(A) According to Faraday's first law of electrolysis,the mass of substance deposited $(m)$ is given by $m = Z \cdot I \cdot t$,where $Z$ is the electr

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(A) According to Faraday's first law of electrolysis,the mass of substance deposited $(m)$ is given by $m = Z \cdot I \cdot t$,where $Z$ is the electrochemical equivalent,$I$ is the current,and $t$ is the time.In the first case,$m_1 = Z \cdot I \cdot t$.In the second case,the current becomes $I' = \frac{I}{4}$ and the time becomes $t' = 4t$.Substituting these values,$m_2 = Z \cdot (\frac{I}{4}) \cdot (4t) = Z \cdot I \cdot t$.Therefore,$m_2 = m_1$,which means the amount of copper deposited remains the same.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

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In a copper voltameter experiment,current is decreased to one-fourth of the initial value but it is passed for four times the earlier duration. The amount of copper deposited will be:

A
Same
B
One-fourth the previous value
C
Four times the previous value
D
$\frac{1}{16}$th of the previous value

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Class 12

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Electrochemistry

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Faraday’s law of electrolysis

By passing electric current,$NaClO_{3}$ is converted into $NaClO_{4}$ according to the following equation:
$NaClO_{3} + H_{2}O \rightarrow NaClO_{4} + H_{2}$
How many moles of $NaClO_{4}$ will be formed when $3 \ F$ of charge is passed through $NaClO_{3}$?

MediumKCET 2017

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Assertion $(A)$: The charge on one mole of electrons is one Faraday.
Reason $(R)$: The quantity of current required to deposit one mole of $Mg$ from $Mg^{2+}$ electrolyte solution is two Faradays.
The correct answer is

EasyAP EAMCET 2018

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To gain $1 \ mole$ $Mg$ from $Mg^{2+}$ and $1 \ mole$ $Al$ from $Al^{3+}$,how many Coulombs of electricity are required respectively?

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On passing $0.5 \ F$ of electricity through $NaCl$,the amount of $Cl_2$ gas deposited at the anode is .............. $g$.

Medium

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If $1$ mole of electrons is passed through the solutions of $AlCl_3$,$AgNO_3$,and $MgSO_4$,in what ratio will $Al$,$Ag$,and $Mg$ be deposited at the electrodes?

EasyGUJCET 2019

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In a copper voltameter experiment,current is decreased to one-fourth of the initial value but it is passed for four times the earlier duration. The amount of copper deposited will be:

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By passing electric current,$NaClO_{3}$ is converted into $NaClO_{4}$ according to the following equation:$NaClO_{3} + H_{2}O \rightarrow NaClO_{4} + H_{2}$How many moles of $NaClO_{4}$ will be formed when $3 \ F$ of charge is passed through $NaClO_{3}$?

Assertion $(A)$: The charge on one mole of electrons is one Faraday.Reason $(R)$: The quantity of current required to deposit one mole of $Mg$ from $Mg^{2+}$ electrolyte solution is two Faradays.The correct answer is

To gain $1 \ mole$ $Mg$ from $Mg^{2+}$ and $1 \ mole$ $Al$ from $Al^{3+}$,how many Coulombs of electricity are required respectively?

On passing $0.5 \ F$ of electricity through $NaCl$,the amount of $Cl_2$ gas deposited at the anode is .............. $g$.

If $1$ mole of electrons is passed through the solutions of $AlCl_3$,$AgNO_3$,and $MgSO_4$,in what ratio will $Al$,$Ag$,and $Mg$ be deposited at the electrodes?

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By passing electric current,$NaClO_{3}$ is converted into $NaClO_{4}$ according to the following equation:
$NaClO_{3} + H_{2}O \rightarrow NaClO_{4} + H_{2}$
How many moles of $NaClO_{4}$ will be formed when $3 \ F$ of charge is passed through $NaClO_{3}$?

Assertion $(A)$: The charge on one mole of electrons is one Faraday.
Reason $(R)$: The quantity of current required to deposit one mole of $Mg$ from $Mg^{2+}$ electrolyte solution is two Faradays.
The correct answer is