JEE Chem — Electrochemistry Deep Dive

Chapter Overview & Weightage

Electrochemistry is a high-yield JEE chapter that combines thermodynamics, kinetics, and equilibrium. JEE Main typically asks $2-3$ questions, JEE Advanced occasionally adds a multi-step problem combining electrolysis with stoichiometry.

Typical JEE weightage: $2-3$ questions in JEE Main, $1-2$ in JEE Advanced.

Year	JEE Main Qs	JEE Advanced Qs
2020	2	1
2021	3	2
2022	2	1
2023	3	1
2024	3	2

Key Concepts You Must Know

Galvanic vs electrolytic cells
Standard electrode potentials and the EMF of a cell
Nernst equation
Faraday’s laws of electrolysis
Conductance: specific, equivalent, molar
Kohlrausch’s law and limiting molar conductivity
Electrochemical series and its applications
Relationship $\Delta G = -nFE$ and $\ln K = nFE^\circ/RT$
Batteries: lead-acid, dry cell, fuel cell

Important Formulas

$E = E^\circ - \frac{0.0591}{n}\log Q \quad \text{(at 298 K)}$

where $Q$ is the reaction quotient and $n$ is the number of electrons transferred.

$\Delta G = -nFE, \quad \Delta G^\circ = -nFE^\circ = -RT\ln K$

$m = ZIt = \frac{MIt}{nF}$

where $m$ is mass deposited, $Z$ is electrochemical equivalent, $I$ is current, $t$ is time, $M$ is molar mass.

$\Lambda_m = \frac{\kappa \times 1000}{c} \,\text{(S cm}^2\text{/mol)}$

with $\kappa$ in S/cm and $c$ in mol/L.

Solved Previous Year Questions

PYQ 1 (JEE Main 2023)

Calculate the EMF of the cell $Zn|Zn^{2+}(0.1\,M) || Cu^{2+}(1.0\,M)|Cu$ . Given $E^\circ_{Zn^{2+}/Zn} = -0.76$ V, $E^\circ_{Cu^{2+}/Cu} = +0.34$ V.

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10$ V.

Apply Nernst with $n = 2$ , $Q = [Zn^{2+}]/[Cu^{2+}] = 0.1/1.0 = 0.1$ .

$E = 1.10 - (0.0591/2)\log 0.1 = 1.10 + 0.0296 = 1.13$ V.

PYQ 2 (JEE Main 2024)

Current of $5$ A passes through molten $AlCl_3$ for $30$ minutes. Find the mass of Al deposited. ( $M_{Al} = 27$ g/mol, $F = 96500$ C)

$Q = It = 5 \times 1800 = 9000$ C.

Moles of electrons = $9000/96500 = 0.0933$ .

Moles of Al = $0.0933/3 = 0.0311$ (since $Al^{3+} + 3e^- \to Al$ ).

Mass = $0.0311 \times 27 = 0.84$ g.

PYQ 3 (JEE Advanced 2022)

For a cell at equilibrium, $E = 0$ . Use this to derive the relation between $E^\circ$ and equilibrium constant $K$ .

At equilibrium: $0 = E^\circ - (0.0591/n)\log K$ .

$\log K = nE^\circ/0.0591$ , or equivalently $\Delta G^\circ = -RT\ln K = -nFE^\circ$ .

Difficulty Distribution

Difficulty	% of JEE Qs	Typical type
Easy	$25\%$	Direct EMF or Faraday plug-ins
Medium	$55\%$	Nernst with non-standard concentrations, mixed electrolyses
Hard	$20\%$	Multi-step problems (electrochem + stoichiometry + buffer)

Expert Strategy

Memorise the electrochemical series order for at least 10 common metals. JEE often asks “which metal can displace which from solution?” — direct application of the series.

For Nernst, always note $n$ from the balanced cell equation. Half-reactions in the series are written for one electron — multiply through to balance the cell.

Use the shortcut $\log K = nE^\circ/0.0591$ at $298$ K. Faster than going through $\Delta G$ for equilibrium-constant questions.

Common Traps

Confusing $E^\circ_{\text{cell}}$ sign convention. $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ where both are reduction potentials. Some students subtract anode oxidation potential — same answer, but easier to mess up signs.

Using $\log Q$ when $Q$ is dimensionally inconsistent (e.g., concentrations in different units). Always use mol/L.

Forgetting to convert minutes/hours to seconds in Faraday’s law. $Q = It$ requires $t$ in seconds.

Treating molten electrolyte and aqueous solution the same. Aqueous electrolysis can deposit $H_2/O_2$ from water instead of the salt’s ions. Always check the standard potentials.