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Electrochemistry appeared 40 times across 3 years — 4.7% of Chemistry. This question is from Batteries and Commercial Cells.

Year 2026 2025 2024 Total
Questions 13 19 8 40

Match List-I with List-II:
List-I (Applications)List-II (Batteries/Cell)
(A) Transistors(I) Anode - Zn/Hg; Cathode - HgO + C
(B) Hearing aids(II) Hydrogen fuel cell
(C) Invertors(III) Anode - Zn; Cathode - Carbon
(D) Apollo space ship(IV) Anode - Pb; Cathode - Pb | PbO₂
Choose the correct answer from the options given below:

Solution & Explanation

Core Logic

Matching applications to their respective electrochemical cells: * Transistors use standard dry cells: Anode is Zn container, Cathode is carbon rod coated with MnO₂ arrow (III). * Hearing aids require compact voltage outputs over time, matching Mercury cells: Anode Zn/Hg, Cathode HgO + C arrow (I). * Invertors utilize rechargeable systems, matching Lead-storage batteries: Anode Pb, Cathode Pb | PbO₂ arrow (IV). * Apollo space ship dynamically powered via Hydrogen-Oxygen Fuel cells arrow (II).

Pattern Recognition

Space missions universally trigger fuel cell pairs in standard test patterns due to the secondary requirement of gathering pure drinking water byproduct.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Reference Study Guides

More Electrochemistry Previous-Year Questions — Page 5

Q36 jee_main_2025_29_jan_evening Products of Electrolysis
O₂ gas will be evolved as a product of electrolysis of: (A) an aqueous solution of AgNO₃ using silver electrodes. (B) an aqueous solution of AgNO₃ using platinum electrodes. (C) a dilute solution of H₂SO₄ using platinum electrodes. (D) a high concentration solution of H₂SO₄ using platinum electrodes. Choose the correct answer from the options given below:
  • A. (B) and (C) only
  • B. (A) and (D) only
  • C. (B) and (D) only
  • D. (A) and (C) only

Solution

Core Logic

Analyzing anodic reactions during electrolysis: * Case (A): With active Ag electrodes, silver oxidation occurs at the anode (Ag arrow Ag⁺ + e⁻). No oxygen is evolved. * Case (B): With inert Pt electrodes, oxidation of water occurs preferentially at the anode over NO₃^- ions:2H₂O arrow O₂ + 4H⁺ + 4e⁻

  • Case (C): In dilute H₂SO₄, water oxidation takes place, releasing O₂ gas at the anode.
  • Case (D): In concentrated H₂SO₄, oxidation of SO₄²⁻ creates peroxodisulphate ions (S₂O₈²⁻), inhibiting oxygen evolution.
Pattern Recognition

Remember that active electrodes participate directly in redox reactions, whereas inert electrodes (Pt, Graphite) yield oxygen gas when water is oxidized in the presence of oxoanions like NO₃^- or dilute SO₄²⁻.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q46 jee_main_2025_28_jan_morning Molar Conductivity and Cell Resistance
Given below is the plot of the molar conductivity vs concentration for KCl in aqueous solution.
Molar conductivity vs root concentration graph for Q46 - JEE Main 2025 Morning
The image features a standard linear plot tracing electrolytic molar conductance trends over root concentration variations.
If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100Ω then the resistance of the same cell with the dilute solution is xΩ The value of x is (Nearest integer)
Numerical Answer. Answer: 150 to 150

Solution

Related Formula

Conductivity relationship with cell parameters:

κ = G · G^* = (G^*)/(R) λm = (κ × 1000)/(C)

where G^* represents the static cell constant.

Step 1: Setting Up Ratios

Using concentration subscripts c (concentrated) and d (dilute):

(κc)/(κd) = (Rd)/(Rc)

Expressing conductivity through molar conductivity values:

κ = (λm · C)/(1000) ((λm · C)c)/((λm · C)d) = (Rd)/(Rc)

Substituting the graphical read coordinates (Cc = 0.15², Cd = 0.1² with scaled λm parameters):

(100 · (0.15)²)/(150 · (0.1)²) = (Rd)/(100) Rd = 150 Ω
Pattern Recognition

Sees: Resistance correlation across specific graph coordinates. Shortcut: Equate cell parameters through κ ∝ (1)/(R) and solve for the target resistance directly.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q45 jee_main_2025_03_april_morning Limiting Molar Conductivity
Correct order of limiting molar conductivity for cations in water at 298 K is:
  • A. H⁺>Na⁺>K⁺>Ca²⁺>Mg²⁺
  • B. H⁺>Ca²⁺>Mg²⁺>K⁺>Na⁺
  • C. Mg²⁺>H⁺>Ca²⁺>K⁺>Na⁺
  • D. H⁺>Na⁺>Ca²⁺>Mg²⁺>K⁺

Solution

Core Logic

Limiting molar conductivity relies heavily on the charge and hydrodynamic radius of the hydrated ion. Let us verify the standard experimental limiting molar ionic conductivities (λ°) at 298 K:

  • H⁺: 349.8 S cm² mol⁻¹ (exhibits Grotthuss proton-hopping conduction mechanism)
  • Ca²⁺: 119.0 S cm² mol⁻¹
  • Mg²⁺: 106.1 S cm² mol⁻¹
  • K⁺: 73.5 S cm² mol⁻¹
  • Na⁺: 50.1 S cm² mol⁻¹
Step 1: Trend Layout

Arranging these values in descending order yields:

H⁺ > Ca²⁺ > Mg²⁺ > K⁺ > Na⁺
Pattern Recognition

Shortcut: H⁺ always has the absolute highest value due to its unique proton-hopping transport system. For metal ions, a higher ionic charge boosts conductivity (M²⁺ > M⁺), and within a group, a smaller hydrated radius (larger bare ion) increases mobility (K⁺ > Na⁺).

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q49 jee_main_2025_04_april_evening Conductance of Electrolytic Solutions
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185~S~cm²mol⁻¹ and the ionic conductance of hydroxyl and chloride ions are 170 and 70~S~cm²mol⁻¹ , respectively. If molar conductance of 0.02~M solution of ammonium hydroxide is 85.5~S~cm²mol⁻¹ , its degree of dissociation is given by x× 10⁻¹ . The value of x is _______. (Nearest integer)
Numerical Answer. Answer: 2.9 to 3.1

Solution

Related Formula
Λm^°(NH₄OH) = λ^°(NH₄^+) + λ^°(OH^-) (Kohlrausch's Law) α = (Λm^c)/(Λm^°)
Core Logic
  • Find the limiting molar conductance of NH₄^+ using the NH₄Cl data:
Λm^°(NH₄Cl) = λ^°(NH₄^+) + λ^°(Cl^-) = 185 λ^°(NH₄^+) = 185 - 70 = 115 ~S · cm² · mol⁻¹
  • Calculate Λm^° for the weak electrolyte ammonium hydroxide (NH₄OH):
Λm^°(NH₄OH) = 115 + 170 = 285 ~S · cm² · mol⁻¹
  • Evaluate the degree of dissociation (α):
α = (85.5)/(285) = 0.3 = 3 × 10⁻¹

Comparing with the format x × 10⁻¹, the value of x is 3.

Pattern Recognition

Kohlrausch's law allows direct algebraic recombination of ion conductances. Always construct the targeted weak base compound value by filtering out the spectator chloride contribution from the initial salt parameters.

Chapter Mix

Class 12 Chemistry: Electrochemistry

More Electrochemistry Questions — jee_main_2025_29_jan_evening

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