JEE Main · Chemistry → Steady

Electrochemistry appeared 40 times across 3 years — 4.7% of Chemistry. This question is from Kohlrausch's Law.

Year 2026 2025 2024 Total
Questions 13 19 8 40

Given below are two statements: Statement I: Mohr's salt is composed of only three types of ions-ferrous, ammonium and sulphate. Statement II: If the molar conductance at infinite dilution of ferrous, ammonium and sulphate ions are x₁, x₂ and x₃ S cm² mol⁻¹, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by x₁ + x₂ + 2x₃. In the light of the given statements, choose the correct answer from the options given below:

Solution & Explanation

Related Formula
λm∞ = ν_+ λ_+∞ + ν_- λ_-∞
Core Logic

Statement I: Mohr's salt is a double salt with chemical formula:

FeSO₄ · (NH₄)₂SO₄ · 6H₂O

When dissolved in water, it completely dissociates into three distinct ionic species:

Fe²⁺ (ferrous), NH₄^+ (ammonium), and SO₄²⁻ (sulphate)

Thus, Statement I is true.

Statement II: According to Kohlrausch's law of independent migration of ions:

λm∞(Mohr's Salt) = 1 · λm∞(Fe²⁺) + 2 · λm∞(NH₄^+) + 2 · λm∞(SO₄²⁻) λm∞ = x₁ + 2x₂ + 2x₃

Statement II claims the expression is x₁ + x₂ + 2x₃ (missing the coefficient 2 for ammonium). Thus, Statement II is false.

Pattern Recognition

Kohlrausch's law matches stoichiometric coefficients directly to the ion quantities released. Mohr's salt formula contains (NH₄)₂, requiring a multiplier of 2 for ammonium ion conductance.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 12 Chemistry: d- and f-Block Elements

Reference Study Guides

More Electrochemistry Previous-Year Questions — Page 8

Q82 jee_main_2024_29_jan_morning Faradays Laws of Electrolysis
The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A for 15 minutes is ______ × 10⁻⁴ g. (Atomic mass of zinc = 65.4 amu)
Numerical Answer. Answer: 45.75 to 46

Solution

Related Formula
W = Z · I · t = (M)/(n · F) · I · t

where, W = mass deposited Z = electrochemical equivalent I = current in amperes t = time in seconds M = molar mass n = n-factor (electrons exchanged) F = Faraday's constant (96500 C/mol)

Core Logic

The electrolysis of zinc sulphate (ZnSO₄) involves the reduction of zinc ions at the cathode:

Zn⁺² + 2e^- arrow Zn

Here, the n-factor (n) is 2.

Step 1: Calculation

Given values: I = 0.015 A t = 15 minutes = 15 × 60 seconds = 900 s M = 65.4 g/mol F ≈ 96500 C

Plugging the values into Faraday's First Law:

W = (65.4)/(2 × 96500) × 0.015 × 15 × 60 W = (65.4)/(193000) × 13.5 W = 3.3886 × 10⁻⁴ × 13.5 W = 45.746 × 10⁻⁴ g

Rounding to two decimal places (or nearest integer depending on convention), we get 45.75 × 10⁻⁴ g.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q80 jee_main_2024_30_january_evening Standard Electrode Potential
Reduction potential of ions are given below: ClO₄^- E° = 1.19V IO₄^- E° = 1.65V BrO₄^- E° = 1.74V The correct order of their oxidising power is:
  • A. ClO₄^- > IO₄^- > BrO₄^-
  • B. BrO₄^- > IO₄^- > ClO₄^-
  • C. BrO₄^- > ClO₄^- > IO₄^-
  • D. IO₄^- > BrO₄^- > ClO₄^-

Solution

Core Logic

The Standard Reduction Potential (E^°) measures a species' tendency to undergo reduction (gain electrons). A higher, more positive E^° value means the species has a stronger tendency to be reduced, which in turn makes it a stronger oxidizing agent.

Comparing the given E^° values: BrO₄^-: 1.74V IO₄^-: 1.65V ClO₄^-: 1.19V

The order of oxidizing power follows the magnitude of the reduction potential: BrO₄^- > IO₄^- > ClO₄^-

Pattern Recognition

Higher +ve Standard Reduction Potential (SRP) = Stronger Oxidising Agent.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 12 Chemistry: The p Block Elements

Q65 jee_main_2024_31_jan_morning Batteries
The metals that are employed in the battery industries are A. Fe B. Mn C. Ni D. Cr E. Cd Choose the correct answer from the options given below:
  • A. B, C and E only
  • B. A, B, C, D and E
  • C. A, B, C and D only
  • D. B, D and E only

Solution

Core Logic

Mn, Ni, and Cd metals are predominantly used in battery industries.

  • Mn is used in dry cells (Leclanche cell).
  • Ni and Cd are used in Nickel-Cadmium (Ni-Cd) rechargeable batteries.
Chapter Mix

Class 12 Chemistry: Electrochemistry

Q67 jee_main_2024_31_jan_morning Electrolytic Conductance
Identify the factor from the following that does not affect electrolytic conductance of a solution.
  • A. The nature of the electrolyte added.
  • B. The nature of the electrode used.
  • C. Concentration of the electrolyte.
  • D. The nature of solvent used.

Solution

Core Logic

Conductivity of an electrolytic cell is affected by the concentration of the electrolyte, the nature of the electrolyte, and the nature of the solvent. It does not depend on the nature of the electrode used.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q90 jee_main_2024_31_jan_morning Faraday's Laws of Electrolysis
One Faraday of electricity liberates x × 10⁻¹ gram atom of copper from copper sulphate, x is
Numerical Answer. Answer: 5 to 5

Solution

Core Logic

The reduction reaction for copper is:

Cu²⁺ + 2e^- arrow Cu

This shows that 2 moles of electrons (2 Faraday) are required to deposit 1 mole (or 1 gram atom) of Cu.

Therefore, 1 Faraday of electricity will deposit:

(1)/(2) = 0.5 moles of Cu
Step 1: Finding x
0.5 mole = 0.5 gram atom = 5 × 10⁻¹ gram atom

Hence, x = 5.

Chapter Mix

Class 12 Chemistry: Electrochemistry

More Electrochemistry Questions — jee_main_2025_07_april_morning

Practice all Electrochemistry previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)