JEE Chemistry — Stoichiometry and Redox Complete Chapter Guide

Chapter Overview & Weightage

Stoichiometry and Redox together form the calculation backbone of chemistry. Mole concept questions test your numerical agility, while redox balancing tests your understanding of electron transfer. Both are essential — and both appear every single year.

Combined weightage is 4-5% in JEE Main. Expect 1-2 calculation-heavy questions from stoichiometry and 1 question from redox reactions per paper. JEE Advanced integrates redox into electrochemistry and analytical chemistry problems.

Year	JEE Main (Q count)	Key Topics Tested
2024	2	Limiting reagent, equivalent weight in redox
2023	2	n-factor determination, % composition
2022	2	Back titration, balancing redox in acidic medium
2021	1	Mole concept with gas law, disproportionation
2020	2	Equivalent weight of oxidising agent, mole ratio

Key Concepts You Must Know

Tier 1 (Core calculation skills)

Mole concept: $n = \frac{m}{M} = \frac{V(L)}{22.4} \text{ (at STP)} = \frac{N}{N_A}$
Molarity ( $M$ ), molality ( $m$ ), normality ( $N$ ), mole fraction
Limiting reagent identification
% composition, empirical and molecular formula
Equivalent weight = $\frac{\text{Molecular weight}}{n\text{-factor}}$

Tier 2 (Redox fundamentals)

Oxidation state rules and calculation for complex species
Identifying oxidising agent (gets reduced) and reducing agent (gets oxidised)
n-factor: change in oxidation state per atom $\times$ number of such atoms
Balancing redox reactions: ion-electron method (half-reaction method)

Tier 3 (Advanced applications)

Back titration and double indicator titration
Disproportionation and comproportionation reactions
Equivalent weight in reactions where the same substance acts differently

Important Formulas

n = \frac{m}{M} \quad \text{(mass and molar mass)}

n = \frac{PV}{RT} \quad \text{(ideal gas at any T, P)}

\text{At STP: } n = \frac{V(\text{L})}{22.4}

\text{Number of particles} = n \times N_A \quad (N_A = 6.022 \times 10^{23})

Term	Formula	Units
Molarity ( $M$ )	$\frac{\text{moles of solute}}{\text{volume of solution (L)}}$	mol/L
Molality ( $m$ )	$\frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$	mol/kg
Normality ( $N$ )	$\frac{\text{gram equivalents}}{\text{volume of solution (L)}}$	eq/L
Relation	$N = M \times n\text{-factor}$	—

For acids: n-factor = number of $H^+$ ions donated (basicity)

For bases: n-factor = number of $OH^-$ ions donated (acidity)

For redox reactions: n-factor = total change in oxidation state per formula unit

Example: In $KMnO_4$ (acidic medium), Mn goes from +7 to +2. n-factor = 5.

Example: In $K_2Cr_2O_7$ (acidic medium), each Cr goes from +6 to +3, and there are 2 Cr atoms. n-factor = $2 \times 3 = 6$ .

At the equivalence point of any reaction:

N_1 V_1 = N_2 V_2

or equivalently:

n_1 \times \text{n-factor}_1 = n_2 \times \text{n-factor}_2

This works for acid-base, redox, and precipitation reactions.

The n-factor approach is the fastest way to solve titration problems in JEE. Instead of writing and balancing the full reaction, just determine the n-factor of each reactant and apply the law of equivalence directly.

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 (January, Shift 1)

Problem: 25 mL of 0.1 M $KMnO_4$ in acidic medium is required to completely react with 25 mL of $FeSO_4$ solution. Find the molarity of $FeSO_4$ .

Solution:

In acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

n-factor of $KMnO_4$ = 5 (Mn: +7 → +2)

$Fe^{2+} \rightarrow Fe^{3+} + e^-$

n-factor of $FeSO_4$ = 1

Using law of equivalence:

M_1 \times n_1 \times V_1 = M_2 \times n_2 \times V_2

0.1 \times 5 \times 25 = M_2 \times 1 \times 25

M_2 = \mathbf{0.5 \text{ M}}

PYQ 2 — JEE Main 2023 (April, Shift 2)

Problem: Find the oxidation state of Cr in $K_2Cr_2O_7$ .

Solution:

Let oxidation state of Cr = $x$ .

$K_2Cr_2O_7$ : $2(+1) + 2(x) + 7(-2) = 0$

$2 + 2x - 14 = 0$

$2x = 12$

$x = \mathbf{+6}$

A common error is forgetting the subscript. There are 2 Cr atoms, so the equation is $2x$ , not just $x$ . Similarly for $S$ in $Na_2S_4O_6$ (tetrathionate), there are 4 sulfur atoms, but they are NOT all in the same oxidation state — this is an advanced trap.

PYQ 3 — JEE Main 2022 (June, Shift 1)

Problem: 10 g of a mixture of $Na_2CO_3$ and $NaHCO_3$ requires 150 mL of 1 M HCl for complete reaction. Find the percentage of $Na_2CO_3$ in the mixture.

Solution:

Let mass of $Na_2CO_3$ = $a$ g, mass of $NaHCO_3$ = $(10 - a)$ g.

$Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$ (n-factor = 2)

$NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$ (n-factor = 1)

Moles of HCl = $0.15 \times 1 = 0.15$ mol

\frac{2a}{106} + \frac{10 - a}{84} = 0.15

\frac{2a}{106} + \frac{10 - a}{84} = 0.15

Multiply through by $106 \times 84 = 8904$ :

2a \times 84 + (10 - a) \times 106 = 0.15 \times 8904

168a + 1060 - 106a = 1335.6

62a = 275.6

a = 4.45 \text{ g}

% of $Na_2CO_3$ = $\frac{4.45}{10} \times 100 = \mathbf{44.5\%}$

Difficulty Distribution

Difficulty	% of Questions	What to Expect
Easy	30%	Oxidation state calculation, direct mole concept
Medium	50%	Limiting reagent, n-factor determination, titration calculation
Hard	20%	Back titration, mixture analysis, disproportionation n-factor

Expert Strategy

Week 1: Drill the mole concept until unit conversions are instant. You should convert between grams, moles, particles, and volume (for gases) without thinking. Speed here directly translates to more time for harder questions.

Week 2: Master n-factor determination. Make a table: common reagents ( $KMnO_4$ , $K_2Cr_2O_7$ , $H_2C_2O_4$ , $Na_2S_2O_3$ , $FeSO_4$ ) with their n-factors in acidic and basic media. This table gets used in every titration and equivalent weight problem.

Week 3: Solve mixture problems and back titrations. These are multi-step calculations where setting up the equations correctly is half the battle. Practice writing the equivalence equation before touching the calculator.

In JEE, stoichiometry questions often have tricky units — mixing mL with L, or giving mass in mg. Before calculating, convert everything to a consistent unit system. This single habit eliminates 80% of silly errors.

Common Traps

Trap 1 — Different n-factors in different media. $KMnO_4$ has n-factor 5 in acidic medium (Mn: +7 → +2), 3 in neutral medium (+7 → +4), and 1 in strongly basic medium (+7 → +6). Always check the medium specified in the question.

Trap 2 — Disproportionation reactions. When the same element is both oxidised and reduced (e.g., $Cl_2 + NaOH$ ), the n-factor calculation requires finding the fraction of atoms oxidised vs reduced, then using the appropriate change. Don’t apply the change to all atoms uniformly.

Trap 3 — STP conditions have changed. IUPAC defines STP as 273.15 K and 1 bar (not 1 atm). At 1 bar, molar volume is 22.7 L (not 22.4 L). However, most JEE problems still use the old STP (1 atm, 22.4 L). Read the question carefully for which convention is used.

Trap 4 — Molarity changes with temperature; molality does not. Since molarity depends on volume (which expands with temperature), it changes. Molality depends on mass of solvent, which doesn’t change. JEE sometimes asks which concentration term is temperature-independent.

Trap 5 — Equivalent weight in reactions where the product varies. The equivalent weight of $H_3PO_4$ is $M/1$ , $M/2$ , or $M/3$ depending on whether it forms $NaH_2PO_4$ , $Na_2HPO_4$ , or $Na_3PO_4$ . The n-factor depends on the reaction, not just the formula.