Question:
The amount of substance deposited due to passage of $1\text{F}$ of electricity is called
The amount of substance deposited due to passage of $1\text{F}$ of electricity is called
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Don't confuse "Equivalent weight" with "Electrochemical equivalent ($z$)."
• Equivalent weight: Deposited by $1\text{ Faraday}$ ($96500\text{ C}$).
• Electrochemical equivalent: Deposited by 1 Coulomb ($1\text{ Ampere}$ for $1\text{ second}$).
• Equivalent weight: Deposited by $1\text{ Faraday}$ ($96500\text{ C}$).
• Electrochemical equivalent: Deposited by 1 Coulomb ($1\text{ Ampere}$ for $1\text{ second}$).
Updated On: Jul 1, 2026
- Atomic weight
- Equivalent weight
- Electrochemical equivalent
- Molecular weight
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The Correct Option is B
Solution and Explanation
This definition arises directly from Faraday's Laws of Electrolysis and the relationship between the Faraday constant and chemical equivalents.
1. Faraday's Second Law:
The law states that when the same quantity of electricity is passed through different electrolytes, the masses of the substances deposited are proportional to their chemical equivalent weights ($E$).
2. Defining the Faraday ($1\text{F}$): $1\text{ Faraday}$ ($96500\text{ C}$) is the charge carried by one mole of electrons.
3. Theoretical Relationship: According to the formula $m = \frac{E \cdot Q}{96500}$: If we pass exactly $Q = 96500\text{ C}$ (which is $1\text{F}$): $$m = \frac{E \cdot 96500}{96500} = E$$ Therefore, the mass of the substance deposited by $1\text{ Faraday}$ of electricity is numerically equal to its
Equivalent weight.
2. Defining the Faraday ($1\text{F}$): $1\text{ Faraday}$ ($96500\text{ C}$) is the charge carried by one mole of electrons.
3. Theoretical Relationship: According to the formula $m = \frac{E \cdot Q}{96500}$: If we pass exactly $Q = 96500\text{ C}$ (which is $1\text{F}$): $$m = \frac{E \cdot 96500}{96500} = E$$ Therefore, the mass of the substance deposited by $1\text{ Faraday}$ of electricity is numerically equal to its
Equivalent weight.
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