Class 11 — Redox Reactions

Chapter Overview & Weightage

Redox Reactions is one of the most scoring chapters in Class 11 chemistry. The concepts (oxidation states, balancing equations, electron transfer) carry directly into electrochemistry in Class 12 — so investing here pays double.

Year	Marks
2024	6
2023	5
2022	7
2021	5

Typical mix: 1-mark MCQ on oxidation states, 2-mark balancing using oxidation number method, 3-mark balancing using ion-electron method, occasional 5-mark mixed question.

Key Concepts You Must Know

Oxidation: loss of electrons / increase in oxidation state.
Reduction: gain of electrons / decrease in oxidation state.
Oxidising agent: gets reduced (gains electrons), causes oxidation in others.
Reducing agent: gets oxidised (loses electrons), causes reduction in others.
Disproportionation: same element gets both oxidised and reduced (e.g., $\text{Cl}_2 + \text{NaOH}$ ).
Comproportionation: opposite — two species of same element with different oxidation states combine to give a single intermediate state.

Important Formulas

O is usually $-2$ (except in peroxides $-1$ , superoxides $-1/2$ , $\text{OF}_2$ where it’s $+2$ ).
H is $+1$ with non-metals, $-1$ with metals.
Halogens are $-1$ (except when bonded to more electronegative elements).
Sum of oxidation states = charge on species.

Write skeletal half-reactions for oxidation and reduction.
Balance atoms other than O and H.
Balance O by adding H₂O.
Balance H by adding H⁺ (acidic) or H₂O / OH⁻ (basic).
Balance charge by adding electrons.
Multiply half-reactions to equalise electrons.
Add and cancel common species.

Solved Previous Year Questions

PYQ 1 (CBSE 2023, 3 marks)

Balance the following equation in acidic medium using ion-electron method:

$\text{MnO}_4^- + \text{Fe}^{2+} \to \text{Mn}^{2+} + \text{Fe}^{3+}$

Solution:

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

Oxidation half: $\text{Fe}^{2+} \to \text{Fe}^{3+} + e^-$

Multiply oxidation by 5:

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$

PYQ 2 (CBSE 2022, 2 marks)

Determine the oxidation state of Mn in $\text{KMnO}_4$ and S in $\text{H}_2\text{SO}_4$ .

Solution:

$\text{KMnO}_4$ : K is +1, O is $-2$ each. So $1 + x + 4(-2) = 0 \implies x = +7$ .

$\text{H}_2\text{SO}_4$ : H is +1 each, O is $-2$ each. So $2(1) + x + 4(-2) = 0 \implies x = +6$ .

PYQ 3 (CBSE 2024, 5 marks)

Identify the oxidising and reducing agents in the reaction:

$3\text{Cl}_2 + 6\text{NaOH} \to 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}$

Solution: Cl in $\text{Cl}_2$ is 0. In NaCl, Cl is $-1$ (reduced). In $\text{NaClO}_3$ , Cl is $+5$ (oxidised).

Same element ( $\text{Cl}_2$ ) acts as both oxidising and reducing agent → disproportionation.

Difficulty Distribution

Easy (50%): Oxidation state assignment for common species.
Medium (35%): Balancing simple redox reactions.
Hard (15%): Disproportionation, balancing in basic medium, mixed-step problems.

Expert Strategy

For oxidation state of oxygen: default is $-2$ . Memorise the exceptions:

Peroxides ( $\text{H}_2\text{O}_2$ ): $-1$
Superoxides ( $\text{KO}_2$ ): $-1/2$
$\text{OF}_2$ : $+2$ (F is more electronegative)
$\text{O}_2\text{F}_2$ : $+1$

These exceptions are favourite trick questions.

For balancing in basic medium, balance as if it’s acidic first, then add equal $\text{OH}^-$ to both sides to neutralise the $\text{H}^+$ . Easier than starting in basic from scratch.

Common Traps

Trap 1: Fractional oxidation states. Some compounds genuinely have fractional oxidation states (e.g., $\text{Fe}_3\text{O}_4$ has Fe at $+8/3$ on average). Don’t reject them as wrong.

Trap 2: $\text{H}_2\text{O}_2$ as both oxidising and reducing agent. Hydrogen peroxide can lose or gain electrons depending on the partner. With KI, it’s an oxidiser; with $\text{KMnO}_4$ , it’s a reducer.

Trap 3: Counting electrons in the half-reaction wrong. Always check: charge on LHS = charge on RHS after adding electrons. If charges don’t match, you’ve miscounted.

CBSE 5-mark questions almost always include either a disproportionation or a balancing in basic medium. Practice both types until they’re automatic.