CBSE Class 12 Chemistry Solutions and Electrochemistry Notes PDF (2025–2026)

Accessing the right CBSE Class 12 Chemistry Solutions and Electrochemistry Notes PDF is the most critical first step for scoring 95+ in the CBSE Class 12 Chemistry theory paper for the 2025–2026 board exam. Physical Chemistry carries 23 marks in total, with Solutions (7 marks) and Electrochemistry (9 marks) contributing over two-thirds of this weightage. These two chapters are heavily conceptual, calculation-intensive, and feature recurring derivation-based and numerical questions. This comprehensive revision guide breaks down all NCERT core definitions, standard formulas, graphical trends, electrochemical equations, and high-frequency CBSE questions.

Key Concepts in Solutions

A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits. For the board exam 12 syllabus, binary solutions (one solute and one solvent) form the primary focus.

1. Expressing Concentration of Solutions

• Molarity (M): Number of moles of solute dissolved per litre of solution. M = (WB × 1000) / (MB × V(mL)). Note: Molarity depends on temperature because liquid volume expands or contracts with temperature variations.
• Molality (m): Number of moles of solute present per 1000 g (1 kg) of solvent. m = (WB × 1000) / (MB × WA(g)). Molality is independent of temperature as mass remains unaffected by thermal changes.
• Mole Fraction (x): Ratio of the number of moles of one component to the total number of moles of all components. xA = nA / (nA + nB) and xA + xB = 1.

2. Henry's Law & Raoult's Law

Henry's Law: At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution. Mathematically, p = KH × x, where KH is Henry's law constant. Higher KH values at a given pressure imply lower gas solubility. Aquatic species are more comfortable in cold water because the solubility of oxygen increases with a decrease in temperature.

Raoult's Law: For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in solution: pA = p°A × xA and pB = p°B × xB. Total vapour pressure: Ptotal = pA + pB = p°A + (p°B - p°A)xB.

3. Ideal vs Non-Ideal Solutions

Property	Ideal Solutions	Positive Deviation	Negative Deviation
Raoult's Law	Obeys Raoult's Law at all concentrations	pA > p°AxA, pB > p°BxB	pA < p°AxA, pB < p°BxB
Intermolecular Forces	A-B interactions = A-A & B-B	A-B interactions < A-A & B-B	A-B interactions > A-A & B-B
ΔHmix and ΔVmix	ΔHmix = 0, ΔVmix = 0	ΔHmix > 0, ΔVmix > 0	ΔHmix < 0, ΔVmix < 0
Azeotrope Type	Do not form azeotropes	Minimum boiling azeotrope	Maximum boiling azeotrope
Example	n-Hexane + n-Heptane, Benzene + Toluene	Ethanol + Acetone, CS2 + Acetone	Chloroform + Acetone, Phenol + Aniline

4. Colligative Properties & Van 't Hoff Factor

Colligative properties depend only on the total number of solute particles present in solution, regardless of their chemical identity:

1. Relative Lowering of Vapour Pressure: (p°A - pA) / p°A = i × xB ≈ i × (WB × MA) / (MB × WA)
2. Elevation of Boiling Point: ΔTb = Tb - T°b = i × Kb × m (where Kb is the molal boiling point elevation constant or ebullioscopic constant).
3. Depression of Freezing Point: ΔTf = T°f - Tf = i × Kf × m (where Kf is the molal freezing point depression constant or cryoscopic constant). Used in anti-freeze solutions (e.g., ethylene glycol in automobile radiators).
4. Osmotic Pressure (Π): Hydrostatic pressure that stops the net flow of solvent into solution through a semipermeable membrane (SPM): Π = i × C R T = i × (nB / V) R T.

Van 't Hoff Factor (i): Ratio of observed colligative property to calculated normal colligative property. For dissociation (where one molecule produces n ions with degree of dissociation α): i = 1 + (n - 1)α. For association of n molecules: i = 1 + (1/n - 1)α.

Key Concepts in Electrochemistry

CBSE Chemistry syllabus treats electrochemistry as the bridge between electrical energy and chemical transformations occurring during spontaneous and non-spontaneous redox processes.

1. Electrochemical Cells & The Nernst Equation

A galvanic cell generates electrical potential via spontaneous chemical reactions. Standard cell potential is defined as E°cell = E°cathode - E°anode = E°right - E°left.

The Nernst Equation calculates the electromotive force (EMF) under non-standard concentration conditions:

Ecell = E°cell - (2.303 R T / n F) log Q
At 298 K (25 °C): Ecell = E°cell - (0.0591 / n) log ([Anode Ion] / [Cathode Ion])

Key thermodynamic relationships frequently tested in board exam numericals:

• Standard Gibbs Free Energy change: ΔG° = -n F E°cell
• Equilibrium Constant relation: log Kc = (n E°cell) / 0.0591 (at 298 K)

2. Conductance, Conductivity, and Kohlrausch's Law

• Conductivity (κ): Inverse of resistivity. κ = G × (l / A) = G × G*, where G* is the cell constant (l/A, unit: cm^-1 or m^-1). SI unit is S m^-1 or S cm^-1.
• Molar Conductivity (Λ_m): Conducting power of all the ions produced by dissolving one mole of an electrolyte in solution: Λm = (1000 × κ) / M (when κ is in S cm^-1 and M is molarity in mol L^-1).
• Concentration Trends:
  • Conductivity (κ) decreases upon dilution because the number of current-carrying ions per unit volume decreases.
  • Molar conductivity (Λm) increases upon dilution because the total volume containing one mole of electrolyte expands. For strong electrolytes, Λ_m increases linearly following the Debye-Hückel-Onsager equation: Λm = Λ°m - A√C. For weak electrolytes, Λ_m increases steeply at high dilution due to increased dissociation.
• Kohlrausch's Law of Independent Migration of Ions: Limiting molar conductivity of an electrolyte is the sum of the individual limiting molar conductivities of its constituent cations and anions: Λ°m = ν+λ°+ + ν-λ°-. Degree of dissociation: α = Λm / Λ°m. Dissociation constant: Ka = (C α2) / (1 - α).

3. Faraday's Laws & Commercial Cells

• Faraday's First Law: m = Z × I × t = (M × I × t) / (n × F) where F = 96500 C mol-1.
• Commercial Batteries: Primary cells (e.g., Dry Leclanché cell, Mercury cell) cannot be recharged. Secondary cells (e.g., Lead-acid storage battery, Ni-Cd battery) are rechargeable through reverse current.
• Fuel Cells: Galvanic cells converting chemical energy of combustion fuels (e.g., H₂, CH₄) directly into electrical energy with ~70% thermodynamic efficiency (e.g., H₂-O₂ fuel cell used in spacecraft).

Important CBSE Questions with Answers

The following problems are drawn directly from the official CBSE Class 12 Chemistry question bank and past board examinations.

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

Answer:
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 cation and anion at infinite dilution.
Mathematically:
Λ°m(AxBy) = x λ°Ay+ + y λ°Bx-
Application Example: Calculating the limiting molar conductivity of weak acetic acid using strong electrolytes:
Λ°m(CH3COOH) = λ°CH3COO- + λ°H+ = 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 is the conductance of a solution of 1 cm (or 1 m) length with an area of cross-section 1 cm² (or 1 m²), which corresponds to the conductance of unit volume of the electrolytic solution.
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 of solution: Λm = κ / C.
Variation with Concentration:

• Conductivity (κ): Decreases with decrease in concentration (dilution) for both strong and weak electrolytes because the number of current-carrying ions per unit volume decreases.
• Molar Conductivity (Λ_m): Increases with decrease in concentration (dilution). In strong electrolytes, interionic interactions weaken, increasing ionic mobility. In weak electrolytes, the degree of ionisation (α) increases exponentially as concentration approaches zero.

Question 3: Calculate the energy required to reduce 1 mole of Al³⁺ to Al at cathode. (E° = 1.5 V, F = 96500 C/mol)

Answer:
Step 1: Write the half-cell reduction equation:
Al3+ + 3e- → Al(s)
The number of electrons transferred per mole of Al³⁺ is n = 3.
Step 2: Apply the electrical energy relationship:
Electrical Energy (E) = n × F × E°
Step 3: Substitute given numerical values:
E = 3 × 96500 C/mol × 1.5 V
E = 434,250 J = 434.25 kJ
Final Answer: The energy required is 434.25 kJ (or 4.34 × 10⁵ J).

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

Answer:
Step 1: Ionic dissociation and cathode reaction:
Al2O3 → 2Al3+ + 3O2-
Al3+ + 3e- → Al (Each mole of Al requires 3 moles of electrons).
Step 2: Calculate moles of Al produced:
Mass of Al = 5.12 kg = 5120 g
Molar mass of Al = 27 g/mol
Moles of Al = 5120 / 27 ≈ 189.63 mol
Step 3: Calculate total electric charge (Q):
Total moles of electrons required = 189.63 × 3 = 568.89 mol e-
Charge (Q) = 568.89 mol × 96500 C/mol = 5.49 × 107 C
Final Answer: The total quantity of electricity required is 5.49 × 10⁷ C.

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

Answer:
Step 1: Determine the overall cell reaction:
Anode (Oxidation): Zn(s) → Zn2+(aq) + 2e-
Cathode (Reduction): Cu2+(aq) + 2e- → Cu(s)
Net Reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) (where n = 2).
Step 2: Apply the Nernst Equation at 298 K:
Ecell = E°cell - (0.0591 / n) log ([Zn2+] / [Cu2+])
Step 3: Substitute the concentrations and solve:
Ecell = 1.10 - (0.0591 / 2) log (0.1 / 0.01)
Ecell = 1.10 - (0.02955) log(10)
Since log(10) = 1:
Ecell = 1.10 - 0.02955 = 1.07045 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:

Parameter	Primary Batteries	Secondary Batteries
Reversibility	Non-rechargeable. Chemical reaction is irreversible; the battery becomes dead after discharging.	Rechargeable. Chemical reaction is reversible by passing an external electric current in the opposite direction.
Energy Conversion	Converts chemical energy to electrical energy in a single cycle.	Acts as a galvanic cell during discharge and as an electrolytic cell during recharge.
Internal Resistance & Cost	Higher internal resistance; inexpensive and lightweight for low-drain devices.	Low internal resistance; higher initial cost but economical over hundreds of duty cycles.
Examples & Applications	Dry Leclanché Cell: Zn anode, MnO2/C cathode; used in torches, clocks (EMF ≈ 1.5 V). Mercury Button Cell: Zn-Hg amalgam anode, HgO cathode; used in hearing aids and watches (EMF = 1.35 V).	Lead Storage Battery: Pb anode, PbO2 cathode, 38% H2SO4 electrolyte; used in automobiles and inverters. Nickel-Cadmium (Ni-Cd) Cell: Cd anode, NiO(OH) cathode; used in portable electronics.

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

Answer:
Corrosion is the slow deterioration and oxidisation of a metal into its oxide, hydroxide, or salt through electrochemical reactions with atmospheric moisture and gases.

Electrochemical Theory of Rusting: Rusting occurs via the formation of micro galvanic cells across the uneven surface of impure iron in contact with water containing dissolved CO₂ and atmospheric O₂.

1. At Anode (Oxidation spot on iron surface): Pure iron oxidises and releases electrons:
   Fe(s) → Fe2+(aq) + 2e-   [E°(Fe2+/Fe) = -0.44 V]
2. At Cathode (Reduction spot on moisture layer): Electrons travel through the metal to an adjacent area where atmospheric O₂ is reduced in the presence of H⁺ ions (from carbonic acid H₂CO₃ formed by CO₂ + H₂O):
   O2(g) + 4H+(aq) + 4e- → 2H2O(l)   [E° = +1.23 V]
3. Overall Electrochemical Cell Reaction:
   2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l)   [E°cell = 1.67 V]
4. Rust Formation: Ferrous ions are further oxidised by atmospheric oxygen to hydrated ferric oxide:
   4Fe2+(aq) + O2(g) + 4H2O(l) → 2Fe2O3(s) + 8H+(aq)
Fe2O3(s) + xH2O(l) → Fe2O3·xH2O (Rust)

Prevention Methods: Barrier protection (painting, oiling), sacrificial protection (galvanising with zinc), and cathodic protection connecting magnesium sacrificial blocks to underground iron pipes.

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

Answer:
A Daniel Cell is a voltaic cell that harnesses the chemical energy liberated during the spontaneous redox displacement between zinc and copper ions to generate electric potential.

1. Construction:

• Anode Compartment: A zinc metallic plate immersed in 1.0 M aqueous ZnSO4 solution (negative terminal).
• Cathode Compartment: A copper metallic plate immersed in 1.0 M aqueous CuSO4 solution (positive terminal).
• Salt Bridge: An inverted U-tube packed with agar-agar gel containing an inert electrolyte (KCl, KNO3, or NH4NO3) bridging both electrolytic solutions.

2. Working & Electrode Half-Reactions:

• At Anode (Oxidation): Zn(s) → Zn2+(aq) + 2e-   [E°ox = +0.76 V]
• At Cathode (Reduction): Cu2+(aq) + 2e- → Cu(s)   [E°red = +0.34 V]
• Net Cell Reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
• Standard Cell Potential: E°cell = E°Cu2+/Cu - E°Zn2+/Zn = +0.34 V - (-0.76 V) = +1.10 V

3. Functions of the Salt Bridge:

1. Completes the electrical circuit by permitting ion mobility without mechanical mixing of the two solutions.
2. Maintains electrical neutrality: K⁺ ions migrate into the cathode compartment to neutralize excess SO₄^2-, while Cl^- ions migrate into the anode compartment to neutralize excess Zn²⁺.
3. Eliminates liquid-junction potential at the interface between the two solutions.

How to Prepare for This Topic

Excelling in Solutions and Electrochemistry for the CBSE Class 12 board examinations requires structured problem-solving and rigorous conceptual clarity:

1. Strictly Verify Numerical Units: In electrochemistry, check whether conductivity (κ) is given in S cm^-1 or S m^-1. When using κ in S cm^-1, calculate molar conductivity with Λm = (1000 × κ) / M to obtain S cm² mol^-1. For colligative properties, always convert solute mass into grams and solvent mass into kilograms.
2. Practice Standard NCERT Graphs: Key visual questions in the 2025–2026 board exam include Raoult's law vapour pressure-composition curves, boiling point elevation and freezing point depression phase curves, and Debye-Hückel-Onsager plots (Λm vs √C) highlighting the steep upward curvature of weak electrolytes (e.g., CH₃COOH) versus strong electrolytes (e.g., KCl).
3. Master Nernst Equation Calculations: Be proficient in identifying n (electrons exchanged), formulating the reaction quotient Q with correct stoichiometric exponents, and computing log Kc and ΔG° values without arithmetic slips.
4. Solve Curated CBSE Chapter-Wise PYQs: Practicing 5-year board question papers under timed constraints builds speed and eliminates common sign errors. Access chapter-wise test papers on the Theorify QPTool Question Paper Generator to test your conceptual depth against official marking schemes.

Where to Practice More

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• Generate custom, syllabus-aligned chapter tests with detailed marking rubrics on the Theorify QPTool Question Bank.
• Practice mock papers, assertion-reason sets, and case-based numerical worksheets designed specifically for CBSE Class 12 Chemistry board exam preparation.

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