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 .

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<p>To determine the reaction potentials at the anode and cathode in the methanol fuel cell, we start with the given standard cell potential:</p> <p>$ E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} $</p> <p>Given:</p> <p><p>$ E_{\text{cell}}^{\circ} = 1.21 \, \text{V} $</p></p> <p><p>$ E_{\text{O}_2/\text{H}_2\text{O}}^{\circ} = 1.229 \, \text{V} $</p></p> <p>Substituting the values,</p> <p>$ 1.21 = 1.229 - E_{\text{anode}}^{\circ} $</p> <p>Solving for $ E_{\text{anode}}^{\circ} $:</p> <p>$ E_{\text{anode}}^{\circ} = 1.229 - 1.21 = 0.019 \, \text{V} $</p> <p>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:</p> <p><p><strong>Anode Reaction:</strong> Oxidation of methanol occurs here, contributing to the fuel cell’s electrical output.</p></p> <p><p><strong>Cathode Reaction:</strong> Reduction of oxygen occurs at the cathode, where the $ E_{\text{O}_2/\text{H}_2\text{O}}^{\circ} $ potential is utilized.</p></p> <p>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.</p>