Assertion (A): \[{\text{1 faraday = 96,500 coulomb}}\] . It is a charge of \[{\text{1 mole}}\] electrons.
Reason (R): \[{\text{1 faraday}}\] charge liberates one-gram equivalent of substance at an electrode.
A) Both (R) and (A) are true and reason is the correct explanation of assertion
B) Both (R) and (A) are true but reason is not correct explanation of assertion
C) Assertion (A) is true but reason (R) is false
D) Assertion (A) and reason (R) both are false
E) Assertion (A) is false but reason (R) is true
Answer
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Hint: There are two laws of faraday for the electrolysis. Out of these two laws, one law is applicable to the assertion and the reason. Faraday gives the charge on a fixed amount of electrons.
Complete step-by-step answer:
According to the faraday’s first law of electrolysis, the mass of ions formed or discharged at electrode during electrolysis is directly proportional to the amount of electricity passed.
\[
\Rightarrow {\text{n = }}\dfrac{{{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right){\text{ }}}}{{\text{F}}} \\
\Rightarrow {\text{N = }}\dfrac{{{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right){\text{ }}}}{{\text{F}}} \times {\text{ mole ratio}} \\
\Rightarrow {\text{W = }}\dfrac{{{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right){\text{ }}}}{{\text{F}}} \times {\text{ mole ratio }} \times {\text{ molecular weight}} \\
\Rightarrow {\text{W = }}\dfrac{{{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right){\text{ }}}}{{\text{F}}} \times {\text{ mole ratio }} \times {\text{ equivalent weight}} \\
\]
Here, n represents the number of moles of electrons passed, N represents the number of moles of substance formed or deposited, W is the mass of substance formed or deposited, \[{\text{I}}\left( {\text{A}} \right)\] is the amount of current passed in amperes, \[{\text{t}}\left( {\text{s}} \right)\] is the time in seconds for which the current is passed and \[{\text{F}}\] is faraday’s constant. The mole ratio gives the ratio of the number of moles of the substance present in the half cell reaction to the number of moles of electrons that participate in the half cell reaction.
Here, \[{\text{1 faraday}}\] represents the charge on one mole of electrons. It is equal to \[{\text{96,500 coulomb}}\] .
When you pass \[{\text{1 faraday}}\]of charge, you will obtain one gram equivalent of substance at an electrode.
This is because when \[{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right) = {\text{F}}\]
Then
\[
\Rightarrow {\text{W = }}\dfrac{{{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right){\text{ }}}}{{\text{F}}} \times {\text{ mole ratio }} \times {\text{ equivalent weight}} \\
\Rightarrow {\text{W = }}\dfrac{{\text{F}}}{{\text{F}}} \times {\text{ mole ratio }} \times {\text{ equivalent weight}} \\
\Rightarrow {\text{W = mole ratio }} \times {\text{ equivalent weight}} \\
\]
Thus, both (R) and (A) are true and reason is the correct explanation of assertion.
Hence, the correct answer is the option (A).
Note: According to faraday’s second law of electrolysis, the amount of substance deposited is proportional to the equivalent weight, when the same amount of electric current is passed through different electrolytes. For example, the equivalent weights of silver and aluminium are 108 and 9 respectively. When one mole of electricity is passed through two separate electrolytes containing silver and aluminium, the number of moles of silver and aluminium deposited are \[108{\text{ g}}\] and \[9{\text{ g}}\] respectively.
Complete step-by-step answer:
According to the faraday’s first law of electrolysis, the mass of ions formed or discharged at electrode during electrolysis is directly proportional to the amount of electricity passed.
\[
\Rightarrow {\text{n = }}\dfrac{{{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right){\text{ }}}}{{\text{F}}} \\
\Rightarrow {\text{N = }}\dfrac{{{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right){\text{ }}}}{{\text{F}}} \times {\text{ mole ratio}} \\
\Rightarrow {\text{W = }}\dfrac{{{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right){\text{ }}}}{{\text{F}}} \times {\text{ mole ratio }} \times {\text{ molecular weight}} \\
\Rightarrow {\text{W = }}\dfrac{{{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right){\text{ }}}}{{\text{F}}} \times {\text{ mole ratio }} \times {\text{ equivalent weight}} \\
\]
Here, n represents the number of moles of electrons passed, N represents the number of moles of substance formed or deposited, W is the mass of substance formed or deposited, \[{\text{I}}\left( {\text{A}} \right)\] is the amount of current passed in amperes, \[{\text{t}}\left( {\text{s}} \right)\] is the time in seconds for which the current is passed and \[{\text{F}}\] is faraday’s constant. The mole ratio gives the ratio of the number of moles of the substance present in the half cell reaction to the number of moles of electrons that participate in the half cell reaction.
Here, \[{\text{1 faraday}}\] represents the charge on one mole of electrons. It is equal to \[{\text{96,500 coulomb}}\] .
When you pass \[{\text{1 faraday}}\]of charge, you will obtain one gram equivalent of substance at an electrode.
This is because when \[{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right) = {\text{F}}\]
Then
\[
\Rightarrow {\text{W = }}\dfrac{{{\text{I}}\left( {\text{A}} \right){\text{ }} \times {\text{ t}}\left( {\text{s}} \right){\text{ }}}}{{\text{F}}} \times {\text{ mole ratio }} \times {\text{ equivalent weight}} \\
\Rightarrow {\text{W = }}\dfrac{{\text{F}}}{{\text{F}}} \times {\text{ mole ratio }} \times {\text{ equivalent weight}} \\
\Rightarrow {\text{W = mole ratio }} \times {\text{ equivalent weight}} \\
\]
Thus, both (R) and (A) are true and reason is the correct explanation of assertion.
Hence, the correct answer is the option (A).
Note: According to faraday’s second law of electrolysis, the amount of substance deposited is proportional to the equivalent weight, when the same amount of electric current is passed through different electrolytes. For example, the equivalent weights of silver and aluminium are 108 and 9 respectively. When one mole of electricity is passed through two separate electrolytes containing silver and aluminium, the number of moles of silver and aluminium deposited are \[108{\text{ g}}\] and \[9{\text{ g}}\] respectively.
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