JEE Main 2025 — Electrochemistry Question with Solution

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.