(B) The formula for the mass deposited is $w = z \cdot i \cdot t$,where $w$ is the mass,$z$ is the electrochemical equivalent,$i$ is the current,and $

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(B) The formula for the mass deposited is $w = z \cdot i \cdot t$,where $w$ is the mass,$z$ is the electrochemical equivalent,$i$ is the current,and $t$ is the time.Given: $w = 0.066 \ g$,$i = 2 \ A$,$t = 100 \ s$.Rearranging the formula to solve for $z$: $z = \frac{w}{i \times t}$.Substituting the values: $z = \frac{0.066}{2 \times 100} = \frac{0.066}{200} = 0.00033 \ g \ C^{-1}$.Therefore,$z = 3.3 \times 10^{-4} \ g \ C^{-1}$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

Medium

$0.066 \ g$ of metal was deposited when a current of $2 \ A$ is passed through a metal ion solution for $100 \ s$. What is the electrochemical equivalent (in $g \ C^{-1}$) of the metal?

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A
$3.3 \times 10^{-6}$
B
$3.3 \times 10^{-4}$
C
$0.033$
D
$3.3$

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Electrochemistry

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Faraday’s law of electrolysis

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In an electroplating experiment,$m \ g$ of silver is deposited when $4 \ A$ of current flows for $2 \ min$. The amount (in $g$) of silver deposited by $6 \ A$ of current flowing for $40 \ s$ will be:

Medium

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Calculate the mass of a divalent metal produced at the cathode by passing $5 \ A$ current through its salt solution for $100 \ minutes$. (Molar mass of metal $= x \ g/mol$)

MediumMHT CET 2022

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When $3 \ F$ of electricity is passed through an aqueous solution of iron $(II)$ bromide,what is the mass of iron metal deposited at the cathode in $g$?

Medium

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Match List $I$ with List $II$.
List $I$ (Conversion) List $II$ (Number of Faraday required)
$A$. $1 \text{ mol of } H_2O \text{ to } O_2$ $I$. $3F$
$B$. $1 \text{ mol of } MnO_4^- \text{ to } Mn^{2+}$ $II$. $2F$
$C$. $1.5 \text{ mol of } Ca \text{ from molten } CaCl_2$ $III$. $1F$
$D$. $1 \text{ mol of } FeO \text{ to } Fe_2O_3$ $IV$. $5F$
Choose the correct answer from the options given below:

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When a current of $X$ Faraday is passed through a solution of aluminium nitrate,$1 \ mol$ of $Al$ is deposited. Then,$X =$ . . . . . . ?

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List $I$ (Conversion)	List $II$ (Number of Faraday required)
$A$. $1 \text{ mol of } H_2O \text{ to } O_2$	$I$. $3F$
$B$. $1 \text{ mol of } MnO_4^- \text{ to } Mn^{2+}$	$II$. $2F$
$C$. $1.5 \text{ mol of } Ca \text{ from molten } CaCl_2$	$III$. $1F$
$D$. $1 \text{ mol of } FeO \text{ to } Fe_2O_3$	$IV$. $5F$

$0.066 \ g$ of metal was deposited when a current of $2 \ A$ is passed through a metal ion solution for $100 \ s$. What is the electrochemical equivalent (in $g \ C^{-1}$) of the metal?

Similar Questions

In an electroplating experiment,$m \ g$ of silver is deposited when $4 \ A$ of current flows for $2 \ min$. The amount (in $g$) of silver deposited by $6 \ A$ of current flowing for $40 \ s$ will be:

Calculate the mass of a divalent metal produced at the cathode by passing $5 \ A$ current through its salt solution for $100 \ minutes$. (Molar mass of metal $= x \ g/mol$)

When $3 \ F$ of electricity is passed through an aqueous solution of iron $(II)$ bromide,what is the mass of iron metal deposited at the cathode in $g$?

When a current of $X$ Faraday is passed through a solution of aluminium nitrate,$1 \ mol$ of $Al$ is deposited. Then,$X =$ . . . . . . ?

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