
A constant current of 30A is passed through an aqueous solution of NaCl for a time of 1 hour. What is the volume of $\mathrm{Cl}_{2}$ gas at STP produced?
A) 30.00 L
B) 25.08 L
C) 12.54 L
D) 1.12 L
Answer
486.6k+ views
Hint: From the question above, it is clear that the flow of current through the external battery circuit fully depends upon how many electrons get transferred from negative electrode or cathode to positive metallic ion or cations. Thus, we will use the first law of Faraday. Q = I x t (t in seconds).
Complete step by step answer:
Faraday’s first law of electrolysis:
This law states that the chemical deposition due to the flow of current through an electrolyte which is directly proportional to the quantity of electricity passed through it.
Mass of chemical deposition, $m \propto Q$
$\implies m=ZQ$ -- (1)
Where, Q = quantity of electricity in Coulombs, and
Z = constant proportionality which is known as the electrochemical equivalent of a substance
Faraday’s second law of electrolysis:
This law states that when the same quantity of electricity passes through a series of electrolytes, then the mass of substance deposited is proportional to their respective chemical equivalent weights.
-Now, let us write the equations as given:
$2 \mathrm{NaCl} \stackrel{\text { Electrolysis }}{\longrightarrow} 2 \mathrm{Na}^{+}+2 \mathrm{Cl}^{-}$
At cathode: $2 \mathrm{Na}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Na}$
At anode: $2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2}+2 \mathrm{e}^{-}$
So 2 Faraday of electricity is required to liberate 1 mole = 22.4 liters of $\mathrm{Cl}_{2}$ at STP.
Amount of current passed
$=30 \times 60 \times 60 = 108000$ coulombs
So $108000$ coulombs will liberate $\dfrac{108000}{96500 \times 2}=0.5596$ moles $\mathrm{Cl}_{2}$
No.of liters of Chlorine liberated at STP
$=0.5596 \times 22.4$
$=12.54$ liters
So, the correct answer is “Option C”.
Note: In chemistry, Faraday’s laws of electrolysis are the quantitative laws used to express magnitudes of electrolytic effects were first described by Michael Faraday. These two laws are applicable for the process of electrolysis of metal salts like metal sulfates in an electrolytic cell.
Complete step by step answer:
Faraday’s first law of electrolysis:
This law states that the chemical deposition due to the flow of current through an electrolyte which is directly proportional to the quantity of electricity passed through it.
Mass of chemical deposition, $m \propto Q$
$\implies m=ZQ$ -- (1)
Where, Q = quantity of electricity in Coulombs, and
Z = constant proportionality which is known as the electrochemical equivalent of a substance
Faraday’s second law of electrolysis:
This law states that when the same quantity of electricity passes through a series of electrolytes, then the mass of substance deposited is proportional to their respective chemical equivalent weights.
-Now, let us write the equations as given:
$2 \mathrm{NaCl} \stackrel{\text { Electrolysis }}{\longrightarrow} 2 \mathrm{Na}^{+}+2 \mathrm{Cl}^{-}$
At cathode: $2 \mathrm{Na}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Na}$
At anode: $2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2}+2 \mathrm{e}^{-}$
So 2 Faraday of electricity is required to liberate 1 mole = 22.4 liters of $\mathrm{Cl}_{2}$ at STP.
Amount of current passed
$=30 \times 60 \times 60 = 108000$ coulombs
So $108000$ coulombs will liberate $\dfrac{108000}{96500 \times 2}=0.5596$ moles $\mathrm{Cl}_{2}$
No.of liters of Chlorine liberated at STP
$=0.5596 \times 22.4$
$=12.54$ liters
So, the correct answer is “Option C”.
Note: In chemistry, Faraday’s laws of electrolysis are the quantitative laws used to express magnitudes of electrolytic effects were first described by Michael Faraday. These two laws are applicable for the process of electrolysis of metal salts like metal sulfates in an electrolytic cell.
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