In $H_{2} - O_{2}$ fuel cell the reaction occuring at cathode is:

2 H_{2} O + O_{2} + 4 e^{-} \to 4 O H^{-}

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2 H_{2} + O_{2} \to 2 H_{2} O (l)

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H^{+} + O H^{-} \to H_{2} O

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H^{+} + e^{-} \to \frac{1}{2} H_{2}

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Solution

The correct option is A $2 H_{2} O + O_{2} + 4 e^{-} \to 4 O H^{-}$
In $H_{2} - O_{2}$ fuel cell, hydrogen is oxidised at the anode and oxygen is reduced at the cathode. The half cell reactions are
At anode: $2 O H^{-} + H_{2} \to 2 H_{2} O + 2 e^{-}$
At cathode : $2 H_{2} O + O_{2} + 4 e^{-} \to 4 O H^{-}$

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Similar questions

In H₂ - O₂ fuel cell the reaction occuring at cathode is :

H_{2} - O_{2}

H_{2} - O_{2}

In $H_{2} - O_{2}$ fuel cell the reaction occuring at cathode is:

A fuel cell is:

I.A voltaic cell in which continuous supply of fuels are sent at anode to perform oxidation.

II. A voltaic cell in which fuels such as :

$C H_{4}, H_{2}$ and CO are used up at anode.

III. One which involves the reaction of $H_{2} O_{2}$ fuel cell such as:

Anode : $2 H_{2} + 4 ⊖ O H \to 4 H_{2} O (l) + 4 e^{-}$

Cathode : $O_{2} + 2 H_{2} O (l) + 4 e^{-} \to 4 ⊖ O H$

IV. The efficiency of $H_{2} - O_{2}$ fuel cell is $70$ to $75 %$ .

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