JAMB 2024 Physics · Question 41

In electrolysis, when same quantity of electricity is passed through different electrolytes, mass of substances deposited is proportional to

Options

  1. A.Molecular weight
  2. B.Chemical equivalent
  3. C.Atomic number
  4. D.Chemical equivalent per valency

Answer: B. Chemical equivalent

Explanation written with AI and checked against the answer key

Analysis

The question asks for the relationship between the mass of substances deposited during electrolysis when the same quantity of electricity is passed through different electrolytes. This is a fundamental principle of Faraday's Laws of Electrolysis.

Concept: Faraday’s Second Law of Electrolysis

Faraday’s Second Law of Electrolysis states that when the same quantity of electricity is passed through different electrolytes connected in series, the masses of the substances deposited (or liberated) at the electrodes are directly proportional to their chemical equivalents (also known as equivalent weights).

The chemical equivalent is calculated as:

Chemical Equivalent=Molar MassValency (or oxidation state)\text{Chemical Equivalent} = \frac{\text{Molar Mass}}{\text{Valency (or oxidation state)}}
Step-by-Step Solution
  1. Understand the Law: Faraday's second law specifically addresses the comparison between different substances. It establishes that the amount of chemical change produced by a given quantity of electricity is proportional to the equivalent weight of the substance.
  2. Evaluate the Options: • A (Molecular weight): This is incorrect because it does not account for the valency of the ions. • B (Chemical equivalent): This matches the definition of Faraday's Second Law perfectly. • C (Atomic number): This is irrelevant to the mass deposited in electrolysis. • D (Chemical equivalent per valency): This is redundant and incorrect, as the chemical equivalent already incorporates the valency.
  3. Conclusion: Since the mass deposited (mm) is related to the quantity of electricity (QQ) and the chemical equivalent (EE) by the formula m=Q×EFm = \frac{Q \times E}{F} (where FF is Faraday's constant), it is clear that m∝Em \propto E.
Correct Answer

The correct option is B.

Exam Tip

To remember this easily, think of the "E" in Electrolysis and Equivalent weight. Faraday's Second Law deals with different substances, and the key factor that distinguishes them in this process is their Equivalent weight.

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