JEE MAIN - Chemistry (2025 - 3rd April Evening Shift - No. 16)
The standard cell potential $\left(\mathrm{E}_{\text {cell }}^{\ominus}\right)$ of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as 1.21 V . The standard half cell reduction potential for $\mathrm{O}_2\left(\mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\circ}\right)$ is 1.229 V .
Choose the correct statement :
Explanation
To determine the reaction potentials at the anode and cathode in the methanol fuel cell, we start with the given standard cell potential:
$ E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} $
Given:
$ E_{\text{cell}}^{\circ} = 1.21 \, \text{V} $
$ E_{\text{O}_2/\text{H}_2\text{O}}^{\circ} = 1.229 \, \text{V} $
Substituting the values,
$ 1.21 = 1.229 - E_{\text{anode}}^{\circ} $
Solving for $ E_{\text{anode}}^{\circ} $:
$ E_{\text{anode}}^{\circ} = 1.229 - 1.21 = 0.019 \, \text{V} $
Since this is a fuel cell based on the oxidation of methanol, methanol undergoes oxidation at the anode, while oxygen is reduced at the cathode. Therefore:
Anode Reaction: Oxidation of methanol occurs here, contributing to the fuel cell’s electrical output.
Cathode Reaction: Reduction of oxygen occurs at the cathode, where the $ E_{\text{O}_2/\text{H}_2\text{O}}^{\circ} $ potential is utilized.
Thus, we verify that the half-cell reduction potential for oxidation of methanol by CO$_2$ is indeed 19 mV, aligning with the calculated $ E_{\text{anode}}^{\circ} $. The processes correctly identify the roles of anode and cathode in this fuel cell.
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