The standard cell potential (E_ cell ^\ominus) 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 O_2(E_O_2 / H_2 O^\circ) is 1.229 V . Choose the correct statement :

The standard half cell reduction potential for the reduction of CO_2(E_CO_2 / CH_3 OH^\circ) is 19 mV

Reactants are fed at one go to each electrode.

Reduction of methanol takes place at the cathode.

Electrochemistry PYQ — JEE Main 2025 · 3 Apr · Shift 2 · Q16 with Solution

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Correct answer: C

Methanol is oxidized by oxygen in air:

$\mathrm{CH_3OH + \frac{3}{2}O_2 \rightarrow CO_2 + 2H_2O}$

So:

Hence immediately:

Option B is false because oxygen is not formed at the anode.
Option D is false because methanol does not get reduced at the cathode; it gets oxidized at the anode.

For a galvanic/fuel cell,

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

Given:

$E^\circ_{\text{cell}} = 1.21\,\text{V}$

and for oxygen reduction,

$E^\circ_{\mathrm{O_2/H_2O}} = 1.229\,\text{V}$

Since oxygen is reduced at the cathode,

$1.21 = 1.229 - E^\circ_{\mathrm{CO_2/CH_3OH}}$

Therefore,

$E^\circ_{\mathrm{CO_2/CH_3OH}} = 1.229 - 1.21 = 0.019\,\text{V}$

So,

$E^\circ_{\mathrm{CO_2/CH_3OH}} = 19\,\text{mV}$

Thus Option C is correct.

A fuel cell works by continuous supply of reactants to the electrodes; reactants are not fed "at one go" as in an ordinary cell.

So Option A is false.

Therefore, the correct answer is:

$\boxed{\text{C}}$