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On passing $C$ ampere of electricity through a electrolyte solution for $t$ second, $m$ gram metal deposits on cathode. The equivalent weight $E$ of the metal is
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The correct answer is:
$E=\frac{96500 \times m}{C \times t}$
Faraday's constant (F) is a fundamental physical constant relating the amount of electric charge transferred per mole of electrons transferred in an electrochemical reaction. Its value is approximately 96500 coulombs per mole (C/mol).
The equation $E=96500 \times m /(C \times t)$ represents one way to express Faraday's law of electrolysis.
In the equation:
- $E$ is the equivalent weight of the metal ( $\mathrm{g} / \mathrm{eq})$
- $m$ is the mass of the metal deposited on the cathode ( $\mathrm{g}$ )
- $\mathrm{C}$ is the current passed through the electrolyte (A)
- $\mathrm{t}$ is the time the current flows ( $\mathrm{s}$ )
So, the equivalent weight (E) of a metal can be calculated by dividing the product of the Faraday's constant (F) and the mass ( $\mathrm{m}$ ) of the metal deposited on the cathode by the product of the current $(\mathrm{C})$ and the time $(\mathrm{t})$ the current flows.
The equation $E=96500 \times m /(C \times t)$ represents one way to express Faraday's law of electrolysis.
In the equation:
- $E$ is the equivalent weight of the metal ( $\mathrm{g} / \mathrm{eq})$
- $m$ is the mass of the metal deposited on the cathode ( $\mathrm{g}$ )
- $\mathrm{C}$ is the current passed through the electrolyte (A)
- $\mathrm{t}$ is the time the current flows ( $\mathrm{s}$ )
So, the equivalent weight (E) of a metal can be calculated by dividing the product of the Faraday's constant (F) and the mass ( $\mathrm{m}$ ) of the metal deposited on the cathode by the product of the current $(\mathrm{C})$ and the time $(\mathrm{t})$ the current flows.
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