Electrochemistry Class 12 Notes - CBSE Chemistry Chapter 3 (Free PDF)

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Chapter summary

Electrochemistry studies the two-way link between chemical reactions and electricity: how a spontaneous redox reaction generates current in a galvanic cell, and how external current drives a non-spontaneous reaction during electrolysis. It builds from conductance, conductivity and molar conductivity (with Kohlrausch’s law) to electrode potentials, the Nernst equation, and the ties between EMF, Gibbs energy and the equilibrium constant. For NEET it is a high-yield, numerical-heavy chapter that feeds into batteries, fuel cells, corrosion and Faraday’s laws of electrolysis.

Chapter notes
🃏 Flash Cards: Electrochemistry

Class 12 Chemistry · Chapter 3 – swipe through all 10 cards to understand the whole chapter.

⚡Start here1/10

Conductance & Conductivity

How easily an electrolyte solution carries current, made independent of the cell’s shape.

G = 1/R · κ = G × (l/A) · cell constant G* = l/A

G in siemens (S); κ in S cm⁻1; cell constant l/A in cm⁻1.

  • Conductance G is just 1/resistance; ions (not electrons) carry the current.
  • Conductivity κ = conductance of a 1 cm3 cube of solution.
  • Cell constant l/A standardises the measurement to the liquid itself.
💧Core idea2/10

Molar Conductivity & Dilution

Conductivity of all the ions from one mole of electrolyte, and how it changes when you dilute.

Λ_m = (κ × 1000) / c (S cm2 mol⁻1)

κ in S cm⁻1, c in mol L⁻1; the ×1000 converts L → cm3.

  • On dilution κ DECREASES (fewer ions per cm3) but Λ_m INCREASES.
  • Strong electrolytes: Λ_m = Λ°_m − A√c, extrapolate to c=0 for Λ°_m.
  • Weak electrolytes shoot up steeply near infinite dilution → can’t extrapolate.
➕Key law3/10

Kohlrausch’s Law

At infinite dilution each ion contributes a fixed, independent share to Λ°_m.

Λ°_m = ν₊ λ°₊ + ν₋ λ°₋

ν are stoichiometric ion counts — e.g. CaCl2: Λ°_m = λ°(Ca2⁺) + 2λ°(Cl⁻).

  • Finds Λ°_m of weak acids: Λ°(CH3COOH) = Λ°(CH3COONa) + Λ°(HCl) − Λ°(NaCl).
  • Never drop the stoichiometric multiplier (the 2 for two Cl⁻).
  • H⁺ and OH⁻ have abnormally high λ° via Grotthuss proton-hopping.
🧪Applied4/10

Degree of Dissociation & Ka

Use Λ°_m to find how much a weak electrolyte has actually ionised.

α = Λ_m / Λ°_m · K_a = c α2 / (1 − α)

Measured Λ_m goes on top; for tiny α, K_a ≈ cα2.

  • α is the fraction of the weak electrolyte that has dissociated.
  • Example: Λ_m = 39.1, Λ°_m = 391 → α = 0.1 (10% dissociated).
  • Feed α into K_a = cα2/(1−α) for the dissociation constant.
🔋Core device5/10

Galvanic Cells & EMF

A spontaneous redox reaction turned into electricity by separating the two half-reactions.

E°_cell = E°_cathode − E°_anode (both as reduction potentials)

Anode = oxidation (−); cathode = reduction (+). SHE is defined as 0 V.

  • Notation: Zn | Zn2⁺ || Cu2⁺ | Cu; || is the salt bridge keeping neutrality.
  • Daniell cell: 0.34 − (−0.76) = 1.10 V, positive → spontaneous.
  • E° is intensive — multiplying the equation does NOT change E°.
📊Reference6/10

Electrochemical Series

Electrodes ranked by standard reduction potential decide who reduces and who oxidises.

F2 (+2.87 V) strongest oxidiser · Li⁺/Li (−3.05 V) strongest reducer

Higher E° = stronger oxidising agent → that species is reduced (cathode).

  • More positive E° → species grabs electrons → becomes the cathode.
  • Most negative E° → gives electrons most easily → strongest reducing agent.
  • Series predicts feasibility, displacement and which ion discharges first.
📉Master equation7/10

Nernst Equation

Corrects cell EMF for real, non-standard concentrations at 298 K.

E_cell = E°_cell − (0.0591/n) log Q

n = electrons from the FULL balanced reaction; Q = [products]/[reactants].

  • EMF rises with more reactant ion, falls as products build up.
  • 0.0591 = 2.303RT/F at 298 K only — invalid at other temperatures.
  • Single electrode: E = E° − (0.0591/n) log(1/[Mⁿ⁺]).
♾️Thermodynamics8/10

ΔG°, K and EMF

Voltage, free energy and equilibrium are one consistent story.

ΔG° = −nFE°_cell · E°_cell = (0.0591/n) log K_c

F = 96500 C mol⁻1; at equilibrium E_cell = 0 and Q = K_c.

  • Positive E°_cell ⇒ negative ΔG° ⇒ spontaneous ⇒ K > 1.
  • At equilibrium the cell is dead: E_cell = 0, Q = K_c.
  • Lets you compute K, ΔG° or Ksp straight from a voltage.
⚙️Quantitative9/10

Electrolysis & Faraday’s Laws

External electricity forces a non-spontaneous redox reaction; charge fixes the mass deposited.

w = M I t / (n F) · 1 F = 96500 C = 1 mol e⁻

In electrolysis cathode is NEGATIVE, anode positive — but anode is still oxidation. t in seconds.

  • First law: w ∝ charge, Q = I t; moles of e⁻ = It/F.
  • Second law: same charge → masses ∝ equivalent masses (M/n).
  • Selective discharge: higher reduction potential discharges first; overpotential can flip it (brine → Cl2).
🛡️Real world10/10

Batteries, Fuel Cells & Corrosion

Where galvanic cells power devices, and where unwanted ones rust iron away.

H2–O2 fuel cell: 2H2 + O2 → 2H2O · η = ΔG/ΔH

Primary = single-use; secondary = rechargeable (lead storage, Ni-Cd).

  • Fuel cell is fed reactant continuously; η = ΔG/ΔH can approach 100%.
  • Rusting is electrochemical: Fe → Fe2⁺ at anode, O2 reduced at cathode → Fe2O3·xH2O.
  • Sacrificial protection: a more negative-E° metal (Zn −0.76 V, Mg) corrodes instead of Fe (−0.44 V).
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📝 Practice Electrochemistry — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
The molar conductivity of a 0.007 M acetic acid solution is 20 S cm² mol⁻¹. Given λ°(H⁺) = 350 and λ°(CH₃COO⁻) = 50 S cm² mol⁻¹, the dissociation constant Ka of acetic acid is:
Correct answer: C. Λ°m = λ°(H⁺) + λ°(CH₃COO⁻) = 350 + 50 = 400. α = Λm/Λ°m = 20/400 = 0.05. Ka = cα²/(1−α) ≈ cα² = 0.007 × (0.05)² = 0.007 × 2.5×10⁻³ = 1.75×10⁻⁵ mol L⁻¹.
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Q2NEET 2021
The molar conductance of NaCl, HCl and CH₃COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm² mol⁻¹ respectively. The molar conductance of CH₃COOH at infinite dilution is:
Correct answer: B. By Kohlrausch’s law, Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl) = 91.0 + 426.16 − 126.45 = 390.71 S cm² mol⁻¹ (the Na⁺ and Cl⁻ cancel, leaving H⁺ + CH₃COO⁻).
🔎 See the full step-by-step solution in the app →
Q3NEET 2020
The number of Faradays (F) required to produce 20 g of calcium from molten CaCl₂ (atomic mass of Ca = 40 g mol⁻¹) is:
Correct answer: D. Ca²⁺ + 2e⁻ → Ca, so 2 F deposit 1 mol (40 g) Ca. Moles of Ca = 20/40 = 0.5. Charge = 0.5 × 2 = 1 F.
🔎 See the full step-by-step solution in the app →
Q4NEET 2020
In a typical fuel cell, the reactants (R) and product (P) are:
Correct answer: B. In the standard H₂–O₂ fuel cell, hydrogen and oxygen are the reactants and water is the product: 2H₂ + O₂ → 2H₂O(l). The chemical energy is converted directly into electrical energy.
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Q5NEET 2019
The standard electrode potentials E° of Al³⁺/Al, Ag⁺/Ag, K⁺/K and Cr³⁺/Cr are −1.66 V, 0.80 V, −2.93 V and −0.74 V respectively. The correct order of decreasing reducing power of these metals is:
Correct answer: B. Reducing power increases as the standard reduction potential becomes more negative (easier to oxidise). Order of E°: K(−2.93) < Al(−1.66) < Cr(−0.74) < Ag(0.80), so decreasing reducing power is K > Al > Cr > Ag.
🔎 See the full step-by-step solution in the app →
Q6NEET 2019
For the cell reaction 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq), E°cell = 0.24 V at 298 K. The standard Gibbs energy (ΔrG°) of the cell reaction is (F = 96500 C mol⁻¹):
Correct answer: D. n = 2 electrons transferred. ΔrG° = −nFE°cell = −2 × 96500 × 0.24 = −46320 J = −46.32 kJ mol⁻¹.
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Q7NEET 2017
In the electrochemical cell Zn | ZnSO₄(0.01 M) || CuSO₄(1.0 M) | Cu, the EMF is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ to 0.01 M, the EMF becomes E₂. The relationship between E₁ and E₂ is (RT/F = 0.059):
Correct answer: C. E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]). For E₁: log(0.01/1) = −2, so E₁ = E° + 0.059. For E₂: log(1/0.01) = +2, so E₂ = E° − 0.059. Therefore E₁ > E₂.
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Q8NEET 2016
The molar conductivity of a 0.5 M aqueous AgNO₃ solution with electrolytic conductivity 5.76 × 10⁻³ S cm⁻¹ at 298 K is:
Correct answer: B. Λm = κ × 1000 / c = (5.76×10⁻³ × 1000) / 0.5 = 5.76 / 0.5 = 11.52 S cm² mol⁻¹.
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Q9NEET 2011
Given Cu²⁺ + e⁻ → Cu⁺, E° = +0.15 V and Cu⁺ + e⁻ → Cu, E° = +0.50 V. The standard electrode potential E° for Cu²⁺ + 2e⁻ → Cu is:
Correct answer: A. Electrode potentials are not additive, but ΔG° is. ΔG°(Cu²⁺→Cu) = ΔG°₁ + ΔG°₂ ⇒ −2FE° = −1F(0.15) − 1F(0.50). So E° = (0.15 + 0.50)/2 = 0.325 V.
🔎 See the full step-by-step solution in the app →
Q10NEET 2010
An increase in the equivalent conductance of a strong electrolyte with dilution is mainly due to:
Correct answer: A. A strong electrolyte is already ~fully ionised, so dilution does not appreciably increase the NUMBER of ions. What rises is ionic mobility: with fewer ions per unit volume the inter-ionic attraction (drag) falls, so each ion moves more freely, raising equivalent conductance.
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Frequently Asked Questions

What is a galvanic cell in electrochemistry?

A galvanic (voltaic) cell converts the chemical energy of a spontaneous redox reaction into electrical energy. Oxidation happens at the anode (negative terminal) and reduction at the cathode (positive terminal), with a salt bridge completing the circuit and keeping the solutions electrically neutral.

What is the Nernst equation and when do you use it?

The Nernst equation, E_cell = E°_cell minus (0.0591/n) log Q at 298 K, corrects the cell EMF for real, non-standard ion concentrations, where n is the number of electrons transferred and Q is the reaction quotient. The 0.0591 factor equals 2.303RT/F and is valid only at 298 K.

How do you calculate molar conductivity from conductivity?

Molar conductivity is Lambda_m = (kappa times 1000) divided by c, where kappa is the conductivity in S per cm and c is the concentration in mol per litre. On dilution kappa decreases because there are fewer ions per unit volume, but molar conductivity increases as the ions spread apart and conduct more freely.

Is Electrochemistry important for NEET and how is it weighted?

Yes, Electrochemistry is part of the NEET Class 12 Chemistry syllabus and is a high-scoring, numerical-rich chapter that usually contributes about one to two questions each year. Faraday’s laws, the Nernst equation and the link between E°cell, Gibbs energy and the equilibrium constant are the most frequently tested ideas.

What is the difference between a galvanic cell and an electrolytic cell?

A galvanic cell uses a spontaneous redox reaction (negative delta G) to produce electricity, while an electrolytic cell uses an external power source to force a non-spontaneous reaction. In a galvanic cell the anode is negative and cathode positive, whereas in electrolysis the cathode is negative and the anode positive, though the anode is still the site of oxidation in both.

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