(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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$0.066 \ g$ 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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