The standard cell potential (E_{text{cell}}^{ | vedclass

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Question

(A) The cell reaction involves the oxidation of methanol at the anode and the reduction of oxygen at the cathode. Given: $E_{	ext{cell}}^{\ominus} =

Vedclass · Accepted Answer

The standard half-cell reduction potential for the reduction of $CO_2$ $(E_{CO_2 / CH_3OH}^0)$ is $19 \ mV$

Vedclass · Answer

(A) The cell reaction involves the oxidation of methanol at the anode and the reduction of oxygen at the cathode. Given: $E_{	ext{cell}}^{\ominus} = 1.21 \ V$ and $E_{	ext{cathode}}^{\ominus} (O_2/H_2O) = 1.229 \ V$. Using the formula: $E_{	ext{cell}}^{\ominus} = E_{	ext{cathode}}^{\ominus} - E_{	ext{anode}}^{\ominus}$. $1.21 \ V = 1.229 \ V - E_{	ext{anode}}^{\ominus}$. $E_{	ext{anode}}^{\ominus} = 1.229 \ V - 1.21 \ V = 0.019 \ V = 19 \ mV$. Since the anode reaction is the oxidation of methanol $(CH_3OH + H_2O ightarrow CO_2 + 6H^+ + 6e^-)$,the reduction potential of the $CO_2/CH_3OH$ couple is $19 \ mV$. Thus,option $A$ is correct.

$LIST$-$I$ (Cell)	$LIST$-$II$ (Use/Property/Reaction)
$A$. Leclanche cell	$I$. Converts energy of combustion into electrical energy
$B$. Ni-Cd cell	$II$. Does not involve any ion in solution and is used in hearing aids
$C$. Fuel cell	$III$. Rechargeable
$D$. Mercury cell	$IV$. Reaction at anode $Zn \rightarrow Zn^{2+} + 2e^-$

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