Question

Consider the following two half-cell reactions along with the standard reduction potential given: \[ \text{CO}_2 + 6H^+ + 6e^- \rightarrow \text{CH}_3\text{OH} + \text{H}_2\text{O} \quad E^\circ_{\text{red}} = 0.02 \, \text{V} \] \[ \frac{1}{2} \text{O}_2 + 2H^+ + 2e^- \rightarrow \text{H}_2\text{O} \quad E^\circ_{\text{red}} = 1.23 \, \text{V} \] A fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 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).}

Answer 1

Correct Answer: 561

Step 1: Calculate the cell potential.
The standard cell potential \( E^\circ_{\text{cell}} \) is given by the difference in the standard reduction potentials of the cathode and anode: \[ E^\circ_{\text{cell}} = E^\circ_{\text{red, cathode}} - E^\circ_{\text{red, anode}} \] Substitute the given values: \[ E^\circ_{\text{cell}} = 1.23 \, \text{V} - 0.02 \, \text{V} = 1.21 \, \text{V} \]
Step 2: Calculate the work done by the cell.
The work done by the cell can be calculated using the equation: \[ W = n F E^\circ_{\text{cell}} \] where: - \( n = 6 \) (the number of moles of electrons), - \( F = 96500 \, \text{C/mol} \) (Faraday constant), - \( E^\circ_{\text{cell}} = 1.21 \, \text{V} \). \[ W = 6 \times 96500 \times 1.21 = 7.01 \times 10^5 \, \text{J} \]
Step 3: Calculate the work done with 80% efficiency.
The actual work done is 80% of the ideal work: \[ W_{\text{actual}} = 0.80 \times 7.01 \times 10^5 = 5.61 \times 10^5 \, \text{J} \]
Step 4: Calculate the change in volume of the gas.
The work done to compress an ideal gas isothermally is given by: \[ W = P \Delta V \] where \( P = 1 \, \text{kPa} = 10^3 \, \text{Pa} \). Solving for \( \Delta V \): \[ \Delta V = \frac{W}{P} = \frac{5.61 \times 10^5}{10^3} = 561 \, \text{m}^3 \]