Electrochemistry for NEET UG

Electrochemistry links chemical reactions with electricity. It explains batteries, corrosion, electroplating and how we measure the strength of oxidising and reducing agents.

ChemistryClass 1275 MCQs3 lessons14 flashcards3 memory tricks

Practise Electrochemistry free

What you'll learn

  1. Galvanic cells and electrode potentials
    Find the EMF of a cell from standard electrode potentials and write the cell notation.
  2. The Nernst equation
    Calculate electrode and cell potentials at non-standard concentrations.
  3. Conductance and electrolysis
    Use molar conductivity, Kohlrausch's law and Faraday's laws of electrolysis.

Key formulas

Cell EMFE°cell = E°cathode − E°anode (both as reduction potentials)
SpontaneousE°cell > 0
Link to Gibbs energyΔG° = −nFE°cell, F = 96,500 C/mol
Nernst equation (25 °C)E = E° − (0.0591/n) log Q
At equilibriumE = 0, so E° = (0.0591/n) log K
Conductivityκ = (1/R)(l/A)
Molar conductivityΛm = κ × 1000/C (S cm² mol⁻¹)
Kohlrausch's lawΛ°m = ν₊λ°₊ + ν₋λ°₋

Memory tricks

Which electrode does whatAn Ox, Red Cat.

Oxidation at the Anode, Reduction at the Cathode, in every kind of cell.

Electrochemical seriesFluorine is the greediest for electrons; lithium is the most generous.

Higher standard reduction potential means a stronger oxidising agent: F₂ (+2.87 V). Most negative means the strongest reducing agent: Li (−3.04 V).

Rusting and galvanisingIron rusts faster in Mumbai and Chennai than in Jaipur. Zinc coating sacrifices itself to save the iron.

Salty, humid air is a better electrolyte, so corrosion is faster. Zinc is more reactive than iron, so it corrodes first (sacrificial protection).

Try 3 NEET-style questions

Q1. The charge required to deposit 1 mol of copper from Cu²⁺ solution is

  1. 1 F
  2. 2 F
  3. 3 F
  4. 0.5 F
Show answer

Answer: (2) 2 F

Cu²⁺ + 2e⁻ → Cu, so 2 mol of electrons = 2 F.

Q2. In the corrosion (rusting) of iron, the reaction at the cathodic region is

  1. Fe(s) → Fe²⁺(aq) + 2e⁻
  2. Fe²⁺ → Fe³⁺ + e⁻ only
  3. 2H₂O → O₂ + 4H⁺ + 4e⁻
  4. O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)
Show answer

Answer: (4) O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)

At the anodic region, iron is oxidised to Fe²⁺. Electrons travel to the cathodic region, where oxygen is reduced in the presence of H⁺: O₂ + 4H⁺ + 4e⁻ → 2H₂O.

Q3. The standard hydrogen electrode is represented as

  1. Zn(s) | Zn²⁺(aq, 1 M)
  2. Pt(s) | H₂(g, 1 bar) | H⁺(aq, 1 M)
  3. Cu(s) | Cu²⁺(aq, 1 M)
  4. Ag(s) | AgCl(s) | Cl⁻(aq)
Show answer

Answer: (2) Pt(s) | H₂(g, 1 bar) | H⁺(aq, 1 M)

The SHE has a platinum electrode coated with platinum black, in contact with H₂ gas at 1 bar and H⁺ ions at 1 M.

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