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

2 H_{2 (g)} + O_{2 (g)} ⟶ 2 H_{2} O_{(l)}

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O_{2 (g)} + 2 H_{2} O_{(l)} + 4 e^{-} ⟶ 4 O H_{(a q)}^{-}

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H^{+} + e^{-} ⟶ \frac{1}{2} H_{2}

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H_{(a q)}^{+} + O H_{(a q)}^{-} ⟶ H_{2} O_{(l)}

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Solution

The correct option is B $O_{2 (g)} + 2 H_{2} O_{(l)} + 4 e^{-} ⟶ 4 O H_{(a q)}^{-}$
In a fuel cell,
Cathode reaction can be written as:

$O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)$

Anode reaction can be written as:

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

Net cell reaction:

$2 H_{2} + O_{2} \to 2 H_{2} O$

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Similar questions

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

H_{2} - O_{2}

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

J

4.5 g

H_{2} O_{2}

1 a t m

25^{o} C

2 H_{2} O_{2} (l) ⟶ O_{2} (g) + H_{2} O (l)

2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(l)};

Δ H = - 571.6 k J

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