Electrochemistry

Electrochemistry ties thermodynamics to redox and conduction. Advanced tests the Nernst equation, cell/electrode potentials, electrolysis with Faraday's laws, and conductance (Kohlrausch). Getting signs and n (electrons) right is the whole game.

Key results

• Cell emf: E°cell = E°cathode − E°anode; spontaneous if E°cell > 0. ΔG° = −nFE°.
• Nernst (298 K): E = E° − (0.059/n) log Q. At equilibrium E = 0, so E° = (0.059/n) log K.
• Electrolysis (Faraday): mass deposited m = (E·I·t)/F, where E = equivalent weight; 1 F = 96500 C = 1 mole of electrons.
• Conductance: molar conductivity Λm = κ×1000/M; Kohlrausch Λm° = Σ ionic contributions; Λm rises on dilution.
• Batteries/corrosion: rusting is electrochemical; standard cells (Daniell, fuel cells) apply the above.

Quantity	Relation
Cell emf	E°cat − E°an
Nernst	E° − (0.059/n)log Q
E° and K	(0.059/n)log K
Faraday deposition	EIt/F

Exam Tricks & Tips

• 🎯 In the Nernst equation, n is the electrons transferred in the balanced cell reaction — balance first, then plug in.
• 🎯 E° = (0.059/n) log K links emf to equilibrium: a positive E° means K > 1 and a spontaneous cell.
• 🎯 Faraday's law: moles deposited = (I·t)/(nF); the same charge deposits equivalents in the same ratio across cells in series.
• 🎯 Molar conductivity increases on dilution (ions move more freely); strong electrolytes rise slightly, weak ones sharply (more dissociation).
• 🎯 Use standard reduction potentials directly — the more positive one is the cathode; do not flip signs when subtracting.
• ❌ Common mistake: using 0.059/n with the wrong n or forgetting that Q uses activities/concentrations of the actual cell reaction, not half-reactions separately.

Expected exam pattern

1–2 questions: a Nernst/emf or Faraday-electrolysis numerical and a conductance (Kohlrausch) or standard-potential conceptual. Series-cell charge problems recur in integer format.

Quick recap

E°cell = E°cat − E°an; ΔG° = −nFE°. Nernst E = E° − (0.059/n)log Q; E° = (0.059/n)log K. Faraday m = EIt/F, 1 F = 96500 C. Molar conductivity rises on dilution.

Cover the answer, recall, then check. 12 cards on electrochemistry for JEE Advanced.

Q1. Standard cell potential from electrode potentials.
A1. E°cell = E°cathode − E°anode (reduction potentials).

Q2. Nernst equation at 298 K.
A2. E = E° − (0.059/n) log Q.

Q3. Relation between E° and the equilibrium constant.
A3. E° = (0.059/n) log K.

Q4. Faraday's first law for mass deposited.
A4. m = (E·I·t)/F, E = equivalent weight; 1 F = 96500 C.

Q5. Charge carried by one mole of electrons.
A5. 1 faraday = 96500 C.

Q6. How does molar conductivity change with dilution?
A6. It increases (ions are freer); weak electrolytes rise sharply due to more dissociation.

Q7. Kohlrausch's law of independent ion migration.
A7. Λm° = sum of the limiting molar conductivities of the cation and anion.

Q8. Link between ΔG° and cell emf.
A8. ΔG° = −nFE°.

Q9. Which electrode is the cathode in a galvanic cell?
A9. The one with the higher (more positive) reduction potential; reduction occurs there.

Q10. Sign of E°cell for a spontaneous cell.
A10. Positive (E°cell > 0).

Q11. In series electrolytic cells, how are amounts deposited related?
A11. The same charge passes, so amounts are in the ratio of their equivalent weights.

Q12. Molar conductivity from conductivity κ and molarity M.
A12. Λm = κ × 1000 / M.

Electrochemistry turns chemical energy into electrical (galvanic cells) and back (electrolysis). JEE Advanced tests the Nernst equation, standard-potential reasoning, conductance, and Faraday's laws — a topic that blends redox bookkeeping with thermodynamics.

Core concept: redox reactions transfer electrons; a galvanic cell harnesses a spontaneous reaction to produce EMF, and E°cell links to ΔG° and the equilibrium constant.

Galvanic cells

Beginner — cell EMF and conventions

A galvanic (voltaic) cell has oxidation at the anode (−) and reduction at the cathode (+). E°cell = E°cathode − E°anode (both as reduction potentials). A positive E°cell means the reaction is spontaneous. Cell notation: anode | anode solution || cathode solution | cathode. The standard hydrogen electrode (SHE) defines the zero of the scale.

Intermediate — Nernst equation and thermodynamic links

Non-standard conditions shift the potential: Nernst E = E° − (0.0591/n)log Q at 25 °C, where n = electrons transferred. Key links: ΔG° = −nFE°cell (F = 96500 C/mol), and E°cell = (0.0591/n)log K at equilibrium (when E = 0, Q = K). Concentration cells (same electrodes, different concentrations) have E° = 0 but nonzero E from the concentration difference.

Advanced — conductance and electrolysis

Molar conductivity Λ_m = κ×1000/C rises on dilution; for strong electrolytes it extrapolates to Λ°_m, while weak electrolytes obey Kohlrausch's law (Λ°_m = sum of ionic contributions), letting you find Λ°_m for acetic acid indirectly. Degree of dissociation α = Λ_m/Λ°_m. Faraday's laws of electrolysis: mass deposited ∝ charge passed, m = (Q/F)(M/n) where Q = It; equal charge deposits equivalents in proportion. Electrolysis products depend on electrode potentials and overpotential (e.g., why O₂ not Cl₂, or H₂ not Na, forms).

Worked example

Calculate the EMF of a Daniell cell at 25 °C with [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.01 M. Standard: E°cell = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn) = 0.34 − (−0.76) = 1.10 V. Reaction Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2, Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10. Nernst: E = 1.10 − (0.0591/2)log10 = 1.10 − 0.0296 = 1.07 V. The lowered [Cu²⁺] reduces the driving force slightly, exactly as Le Chatelier predicts — the Nernst equation quantifies it.

How JEE Advanced tests this

Nernst-equation numericals and concentration cells; predicting spontaneity and feasibility from E° values; ΔG° and K from E°cell; Kohlrausch's law and degree of dissociation of weak electrolytes; Faraday's-law electrolysis mass/charge calculations; and identifying electrolysis products.

Exam tricks & shortcuts

• E°cell = E°cathode − E°anode (reduction potentials); positive ⇒ spontaneous.
• Nernst at 25 °C: E = E° − (0.0591/n)log Q.
• ΔG° = −nFE°; log K = nE°/0.0591.
• Faraday: 1 mole electrons = 96500 C = 1 F; m = (ItM)/(nF).
• Mnemonic: "An Ox, Red Cat" (Anode Oxidation, Reduction Cathode).

Key formulas

• Cell emf: E°_cell = E°_cathode − E°_anode; ΔG° = −nFE° (F = 96500).
• Nernst: E = E° − (0.059/n) log Q; at equilibrium E = 0, log K = nE°/0.059.
• Conductivity: κ = (1/R)(l/A); molar Λ_m = 1000κ/M.
• Kohlrausch: Λ°_m = ν₊λ°₊ + ν₋λ°₋; α = Λ_m/Λ°_m.
• Faraday: m = (M I t)/(nF); 1 F = 96500 C.

The Nernst equation gives emf at any concentration; Faraday's laws quantify electrolysis deposits.

Worked Example

Problem: A galvanic cell has a standard cell potential E° = 0.46 V and involves the transfer of n = 2 electrons. At 298 K, find (a) the standard Gibbs energy change and (b) the equilibrium constant K of the cell reaction. (F = 96500 C/mol.)

Solution:
(a) Standard Gibbs energy:
ΔG° = −nFE° = −(2)(96500)(0.46) = −88 780 J = −88.8 kJ.
The large negative value shows the reaction is strongly product-favoured.

(b) Relate E° to K using ΔG° = −RT ln K = −nFE°, or the convenient form at 298 K:
log K = nE°/0.059 = (2)(0.46)/0.059 = 0.92/0.059 = 15.6.
K = 10^15.6 ≈ 3.9 × 10¹⁵.

Answer: (a) ΔG° ≈ −88.8 kJ; (b) K ≈ 3.9 × 10¹⁵.