JEE Chem — Redox Reactions

Chapter Overview & Weightage

Redox is foundational — appears directly in JEE Main and indirectly in Electrochemistry, Inorganic Qualitative Analysis, and Stoichiometry. The chapter itself is short, but its concepts power 12-15 marks across the JEE Chemistry paper.

Year	Direct Qs (Main)	Indirect Qs
2024	1	4
2023	2	3
2022	1	5
2021	2	4
2020	1	4

Redox is the chapter you absolutely cannot skip. Mastering oxidation numbers and balancing reactions in acidic/basic media unlocks easy marks across electrochemistry and inorganic.

Key Concepts You Must Know

Oxidation number rules: free element 0, group 1 metal +1, group 2 +2, F always -1, O usually -2 (peroxides -1, OF $_2$ +2), H usually +1 (hydrides -1).
Oxidation vs reduction: loss of e $^-$ vs gain of e $^-$ .
Oxidising agent (oxidant): gets reduced; reducing agent gets oxidised.
Disproportionation: same species both oxidised and reduced. Example: $2\text{Cu}^+ \to \text{Cu} + \text{Cu}^{2+}$ .
Comproportionation: opposite of disproportionation.
Balancing in acidic medium: add $\text{H}^+$ and $\text{H}_2\text{O}$ .
Balancing in basic medium: add $\text{OH}^-$ and $\text{H}_2\text{O}$ .
Equivalent concept: $n$ -factor for redox = electrons transferred per formula unit.

Important Formulas

$\text{Eq. wt.} = \frac{\text{Molar mass}}{n\text{-factor}}$

For redox, $n$ -factor = total change in oxidation number per formula unit.

Agent	Half-reaction	$n$ -factor
KMnO $_4$ (acidic)	$\text{Mn}^{7+} \to \text{Mn}^{2+}$	5
KMnO $_4$ (neutral/basic)	$\text{Mn}^{7+} \to \text{Mn}^{4+}$	3
K $_2$ Cr $_2$ O $_7$ (acidic)	$\text{Cr}^{6+} \to \text{Cr}^{3+}$	6
H $_2$ O $_2$ (oxidising)	$\text{O}^{-1} \to \text{O}^{-2}$	2
H $_2$ O $_2$ (reducing)	$\text{O}^{-1} \to \text{O}^{0}$	2

Solved Previous Year Questions

PYQ 1 (JEE Main 2024)

Find the oxidation number of S in $\text{S}_2\text{O}_3^{2-}$ .

Let S = $x$ . $2x + 3(-2) = -2 \implies 2x = 4 \implies x = +2$ .

Note: this is the average. Two of the S atoms have different actual oxidation states ( $-2$ and $+6$ ), but for stoichiometric purposes the average $+2$ works.

PYQ 2 (JEE Advanced 2023)

Balance: $\text{MnO}_4^- + \text{C}_2\text{O}_4^{2-} \to \text{Mn}^{2+} + \text{CO}_2$ in acidic medium.

Half-reactions:

Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O}$
Oxidation: $\text{C}_2\text{O}_4^{2-} \to 2\text{CO}_2 + 2e^-$

Multiply first by 2, second by 5:

$2\text{MnO}_4^- + 16\text{H}^+ + 5\text{C}_2\text{O}_4^{2-} \to 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 10\text{CO}_2$

PYQ 3 (JEE Main 2022)

$2.5\text{ g}$ of an iron sample is dissolved and oxidised to Fe $^{3+}$ . The solution requires $25\text{ mL}$ of $0.1\text{ M}$ KMnO $_4$ in acidic medium for titration. Find % Fe in the sample.

Half-reactions: Fe $^{2+} \to$ Fe $^{3+}$ ( $n = 1$ ); Mn $^{7+} \to$ Mn $^{2+}$ ( $n = 5$ ).

Moles of KMnO $_4$ = $0.025 \times 0.1 = 2.5 \times 10^{-3}$ . Equivalents = $5 \times 2.5\times 10^{-3} = 0.0125$ .

Same equivalents of Fe = 0.0125. Mass of Fe = $0.0125 \times 56 = 0.7\text{ g}$ .

% Fe = $(0.7/2.5) \times 100 = 28\%$ .

Difficulty Distribution

Sub-topic	Easy	Medium	Hard
Oxidation numbers	70%	25%	5%
Balancing	30%	50%	20%
$n$ -factor	40%	40%	20%
Titrations	20%	50%	30%

Expert Strategy

Memorise the standard $n$ -factors for KMnO $_4$ , K $_2$ Cr $_2$ O $_7$ , and H $_2$ O $_2$ . Half of redox titration MCQs are solved in 30 seconds with these.

For balancing, always do half-reactions separately. Never try to balance the molecular equation directly — too many errors.

When stuck, fall back to “equivalents on both sides are equal.” This single principle solves stoichiometry without molar arithmetic.

Common Traps

Wrong $n$ -factor for KMnO $_4$ in neutral/basic medium. It is 3 (Mn $^{7+} \to$ Mn $^{4+}$ , MnO $_2$ formed), not 5. JEE often specifies “neutral medium” to test this.

For peroxides, oxygen is $-1$ , not $-2$ . Mistakes here cascade into wrong $n$ -factor and wrong product.

Forgetting to add $\text{H}_2\text{O}$ when balancing — students balance atoms but forget the medium. Always check that both sides have equal H and O atoms.