CBSE Class 12 Chemistry Electrochemistry Numericals Solved 2026

Mastering CBSE Class 12 Chemistry is essential for scoring a 95+ in your 2026 board examination. Electrochemistry stands as one of the highest-weightage chapters in Physical Chemistry, typically contributing 8 to 9 marks in the CBSE Class 12 Chemistry paper. Scoring full marks in this unit requires both conceptual precision in electrochemical mechanisms and step-by-step numerical problem-solving skills. In this comprehensive guide, we unpack the fundamental formulas, provide NCERT-aligned solutions to official CBSE board questions, and present a structured strategy to master every numerical variation for the 2026 board exams.

Key Concepts and Master Formula Sheet

Before attempting numerical problems, review the essential electrochemical definitions, laws, and governing equations prescribed by the CBSE syllabus.

1. Electrochemical Cell Potential and the Nernst Equation

For a general redox reaction occurring in a galvanic cell: aA + bB → cC + dD, the standard cell electromotive force (EMF) and non-standard potential at 298 K are given by:

• Standard Cell Potential: E°cell = E°cathode - E°anode = E°right - E°left (using standard reduction potentials)
• Nernst Equation at 298 K: Ecell = E°cell - (0.0591 / n) × log Q, where Q = [C]c[D]d / [A]a[B]b
• Equilibrium Constant Relation: E°cell = (0.0591 / n) × log Kc
• Standard Gibbs Free Energy Change: ΔG° = -nFE°cell (where F = 96,500 C mol^-1)

2. Electrolytic Conductance, Conductivity, and Molar Conductivity

Electrolytic solutions conduct electricity through ion migration. The quantitative metrics include:

• Resistance (R): R = ρ × (l / A) (measured in Ω)
• Conductance (G): G = 1 / R = (1 / ρ) × (A / l) = κ × (A / l) (measured in Siemens, S or Ω^-1)
• Cell Constant (G*): G* = l / A = R × κ (measured in cm^-1 or m^-1)
• Conductivity / Specific Conductance (κ): κ = G × (l / A) = (1 / R) × G* (measured in S cm^-1 or S m^-1)
• Molar Conductivity (Λ_m): Λm = (κ × 1000) / Molarity (in mol L-1) [Unit: S cm² mol^-1]
• Degree of Dissociation (α) for Weak Electrolytes: α = Λm / Λ°m
• Dissociation Constant (K_a or K_c): Ka = (C α2) / (1 - α)

3. Faraday's Laws of Electrolysis

• Faraday's First Law: Mass deposited m = Z × Q = Z × I × t, where electrochemical equivalent Z = M / (n × F)
• Total Charge: Q = I × t = n × F × (moles of substance reacted)

Important CBSE Questions with Step-by-Step Answers

The following problems are drawn directly from official CBSE question banks and recent sample papers. Study the step-by-step marking scheme solutions carefully.

Question 1: State Kohlrausch's Law of independent migration of ions.

Answer:

Statement: Kohlrausch's Law of independent migration of ions states that the limiting molar conductivity (Λ°_m) of an electrolyte can be represented as the sum of the individual limiting molar ionic conductivities of its constituent cations and anions at infinite dilution.

Mathematical Expression:

Λ°m (AxBy) = x λ°+ + y λ°-

Where:

• Λ°m = Limiting molar conductivity of the electrolyte
• λ°+ and λ°- = Limiting molar conductivities of the cation and anion respectively
• x and y = Number of cations and anions produced per formula unit of the electrolyte

Primary Application: It is used to calculate the limiting molar conductivity of weak electrolytes (such as CH₃COOH) by combining the limiting molar conductivities of strong electrolytes: Λ°m(CH3COOH) = Λ°m(CH3COONa) + Λ°m(HCl) - Λ°m(NaCl).

Question 2: Define conductivity and molar conductivity of a solution. How does each vary with concentration?

Answer:

1. Conductivity (κ): Conductivity (or specific conductance) is the conductance of a solution of 1 cm (or 1 m) length with an area of cross-section equal to 1 cm² (or 1 m²). In other words, it is the conductance of unit volume (1 cm³ or 1 m³) of an electrolytic solution.

Variation with Concentration: Conductivity decreases with dilution (decreasing concentration) for both strong and weak electrolytes. This occurs because the number of current-carrying ions present per unit volume of the solution decreases as the solution is diluted.

2. Molar Conductivity (Λ_m): Molar conductivity is the conducting power of all the ions produced by dissolving one mole of an electrolyte in a given volume V of solution placed between two large electrodes separated by unit distance.

Λm = κ × V = (κ × 1000) / C

Variation with Concentration: Molar conductivity increases with dilution (decreasing concentration) for both strong and weak electrolytes:

• Strong Electrolytes: Interionic attractions decrease upon dilution, allowing ions to move with greater mobility. Λ_m increases linearly following the Debye-Hückel-Onsager equation: Λm = Λ°m - A√C.
• Weak Electrolytes: Dilution significantly increases the degree of dissociation (α → 1 as concentration approaches zero), resulting in a steep non-linear increase in Λ_m at high dilutions.

Question 3: Calculate the energy required to reduce 1 mole of Al³⁺ to Al at the cathode. (E° = 1.5 V, F = 96,500 C/mol)

Solution:

Step 1: Write the cathodic reduction half-reaction:

Al3+ + 3e- → Al (s)

From the stoichiometric equation, the number of moles of electrons transferred per mole of Al is n = 3.

Step 2: State the given parameters:

• Moles of Al³⁺ = 1 mol
• Number of electrons transferred (n) = 3
• Standard potential (E°) = 1.5 V
• Faraday constant (F) = 96,500 C mol^-1

Step 3: Calculate electrical energy (Work done / Free energy change magnitude):

Electrical Energy (E) = n × F × E°

E = 3 × 96,500 C × 1.5 V

E = 434,250 J = 434.25 kJ

Final Answer: The energy required to reduce 1 mole of Al³⁺ to Al is 434.25 kJ (or 4.3425 × 10⁵ J).

Question 4: How much electricity (in coulombs) is required to produce 5.12 kg of Al from Al₂O₃?

Solution:

Step 1: Identify the dissociation and electrode reaction:

Al2O3 → 2Al3+ + 3O2-

Cathodic reduction: Al3+ + 3e- → Al

1 mole of Al (atomic mass = 27 g mol^-1) requires 3 moles of electrons = 3 × 96,500 C of charge.

Step 2: Convert mass of Al to moles:

Mass of Al = 5.12 kg = 5,120 g

Moles of Al = Mass / Molar mass = 5120 g / 27 g mol-1 ≈ 189.63 mol

Step 3: Calculate total moles of electrons and required charge (Q):

Total moles of electrons required = 189.63 mol × 3 = 568.89 mol e-

Total Charge (Q) = Moles of electrons × F

Q = 568.888 × 96,500 C = 54,897,778 C ≈ 5.49 × 107 C

Final Answer: The electricity required to produce 5.12 kg of Al from Al₂O₃ is 5.49 × 10⁷ Coulombs.

Question 5: Calculate the EMF of the cell: Zn | Zn²⁺(0.1 M) || Cu²⁺(0.01 M) | Cu at 298 K. (Given: E°_cell = 1.10 V)

Solution:

Step 1: Write the overall cell reaction:

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Here, the number of electrons involved in the redox process is n = 2.

Step 2: Apply the Nernst equation:

Ecell = E°cell - (0.0591 / n) × log([Zn2+] / [Cu2+])

Step 3: Substitute the given values:

• E°_cell = 1.10 V
• n = 2
• [Zn²⁺] = 0.1 M = 10^-1 M
• [Cu²⁺] = 0.01 M = 10^-2 M

[Zn2+] / [Cu2+] = 0.1 / 0.01 = 10

log(10) = 1

Step 4: Compute E_cell:

Ecell = 1.10 - (0.059 / 2) × log(10)

Ecell = 1.10 - 0.0295 × (1) = 1.0705 V ≈ 1.07 V

Final Answer: The EMF of the cell is 1.07 V.

Question 6: Explain the difference between primary and secondary batteries with examples.

Answer:

Batteries are galvanic cells arranged in series to deliver commercial electrical energy. They are categorized based on their reversibility:

Question 7: What is corrosion? Explain the electrochemical theory of rusting of iron.

Answer:

Definition: Corrosion is the slow, spontaneous degradation and deterioration of metals into undesirable compounds (oxides, sulfides, or carbonates) due to chemical or electrochemical reactions with their surrounding atmospheric gases, moisture, and electrolytes.

Electrochemical Mechanism of Rusting of Iron:

Rusting of iron takes place via localized microscopic electrochemical cells formed on the surface of iron in the presence of water containing dissolved oxygen and CO₂ (forming acidic H⁺ ions).

1. At Anode (Oxidation Region): Pure iron acts as the anode and undergoes oxidation:
   Fe(s) → Fe2+(aq) + 2e-   [E°Fe2+/Fe = -0.44 V]
2. Electron Migration: Electrons released at the anodic spot travel through the metal to another surface area with high moisture and dissolved oxygen.
3. At Cathode (Reduction Region): Atmospheric oxygen is reduced to water in the presence of H⁺ ions:
   O2(g) + 4H+(aq) + 4e- → 2H2O(l)   [E° = +1.23 V]
4. Overall Electrochemical Cell Reaction:
   2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l)   [E°cell = 1.67 V]
5. Formation of Rust: The ferrous ions (Fe²⁺) are further oxidized by atmospheric oxygen to ferric ions (Fe³⁺) and precipitate out as hydrated ferric oxide (rust):
   4Fe2+(aq) + O2(g) + 4H2O(l) → 2Fe2O3(s) + 8H+(aq)
Fe2O3(s) + xH2O(l) → Fe2O3·xH2O (s)   [Hydrated Ferric Oxide / Rust]

Question 8: Describe the construction and working of a Daniel cell with a labeled diagram.

Answer:

1. Construction of Daniel Cell:

• Anode Compartment: A zinc (Zn) metal rod immersed in an aqueous 1.0 M zinc sulfate (ZnSO₄) solution. It serves as the negative terminal where oxidation occurs.
• Cathode Compartment: A copper (Cu) metal rod immersed in an aqueous 1.0 M copper sulfate (CuSO₄) solution. It serves as the positive terminal where reduction occurs.
• External Circuit: The zinc and copper electrodes are connected through a conducting metallic wire via a voltmeter and a plug key to measure the potential difference.
• Salt Bridge: An inverted U-tube containing an agar-agar gel mixed with an inert electrolyte (such as KCl, KNO₃, or NH₄NO₃). It connects the two electrolytic solutions without allowing them to mix mechanically.

2. Electrode Reactions and Working:

• At Anode (Oxidation): Zinc atoms lose two electrons each to form Zn²⁺ ions that enter the solution:
  Zn(s) → Zn2+(aq) + 2e-   [E°Zn2+/Zn = -0.76 V]
• At Cathode (Reduction): Cu²⁺ ions from the CuSO₄ solution gain two electrons at the copper electrode surface and deposit as metallic copper:
  Cu2+(aq) + 2e- → Cu(s)   [E°Cu2+/Cu = +0.34 V]
• Net Cell Reaction:
  Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
• Standard Cell EMF:
  E°cell = E°cathode - E°anode = 0.34 V - (-0.76 V) = +1.10 V

3. Functions of the Salt Bridge:

1. It completes the internal electric circuit by allowing ion flow between the two half-cells.
2. It maintains electrical neutrality in both half-cell solutions by releasing balancing counter-ions (K⁺ towards cathode and Cl^- towards anode) as charges accumulate during cell operation.
3. It prevents liquid junction potential between the two liquid phases.

How to Prepare for Electrochemistry Numericals in CBSE 2026

To secure a 100% accuracy rate in the numerical section of the CBSE Class 12 Chemistry board exam, follow these proven preparation steps:

1. Standardize Reduction Potentials: CBSE questions often provide oxidation potentials to test students (e.g., E°Zn/Zn2+ = +0.76 V). Always convert them to Standard Reduction Potentials first: E°red = -E°ox, and then use E°cell = E°cathode - E°anode.
2. Master Unit Conversions in Conductance:
   • If κ is given in S cm-1 and molarity C in mol L-1, use: Λm = (κ × 1000) / C (Result in S cm² mol^-1).
   • If κ is given in SI units (S m-1) and concentration in mol m-3, use: Λm = κ / C (Result in S m² mol^-1). Remember: 1 S m2 mol-1 = 104 S cm2 mol-1.
3. Track Number of Electrons (n) in Redox Half-Reactions: Carefully balance equations before using the Nernst equation or Faraday's laws. For example, in the reaction between Cr₂O₇^2- and Fe²⁺, n = 6 electrons.
4. Memorize Key Logarithm Values: Keep these standard log approximations handy for rapid non-programmable calculations: log 2 ≈ 0.3010, log 3 ≈ 0.4771, log 5 ≈ 0.6990, and log 7 ≈ 0.8451.

Where to Practice More

Practicing authentic past-year questions, chapter-wise mock tests, and official sample question papers is the single most effective way to eliminate step-marking mistakes. Access structured, syllabus-aligned question banks and auto-graded mock assessments on the Theorify QPTool Platform.

You can generate customized chapter tests for Physical Chemistry on the QPTool Class 12 Chemistry Test Generator, or explore full-length solved model question papers at Theorify QPTool CBSE Resources. Consistent numerical practice will ensure you achieve a perfect score in your 2026 board examinations!