Loading…
Loading…
JEE Main · Chemistry · Class 12
Electrochemistry carries up to two of the 25 JEE Main Chemistry questions in our reading of past papers, and they cluster around three things: the Nernst equation, molar conductivity on dilution, and Faraday’s laws. Each is a plug-in question once the number of electrons transferred is right.
Quick answer
Questions per paper
0–2
Our analysis of past papers
Marks at stake
Up to 8
At +4 / −1 per question
Difficulty
Moderate
Numerical, rarely conceptual
Class
12
Pairs with chemical thermodynamics
BrainStride analysis of past papers
A galvanic cell turns a spontaneous reaction into current; an electrolytic cell uses current to force a non-spontaneous one. Oxidation is at the anode in both, but the anode is negative in a galvanic cell and positive in an electrolytic one.
Tabulated values are reduction potentials against the standard hydrogen electrode. E°_cell = E°_cathode − E°_anode, using both as reduction potentials — never reverse a sign and subtract as well.
Concentration changes shift the cell potential: E = E° − (0.0591/n) log Q at 298 K, where Q is the reaction quotient written for the balanced cell reaction. At equilibrium Q = K and E = 0, which is a flat battery.
ΔG = −nFE and ΔG° = −nFE° = −2.303RT log K. These two lines connect a measured voltage to spontaneity and to the equilibrium constant, and questions routinely ask you to travel between them.
Conductivity κ falls on dilution because there are fewer ions per unit volume. Molar conductivity Λ_m = 1000κ/c rises, because it counts the conduction per mole of electrolyte, and every ion is freer to move.
At infinite dilution each ion contributes independently: Λ°_m = ν₊λ°₊ + ν₋λ°₋. This is how Λ°_m for a weak acid is obtained from strong-electrolyte data, and it leads to α = Λ_m/Λ°_m and Ka = cα²/(1 − α).
E°_cell = E°_cathode − E°_anode
Both taken as reduction potentials.
E = E° − (0.0591/n) log Q
n = moles of electrons in the balanced cell reaction.
ΔG = −nFE · ΔG° = −nFE°
F = 96500 C mol⁻¹; a positive E means a spontaneous cell.
E°_cell = (0.0591/n) log K
From ΔG° = −2.303RT log K at 298 K.
κ = G × (l/A)
l/A is the cell constant, in cm⁻¹.
Λ_m = 1000 κ / c
κ in S cm⁻¹ and c in mol L⁻¹ give Λ_m in S cm² mol⁻¹.
Λ°_m = ν₊ λ°₊ + ν₋ λ°₋
Ions contribute independently at infinite dilution.
α = Λ_m / Λ°_m · Ka = cα²/(1 − α)
For weak electrolytes only.
m = (M / nF) × I × t
Charge Q = It; n is the electrons per ion discharged.
Λ_m = Λ°_m − A√c
Debye–Hückel–Onsager: a gentle, linear-in-√c rise on dilution.
Why it happens
n is read off the half-reaction rather than the balanced overall cell reaction.The fix
Balance the full cell reaction first, then count the electrons that actually cancel. For Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2, not 1.Why it happens
Students remember that the anode is oxidation and flip its tabulated value, then still use E°_cathode − E°_anode.The fix
Keep both values as reduction potentials and subtract once. Reverse the sign only if you are adding half-cell potentials instead.Why it happens
The reaction quotient is written from memory rather than from the balanced equation.The fix
Write Q = products/reactants for the cell reaction as balanced, with solids and pure liquids left out.Why it happens
Both are called “conductivity” in a hurry, and they move in opposite directions.The fix
κ falls on dilution (fewer ions per cm³), Λ_m rises (more conduction per mole). Say which one the question asks for before answering.Why it happens
The formula is remembered as m = MIt/F with n dropped.The fix
Use m = (M/nF)It. Depositing one mole of Cu from Cu²⁺ needs 2F, and one mole of Al from Al³⁺ needs 3F.5 questions
Question 1
How much charge is needed to deposit one mole of aluminium from molten Al₂O₃?
Question 2
When a galvanic cell reaches equilibrium, its cell potential is:
Question 3
On dilution, the molar conductivity of a weak electrolyte increases sharply mainly because:
Question 4
For the cell Zn | Zn²⁺ || Cu²⁺ | Cu, given E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V, the standard cell potential is:
Question 5
A 0.1 M solution has conductivity 0.0129 S cm⁻¹. Its molar conductivity is:
A test on one chapter alone tells you little — the exam never asks Electrochemistry in isolation. So this builds a full JEE Main Chemistry paper that leans on it. Weighted like the exam, a 25-question paper gives Electrochemistry 1 question; marked weak, it gets 2, and the rest of the paper keeps the other chapters honest. Fill the plan from previous-year questions or your module.
Your paper
Press Generate paper plan for a chapter-by-chapter blueprint: questions per chapter, numericals, marks and time — ready to print and fill from your PYQs, modules or the BrainStride question bank.
Chapter weights are BrainStride’s reading of past papers, not official data. For any other chapter mix, use the JEE Main Chemistry test generator.
Nernst equation · 3 marks · moderate
For the cell Zn(s) | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu(s), with E°_cell = 1.10 V, calculate the cell potential at 298 K.
Answer: E ≈ 1.07 V.
Faraday’s laws · 3 marks · easy
A current of 5 A is passed through a CuSO₄ solution for 30 minutes. Calculate the mass of copper deposited. (Cu = 63.5 g mol⁻¹)
Answer: About 2.96 g of copper is deposited.
Kohlrausch’s law · 4 marks · moderate
The molar conductivity of 0.01 M acetic acid is 16.3 S cm² mol⁻¹, and Λ°_m for acetic acid is 390.5 S cm² mol⁻¹. Find the degree of dissociation and the dissociation constant.
Answer: α ≈ 0.042 (4.2% dissociated); Ka ≈ 1.8 × 10⁻⁵.
In BrainStride’s analysis of past JEE Main papers, electrochemistry contributes up to about two of the 25 Chemistry questions, and it is grouped with chemical thermodynamics as a two-to-three question block. NTA publishes no official chapter weightage, so treat these as past-paper ranges rather than guarantees.
Taking n from a half-reaction instead of the balanced overall cell reaction. For Zn + Cu²⁺ → Zn²⁺ + Cu, n is 2, and using 1 doubles the correction term. Balance the full cell reaction before writing the Nernst equation, then read n from the electrons that cancel.
Conductivity κ counts ions in a fixed volume, so diluting a solution lowers it. Molar conductivity divides by concentration, Λ_m = 1000κ/c, and so counts conduction per mole of electrolyte. On dilution the ions move more freely and, for weak electrolytes, more of them exist, so Λ_m rises.
No. Questions supply the values they need. What you must know cold is how to use them: E°_cell = E°_cathode − E°_anode with both as reduction potentials, and that the species with the higher reduction potential is the one reduced.
Through ΔG = −nFE. A question can give a cell potential and ask for the free energy change, or give ΔG° and ask for the equilibrium constant via ΔG° = −2.303RT log K. Practising both directions once removes the hesitation that costs time in the paper.
The 20-minute diagnostic tests this chapter with the rest of physical chemistry and names the mistake pattern behind each wrong answer.