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:
Answer & explanation
Correct answer: option 2
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