
A factory produces \[40Kg\] of calcium in two hours by electrolysis. How much aluminium can be produced by the same current in 2 hours if current efficiency is \[50\% \] ?
A. \[22Kg\]
B. \[18Kg\]
C. \[9Kg\]
D. \[27Kg\]
Answer
562.5k+ views
Hint: Faraday’ laws of electrolysis will help to get the answer. You must have detailed knowledge of electrolysis. Electrolysis is the process of separating pure elements from its impure form.
Complete step by step answer:
To start with, let’s start with revising about electrolysis.
Electrolysis is the process in which direct electric current (DC) to drive an otherwise non-spontaneous reaction is used to separate elements from its impure form like ores using an electrolytic cell.
The above question can be solved by using the Faraday’s law of electrolysis, especially the second law of electrolysis. According to this law, “When the same quantity of electricity passes through solutions of different electrolytes, the amounts of the substances liberated at the electrodes are directly proportional to their chemical equivalents.” The quantity needed to liberate one gram equivalent of a substance is known as ‘Faraday’ F.
According to second law of Faraday:
\[\dfrac{{Mass \;of\; Calcium}}{{Mass\; of\; Aluminium}} = \dfrac{{Equi.mass\;of\;Calcium}}{{Equi.mass\;of\;Aluminium}}\]
\[\dfrac{{{W_{Ca}}}}{{{E_{Ca}}}} = \dfrac{{{W_{Al}}}}{{{E_{Al}}}}\]
\[\dfrac{{40}}{{20}} = \dfrac{{{W_{Al}}}}{9}\]
${{W_{Al}}}$ = 18 Kg
As current efficiency is \[50\% \] so \[{W_{Al}} = 9Kg\] .
Hence, the correct option is C.
Additional Information:
Michael Faraday in 1833 wrote the laws of electrolysis for quantitative relationships. He wrote two laws. The first law is that the mass ( \[m\] ) of elements deposited at an electrode in g is directly proportional to the charge \[Q\] in Columbus and the constant of proportionality \[Z\] is called Electro Chemical Equivalent of the substance. The second law is as stated above in the answer. These two laws were written by Faraday to ease the calculations.
Note:
Faraday’s laws of electrolysis helps in solving many questions but we should have knowledge on where to apply which formula.\[\]
Complete step by step answer:
To start with, let’s start with revising about electrolysis.
Electrolysis is the process in which direct electric current (DC) to drive an otherwise non-spontaneous reaction is used to separate elements from its impure form like ores using an electrolytic cell.
The above question can be solved by using the Faraday’s law of electrolysis, especially the second law of electrolysis. According to this law, “When the same quantity of electricity passes through solutions of different electrolytes, the amounts of the substances liberated at the electrodes are directly proportional to their chemical equivalents.” The quantity needed to liberate one gram equivalent of a substance is known as ‘Faraday’ F.
According to second law of Faraday:
\[\dfrac{{Mass \;of\; Calcium}}{{Mass\; of\; Aluminium}} = \dfrac{{Equi.mass\;of\;Calcium}}{{Equi.mass\;of\;Aluminium}}\]
\[\dfrac{{{W_{Ca}}}}{{{E_{Ca}}}} = \dfrac{{{W_{Al}}}}{{{E_{Al}}}}\]
\[\dfrac{{40}}{{20}} = \dfrac{{{W_{Al}}}}{9}\]
${{W_{Al}}}$ = 18 Kg
As current efficiency is \[50\% \] so \[{W_{Al}} = 9Kg\] .
Hence, the correct option is C.
Additional Information:
Michael Faraday in 1833 wrote the laws of electrolysis for quantitative relationships. He wrote two laws. The first law is that the mass ( \[m\] ) of elements deposited at an electrode in g is directly proportional to the charge \[Q\] in Columbus and the constant of proportionality \[Z\] is called Electro Chemical Equivalent of the substance. The second law is as stated above in the answer. These two laws were written by Faraday to ease the calculations.
Note:
Faraday’s laws of electrolysis helps in solving many questions but we should have knowledge on where to apply which formula.\[\]
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