02\text{ V}
  2. 12O2+2H++2e−⟶H2OE2∘=1.23 V\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \longrightarrow \text{H}_2\text{O} \quad E^\circ_2 = 1.23\text{ V}

In the fuel cell, CH3OH\text{CH}_3\text{OH} undergoes oxidation at the anode, while O2\text{O}_2 is reduced at the cathode.

1. Calculate the standard cell potential (Ecell∘E^\circ_{\text{cell}}): Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} Ecell∘=1.23 V−0.02 V=1.21 VE^\circ_{\text{cell}} = 1.23\text{ V} - 0.02\text{ V} = 1.21\text{ V}

2. Calculate the maximum theoretical work (WmaxW_{\text{max}}) for 1 mol1\text{ mol} of CH3OH\text{CH}_3\text{OH}: From the half-reaction for the oxidation of 1 mol1\text{ mol} of CH3OH\text{CH}_3\text{OH}, the number of moles of electrons transferred is n=6n = 6.

Wmax=−ΔG∘=nFEcell∘W_{\text{max}} = -\Delta G^\circ = n F E^\circ_{\text{cell}} Wmax=6×96500 C mol−1×1.21 VW_{\text{max}} = 6 \times 96500\text{ C mol}^{-1} \times 1.21\text{ V} Wmax=700,590 JW_{\text{max}} = 700,590\text{ J}

3. Calculate the actual work derived from the fuel cell: Given that the cell operates with an efficiency of 80%80\%: Wderived=Efficiency×WmaxW_{\text{derived}} = \text{Efficiency} \times W_{\text{max}} Wderived=0.80×700,590 J=560,472 JW_{\text{derived}} = 0.80 \times 700,590\text{ J} = 560,472\text{ J}

4. Calculate the change in volume (ΔV\Delta V) of the ideal gas: The work derived is used to compress the gas against a constant external pressure (PextP_{\text{ext}}) of 1 kPa=1000 Pa1\text{ kPa} = 1000\text{ Pa}.

The magnitude of work required to compress the gas against constant pressure is: Wderived=Pext×∣ΔV∣W_{\text{derived}} = P_{\text{ext}} \times |\Delta V| 560,472 J=1000 Pa×∣ΔV∣560,472\text{ J} = 1000\text{ Pa} \times |\Delta V| ∣ΔV∣=560,4721000 m3=560.472 m3|\Delta V| = \frac{560,472}{1000}\text{ m}^3 = 560.472\text{ m}^3

Rounding off to the nearest integer gives: ΔV≈560 m3\Delta V \approx 560\text{ m}^3

Calculate Volume Change of Ideal Gas Compressed by Fuel Cell Work | Chemistry PYQ Solution - JEE Challenger