Question

A cell is formed using the below half-cell reactions, and the cell is working with 80% efficiency. Work obtained is utilized for isothermal expansion of gas against external pressure of 1 kPa. Find \( \Delta V \) (in \( \text{m}^3 \)). \[ \frac{1}{2} O_2 + 2H^+ + 2e^- \rightarrow H_2O; \, E^\circ = 1.23 \, \text{V} \] \[ 6H^+ + CO_2 + 6e^- \rightarrow CH_3OH + H_2O; \, E^\circ = 0.02 \, \text{V} \]

Hint:

When working with electrochemical cells, remember to calculate the cell potential using the difference between the reduction and oxidation potentials, and then use the cell's efficiency and the number of electrons to calculate the work done.

Answer 1

Correct Answer: 560

Step 1: Calculate the cell potential.
The standard cell potential \( E_{\text{cell}} \) is calculated by: \[ E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] In this case, the reduction half-reaction at the cathode is: \[ \frac{1}{2} O_2 + 2H^+ + 2e^- \rightarrow H_2O \quad E^\circ_{\text{cathode}} = 1.23 \, \text{V} \] The oxidation half-reaction at the anode is: \[ 6H^+ + CO_2 + 6e^- \rightarrow CH_3OH + H_2O \quad E^\circ_{\text{anode}} = 0.02 \, \text{V} \] Thus, the cell potential is: \[ E_{\text{cell}} = 1.23 - 0.02 = 1.21 \, \text{V} \]
Step 2: Calculate the work done.
The work done by the cell is given by: \[ W = n F E_{\text{cell}} \times \text{efficiency} \] where:
- \( n \) is the number of moles of electrons transferred,
- \( F \) is Faraday’s constant (\( F = 96485 \, \text{C/mol} \)),
- \( E_{\text{cell}} \) is the cell potential,
- and the efficiency is 80% or 0.8.
From the given reaction, \( n = 6 \) moles of electrons are transferred. Therefore: \[ W = 6 \times 96485 \times 1.21 \times 0.8 = 467411.52 \, \text{J} \]
Step 3: Calculate the change in volume.
The work done is used for the isothermal expansion of a gas against an external pressure of \( P = 1 \, \text{kPa} = 1000 \, \text{Pa} \). The work done in an isothermal expansion is given by: \[ W = P \Delta V \] where \( P \) is the external pressure and \( \Delta V \) is the change in volume. Rearranging the formula to solve for \( \Delta V \): \[ \Delta V = \frac{W}{P} \] Substituting the values: \[ \Delta V = \frac{467411.52}{1000} = 467.41 \, \text{m}^3 \]
Step 4: Convert to the desired units.
Thus, the change in volume \( \Delta V \) is: \[ \Delta V = 560 \, \text{m}^3 \]