Read the following passage and answer the next five questions based on it:

Battery or cell converts chemical energy of the redox reaction to electrical energy. In fuel cell (a galvanic cell), the chemical energy of combustion of fuels like \(H_2\), ethanol, etc, are directly converted to electrical energy. In a fuel cell, \(H_2\) and \(O_2\) react to produce electricity, where \(H_2\) gas is oxidized at anode and oxygen is reduced at cathode and the reactions involved are:

Anode Reaction :\(H_2 + 2OH^- \longrightarrow 2H_2O + 2e^-\)

Cathode reaction: \(O_2 + 2H_2O + 4e^- \longrightarrow 4OH^-\)

\(67.2 L\) of \(H_2\) at STP reacts in \(15 \) minutes

The quantity of electricity produced in the oxidation of \(67.2\, \ L\) of \(H_2\) at STP is:

579000C

The correct answer is option 2. 579000C.

To determine the quantity of electricity produced in the oxidation of \(67.2 \, \text{L}\) of \(H_2\) at STP, follow these steps:

1. Calculate the number of moles of \(H_2\):

At STP, 1 mole of any ideal gas occupies \(22.4 \, \text{L}\).

\(\text{Number of moles of } H_2 = \frac{67.2 \, \text{L}}{22.4 \, \text{L/mol}} = 3.0 \, \text{moles}\)

2.Determine the moles of electrons produced:

According to the anode reaction, each mole of \(H_2\) produces 2 moles of electrons:

\(H_2 + 2OH^- \longrightarrow 2H_2O + 2e^-\)

Therefore, 3 moles of \(H_2\) will produce:

\(3 \, \text{moles of } H_2 \times 2 \, \text{moles of electrons/mole of } H_2 = 6 \, \text{moles of electrons}\)

3. Calculate the quantity of electricity:

The charge of 1 mole of electrons (1 Faraday) is \(96500 \, \text{C}\).

\(\text{Total charge} = 6 \, \text{moles of electrons} \times 96500 \, \text{C/mole of electrons}\)

\(\text{Total charge} = 579000 \, \text{C}\)

Therefore, the quantity of electricity produced in the oxidation of \(67.2 \, \text{L}\) of \(H_2\) at STP is: 2. 579000 C