(C) According to Faraday's first law of electrolysis,the mass of the substance deposited $(m)$ is given by the formula: $m = \frac{E \times I \times t

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$E = \frac{96500 \times m}{C \times t}$

(C) According to Faraday's first law of electrolysis,the mass of the substance deposited $(m)$ is given by the formula: $m = \frac{E \times I \times t}{F}$,where $E$ is the equivalent weight,$I$ is the current in amperes,$t$ is the time in seconds,and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.Given $I = C$,we have $m = \frac{E \times C \times t}{96500}$.Rearranging the formula to solve for $E$: $E = \frac{m \times 96500}{C \times t}$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

Easy

On passing $C$ ampere of electricity through an electrolyte solution for $t$ seconds,$m$ grams of metal are deposited on the cathode. The equivalent weight $E$ of the metal is:

A
$E = \frac{C \times t}{m \times 96500}$
B
$E = \frac{C \times m}{t \times 96500}$
C
$E = \frac{96500 \times m}{C \times t}$
D
$E = \frac{C \times t \times 96500}{m}$

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Electrochemistry

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Faraday’s law of electrolysis

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On passing $C$ ampere of electricity through an electrolyte solution for $t$ seconds,$m$ grams of metal are deposited on the cathode. The equivalent weight $E$ of the metal is:

Similar Questions

How many $L$ of chlorine gas will be deposited at $NTP$ when a current of $100 \, A$ is passed through molten $NaCl$ for $5 \, hours$?

What is the time (in $sec$) required for depositing all the silver present in $125 \ mL$ of $1 \ M \ AgNO_3$ solution by passing a current of $241.25 \ A$? $(1 \ F = 96500 \ C)$

When the same quantity of electricity is passed through aqueous solutions of $AgNO_3$ and $CuSO_4$,if the number of atoms of $Ag$ and $Cu$ deposited are $x$ and $y$ respectively,then determine the correct relationship.

In the electrolysis of a fused salt,the weight of the deposit on an electrode will not depend on:

For the reaction $Al^{3+} + 3e^{-} \rightarrow Al$ (atomic mass of $Al = 27 \ g \ mol^{-1}$),calculate the quantity of electricity in coulombs required to produce $5.12 \ kg$ of $Al$.

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