The overall reaction of a hydrogen-oxygen fuel cell is:

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

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

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

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

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Solution

The correct option is A
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$

In general, the chemical energy (heat of combustion) of fossil fuels (coal, gas, or oil) is first used for converting water into high pressure steam and then used to run a turbine to produce electricity.

In fuel cell, the chemical energy of combustion of fuels like $H_{2}, C H_{4}, C H_{3} O H$ etc. are directly coverted into electrical energy.
Fuel cell is a galvanic cell.

In hydrogen-oxygen fuel cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution. Here, finely divided platinum or palladium metal is used as catalyst.

At the anode:
$H_{2} (g) + 2 O H^{-} (a q) \to 2 H_{2} O (l) + 2 e^{-}$

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

Overall reaction is:
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$

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

Δ G_{f}^{o} (H_{2} O) = - 237.2 k J {m o l}^{- 1}

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