Read the following passage and answer the next five questions based on it.(32-36)

Battery or cell converts chemical energy of the redox reaction to electrical energy. In fuel cells (a galvanic cell), the chemical energy of combustion of fuels like H2, ethanol, etc., is directly converted to electrical energy. In a fuel cell, H2 and O2 react to produce electricity, where H2
gas is oxidized at the anode and oxygen is reduced at the cathode, and the reactions involved are:

Anode reaction:H₂ + 2OH⁻ → 2H₂O+2e⁻
Cathode reaction:O₂ + 2H₂O +4e⁻ → 4OH⁻

67.2 L of H2 at STP reacts in 15 minutes.

The quantity of electricity produced in the oxidation of 67.2 L of H2 at STP is:

The correct option is 579000 C.

First, we calculate the number of moles of hydrogen gas ($H_2$) reacting at STP (Standard Temperature and Pressure).
At STP, 1 mole of any ideal gas occupies a volume of 22.4 liters (L).
Using this relationship, the number of moles of hydrogen gas is:
$\text{Moles of}{H}_2=\frac{\text{Given Volume}}{\text{Molar Volume at STP}}$
$\text{Moles of}{H}_2=\frac{67.2\text{L}}{22.4\text{L/mol}}=3\text{moles}$

Next, we determine the number of moles of electrons produced from the oxidation reaction of hydrogen gas.
The half-reaction at the anode is given as:
$H_2+2{\text{OH}}^{-}\to 2H_2\text{O}+2{\text{e}}^{-}$
From this balanced equation, we can observe that 1 mole of $H_2$ gas releases 2 moles of electrons during oxidation.
Therefore, for 3 moles of $H_2$, the total number of moles of electrons ($n$) released is:
$n=3\times 2=6\text{moles of electrons}$

Finally, we calculate the quantity of electricity (electric charge, $Q$) produced.
The charge of 1 mole of electrons is equivalent to 1 Faraday ($F$), which is approximately equal to 96500 Coulombs (C):
$1\text{F}=96500\text{C/mol}$
Using the formula:
$Q=n\times F$
$Q=6\text{moles}\times 96500\text{C/mol}$
$Q=579000\text{C}$

Thus, the quantity of electricity produced in the oxidation of 67.2 L of hydrogen gas at STP is 579000 C.