Redox Reactions JEE Notes 2026 | Formulas, PYQs & Trends for JEE Main & Advanced

1

Classical & Electronic Concept of Redox Reactions

Redox reactions form the foundation of electrochemistry, metallurgy, and many biological processes. The term "redox" comes from two concepts — reduction and oxidation. Understanding these dual processes is essential for JEE Chemistry as they appear in inorganic, organic, and physical chemistry questions alike.

1.1 Classical Concept of Oxidation and Reduction

The classical definitions were based on the addition or removal of oxygen and hydrogen from substances — observed long before the discovery of electrons.

Classical Definitions

Oxidation (Classical)

Addition of oxygen
Removal of hydrogen
Removal of electropositive element
Addition of electronegative element

Example: 2Mg + O₂ → 2MgO

Reduction (Classical)

Removal of oxygen
Addition of hydrogen
Addition of electropositive element
Removal of electronegative element

Example: CuO + H₂ → Cu + H₂O

Important for JEE — Oxidizing & Reducing Agents

Oxidizing Agent (Oxidant): The substance that oxidizes another substance and gets itself reduced. It accepts electrons.

Common oxidizing agents: O₂, O₃, H₂O₂, KMnO₄, K₂Cr₂O₇, HNO₃, H₂SO₄ (conc.), halogens (F₂, Cl₂, Br₂, I₂)

Reducing Agent (Reductant): The substance that reduces another substance and gets itself oxidized. It donates electrons.

Common reducing agents: H₂, C, CO, H₂S, SO₂, HI, SnCl₂, Na, K, LiAlH₄, NaBH₄, metals (Zn, Fe, Al)

1.2 Electronic Concept of Redox (Modern)

The electronic concept, based on electron transfer, was developed after the discovery of electrons. This is the most fundamental and widely used definition.

Most Repeated PYQ Foundation

Oxidation (Electronic)

\text{Loss of electrons} \Rightarrow \text{Oxidation}

M → Mⁿ⁺ + ne⁻ (Oxidation half-reaction)

Increase in oxidation number

Reduction (Electronic)

\text{Gain of electrons} \Rightarrow \text{Reduction}

X + ne⁻ → Xⁿ⁻ (Reduction half-reaction)

Decrease in oxidation number

Memory Trick — OIL RIG

OIL RIG:

Oxidation Is Loss of electrons
Reduction Is Gain of electrons

LEO says GER: Loss of Electrons = Oxidation; Gain of Electrons = Reduction

1.3 Redox Reactions — Both Occur Simultaneously

In any redox reaction, oxidation and reduction always occur together. The number of electrons lost in oxidation equals the number of electrons gained in reduction — this is the basis of balancing redox equations.

Complete Redox Reaction Example

\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}

Oxidation Half:

\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-

Zn is the reducing agent (gets oxidized)

Reduction Half:

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}

Cu²⁺ is the oxidizing agent (gets reduced)

PYQ — JEE Main 2023

Question: In the reaction:
2Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2NaI
Which species acts as the reducing agent?

Solution:

Step 1: Assign oxidation numbers

In Na₂S₂O₃: Na = +1, O = −2, so 2(+1) + 2S + 3(−2) = 0 → S = +2

In Na₂S₄O₆: Na = +1, O = −2, so 2(+1) + 4S + 6(−2) = 0 → S = +2.5

In I₂: I = 0; In NaI: I = −1

Step 2: Identify changes

S: +2 → +2.5 (increase → oxidation)

I: 0 → −1 (decrease → reduction)

Step 3: The species that gets oxidized is the reducing agent

\text{Na}_2\text{S}_2\text{O}_3 \text{ (sodium thiosulfate) is the reducing agent}

Answer: Na₂S₂O₃ (Sodium thiosulfate)

2

Oxidation Number — The Most Important Topic

Oxidation number (or oxidation state) is the single most tested concept from this chapter. Almost every JEE Main session has at least one question involving calculation or comparison of oxidation numbers. Master this section for guaranteed marks.

2.1 Definition of Oxidation Number

Most Important — Highest PYQ Frequency

Oxidation Number (O.N.): The charge an atom would possess if all bonds to atoms of different elements were 100% ionic (i.e., if electrons in each bond were assigned to the more electronegative atom).

\text{Oxidation Number} = \text{Valence electrons} - \text{Assigned electrons}

It can be positive, negative, zero, or fractional.

2.2 Rules for Assigning Oxidation Numbers

Must-Memorize Rules — 90% of PYQs Based on These

Rule 1: O.N. of an element in free state = 0 (e.g., Na, O₂, P₄, S₈, He)

Rule 2: O.N. of monoatomic ion = charge on ion (e.g., Na⁺ = +1, Cl⁻ = −1, Fe³⁺ = +3)

Rule 3: O.N. of H = +1 in compounds (exception: metal hydrides like NaH, CaH₂ where H = −1)

Rule 4: O.N. of O = −2 in compounds (exceptions: peroxides H₂O₂, Na₂O₂ where O = −1; superoxides KO₂ where O = −½; OF₂ where O = +2; O₂F₂ where O = +1)

Rule 5: O.N. of F = −1 always (most electronegative element)

Rule 6: O.N. of alkali metals = +1, alkaline earth metals = +2, Al = +3 in compounds

Rule 7: Sum of O.N. of all atoms in a neutral compound = 0

Rule 8: Sum of O.N. in a polyatomic ion = charge on the ion

Exception Alert (JEE Favorite!)

Oxygen positive: In OF₂, O = +2 (F is more electronegative)
Oxygen positive: In O₂F₂, O = +1
Peroxides: H₂O₂, Na₂O₂, BaO₂ → O = −1
Superoxides: KO₂, RbO₂, CsO₂ → O = −½
Hydrogen negative: Metal hydrides (NaH, LiH, CaH₂) → H = −1
Fractional O.N.: In Fe₃O₄, Fe = +8/3; In Na₂S₂O₃, S = +2 (avg); In C₃O₂, C = +4/3

2.3 Difference Between Oxidation Number and Valency

Property	Valency	Oxidation Number
Sign	Always positive	Can be +ve, −ve, or 0
Nature	Combining capacity	Apparent charge
Depends on	Element only	Electronegativity of bonded atom
Fractional?	Always integer	Can be fractional
Example (N in NH₃)	3	−3
Example (N in NO₂)	4	+4

PYQ — JEE Main 2024

Question: The oxidation number of oxygen in OF₂ and O₂F₂ respectively are:

(a) −2, −1 (b) +2, +1 (c) −1, −2 (d) +1, +2

Solution:

For OF₂:

F is the most electronegative element, so F = −1

Compound is neutral: O + 2(−1) = 0

\[O = +2\]

For O₂F₂:

F = −1 (4 fluorine atoms = 4 × −1 = −4)

2O + 4(−1) = 0 → 2O = +4

O = +2 \text{ (but actually } +1 \text{ each)}

Wait — let's recalculate. O₂F₂ has 2 O and 2 F atoms:

2O + 2(−1) = 0 → 2O = +2

O = +1 \text{ in } O_2F_2, \quad O = +2 \text{ in } OF_2

Answer: (b) +2, +1

2.4 Fractional Oxidation Numbers

Fractional O.N. Examples

Fe₃O₄ (Magnetite):

3 \times Fe + 4 \times (-2) = 0 \Rightarrow Fe = +\frac{8}{3}

Actually a mixed oxide: FeO·Fe₂O₃ (one Fe²⁺ and two Fe³⁺)

Na₂S₄O₆ (Sodium tetrathionate):

2(+1) + 4S + 6(-2) = 0 \Rightarrow S = +2.5

Actually: 2 terminal S = +5, 2 middle S = 0 (average = +2.5)

C₃O₂ (Carbon suboxide):

3C + 2(-2) = 0 \Rightarrow C = +\frac{4}{3}

PYQ — JEE Main 2022

Question: The oxidation state of nitrogen in N₃H (hydrazoic acid) is:

(a) −3 (b) +3 (c) −⅓ (d) +⅓

Solution:

For N₃H (hydrazoic acid, HN₃):

H = +1 (bonded to non-metal N)

Sum of O.N. = 0 (neutral compound)

\[3N + (+1) = 0\]

\[3N = -1\]

N = -\frac{1}{3}

This is a fractional oxidation state — average of three nitrogen atoms with different individual O.N.

Answer: (c) −⅓

JEE Shortcut — Quick O.N. Calculation

Always start with elements having fixed O.N. (H, O, F, Group 1, Group 2)
For complex ions, use: Σ(O.N.) = charge on ion
For neutral compounds, use: Σ(O.N.) = 0
If O appears with F, O is positive (F always −1)
Check for peroxide/superoxide before assigning O = −2

3

Balancing Redox Equations

Balancing redox equations is a high-weightage topic in JEE. Two methods are commonly used: the Oxidation Number Method and the Half-Reaction (Ion-Electron) Method. Both methods are based on the principle that electrons lost = electrons gained.

3.1 Oxidation Number Method

Important — Frequently Asked in JEE Main

Steps to balance by Oxidation Number Method:

Write the skeletal equation with products
Assign oxidation numbers to all atoms
Identify atoms whose O.N. changed (oxidized & reduced)
Calculate total increase and decrease in O.N.
Balance the increase and decrease by multiplying with appropriate coefficients
Balance all atoms except O and H
Balance O by adding H₂O
Balance H by adding H⁺ (acidic) or OH⁻ (basic)
Verify charge balance

Worked Example: Balance P₄ + OH⁻ → PH₃ + HPO₃²⁻

Step 1: Assign O.N.

P₄ (0) → PH₃ (P = −3) and HPO₃²⁻ (P = +5)

Step 2: Calculate change

Reduction: P (0) → P (−3), change = 3 per P atom

Oxidation: P (0) → P (+5), change = 5 per P atom

Step 3: Equalize total change

LCM of 3 and 5 = 15

So 5 P atoms get reduced (→ 5 PH₃) and 3 P atoms get oxidized (→ 3 HPO₃²⁻)

Step 4: Total P atoms = 5 + 3 = 8, so use 2P₄

2P_4 + OH^- \rightarrow 5PH_3 + 3HPO_3^{2-}

Step 5: Balance O and H, then charge

2P_4 + 3OH^- + 9H_2O \rightarrow 5PH_3 + 3HPO_3^{2-}

3.2 Half-Reaction (Ion-Electron) Method

Most Repeated Balancing Method in JEE

Steps for Half-Reaction Method (Acidic Medium):

Write the skeletal equation in ionic form
Split into two half-reactions: oxidation and reduction
Balance atoms other than O and H in each half-reaction
Balance oxygen by adding H₂O
Balance hydrogen by adding H⁺
Balance charge by adding electrons (e⁻)
Multiply half-reactions so electrons lost = electrons gained
Add the two half-reactions and simplify

JEE Pro Tip — Basic Medium Balancing

For balancing in basic medium, first balance as in acidic medium, then:

Add OH⁻ to both sides equal to the number of H⁺ present
On the side with both H⁺ and OH⁻, combine to form H₂O
Cancel any common H₂O molecules on both sides

Shortcut: For every H⁺ needed, add H₂O to that side and OH⁻ to the other side.

Worked Example: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (Acidic)

Reduction half: MnO₄⁻ → Mn²⁺

Balance O: MnO₄⁻ → Mn²⁺ + 4H₂O

Balance H: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balance charge: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxidation half: Fe²⁺ → Fe³⁺

Balance charge: Fe²⁺ → Fe³⁺ + e⁻

Equalize electrons: Multiply oxidation by 5

MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O

PYQ — JEE Main 2024

Question: Balance the following redox reaction in acidic medium:
Cr₂O₇²⁻ + C₂H₅OH → Cr³⁺ + CO₂

Solution:

Reduction half: Cr₂O₇²⁻ → Cr³⁺

Balance Cr: Cr₂O₇²⁻ → 2Cr³⁺

Balance O: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O

Balance H: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O

Balance charge: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Oxidation half: C₂H₅OH → CO₂

Balance C: C₂H₅OH → 2CO₂

Balance O: C₂H₅OH + 3H₂O → 2CO₂

Balance H: C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺

Balance charge: C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺ + 12e⁻

Equalize electrons: Multiply reduction by 2, oxidation by 1

2 × (Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O)

1 × (C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺ + 12e⁻)

2Cr_2O_7^{2-} + 16H^+ + C_2H_5OH \rightarrow 4Cr^{3+} + 2CO_2 + 11H_2O

Common Mistakes

Forgetting to balance O and H separately (do O first with H₂O, then H with H⁺)
Not equalizing electrons before adding half-reactions
Using H⁺ in basic medium (must convert to OH⁻)
Not checking final charge balance
Forgetting that H₂O can appear on both sides (simplify if needed)

4

Types of Redox Reactions

Redox reactions can be classified into several types based on how oxidation and reduction occur. JEE frequently tests your ability to identify the type and the species undergoing oxidation/reduction.

4.1 Intermolecular Redox Reactions

Both oxidation and reduction occur in different substances. This is the most common type.

2Al + Fe_2O_3 \rightarrow 2Fe + Al_2O_3

Al (0 → +3) is oxidized; Fe (+3 → 0) is reduced. Different substances undergo oxidation and reduction.

4.2 Intramolecular Redox Reactions

Both oxidation and reduction occur in the same substance but in different elements.

2KClO_3 \xrightarrow{\Delta} 2KCl + 3O_2

Cl (+5 → −1) is reduced; O (−2 → 0) is oxidized. Both occur in KClO₃.

4.3 Disproportionation Reactions

High PYQ Frequency — JEE Favorite!

Disproportionation: The same element in the same species is simultaneously oxidized and reduced. The element must be in an intermediate oxidation state.

2H_2O_2 \rightarrow 2H_2O + O_2

O in H₂O₂ (−1) → O in H₂O (−2) [reduced] and O in O₂ (0) [oxidized]

Common Disproportionation Reactions

Halogens in alkaline medium:

Cl_2 + 2NaOH \rightarrow NaCl + NaClO + H_2O

Cl₂ (0) → Cl⁻ (−1) + ClO⁻ (+1)

Copper(I) in water:

2Cu^+ \rightarrow Cu^{2+} + Cu

Cu⁺ (+1) → Cu²⁺ (+2) + Cu (0)

Phosphorus in alkali:

P_4 + 3NaOH + 3H_2O \rightarrow 3NaH_2PO_2 + PH_3

P₄ (0) → H₂PO₂⁻ (+1) + PH₃ (−3)

Nitrogen compounds:

NH_4NO_2 \rightarrow N_2 + 2H_2O

N (−3 in NH₄⁺) + N (+3 in NO₂⁻) → N₂ (0)

How to Check if Disproportionation is Possible

An element in oxidation state X can disproportionate only if:

The element has a stable state above X (higher O.N.)
AND a stable state below X (lower O.N.)
i.e., X is in an intermediate oxidation state

Example: Cl in ClO⁻ (+1) can disproportionate because Cl can be −1 and +5 (and +7).

Cannot disproportionate: Cl in Cl⁻ (−1) — it's the lowest state, can only be oxidized.

4.4 Comproportionation (Synproportionation)

The reverse of disproportionation: Two species containing the same element in different oxidation states react to form a product with an intermediate oxidation state.

Ag^{2+} + Ag \rightarrow 2Ag^+

Ag²⁺ (+2) and Ag (0) → Ag⁺ (+1) — intermediate state

HClO_3 + 5HCl \rightarrow 3Cl_2 + 3H_2O

Cl (+5 in ClO₃⁻) and Cl (−1 in HCl) → Cl₂ (0)

PYQ — JEE Main 2023

Question: Which of the following is a disproportionation reaction?

(a) 2FeCl₃ + SnCl₂ → 2FeCl₂ + SnCl₄

(b) 2H₂O₂ → 2H₂O + O₂

(c) Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

(d) Zn + 2HCl → ZnCl₂ + H₂

Solution:

For disproportionation: same element must be both oxidized and reduced

(a) Fe (+3 → +2, reduced); Sn (+2 → +4, oxidized) — different elements → NOT disproportionation

(b) O in H₂O₂: −1 → −2 (in H₂O) and −1 → 0 (in O₂) — SAME element both oxidized and reduced → DISPROPORTIONATION ✓

(c) Cu (0 → +2); Ag (+1 → 0) — different elements → NOT disproportionation

(d) Zn (0 → +2); H (+1 → 0) — different elements → NOT disproportionation

\text{Answer: (b) } 2H_2O_2 \rightarrow 2H_2O + O_2

Answer: (b)

4.5 Auto-oxidation and Auto-reduction

Some substances can undergo self-oxidation or self-reduction without an external agent:

Auto-oxidation example:

2PbO_2 \rightarrow 2PbO + O_2

Pb (+4 → +2) is reduced, O (−2 → 0) is oxidized — both in same compound

5

Redox Titrations

Redox titrations are quantitative analytical methods based on redox reactions. The most important titrations for JEE involve KMnO₄ (permanganometric) and K₂Cr₂O₇ (dichromatometric) as oxidizing agents.

5.1 Potassium Permanganate (KMnO₄) Titrations

Most Repeated Titration in JEE

KMnO₄ acts as a self-indicator (purple → colorless at endpoint).

Acidic Medium

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

n-factor = 5, Eq. wt = 158/5 = 31.6

Medium: Dil. H₂SO₄

Neutral/Weak Acidic

MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

n-factor = 3, Eq. wt = 158/3 = 52.67

Strongly Alkaline

MnO_4^- + e^- \rightarrow MnO_4^{2-}

n-factor = 1, Eq. wt = 158/1 = 158

Critical Note — Why HCl and HNO₃ Cannot Be Used

HCl: Gets oxidized by KMnO₄ to Cl₂ (interferes with results)
HNO₃: Itself a strong oxidizing agent (would oxidize the reducing agent being titrated)
Only dilute H₂SO₄ is suitable — neither oxidized nor reduced by KMnO₄

5.2 Potassium Dichromate (K₂Cr₂O₇) Titrations

K₂Cr₂O₇ in Acidic Medium

Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

n-factor = 6

Equivalent weight = 294/6 = 49

Indicator: Diphenylamine or potassium ferricyanide (external)

Advantage over KMnO₄: K₂Cr₂O₇ is a primary standard (stable, can be weighed directly)

5.3 Iodometric and Iodimetric Titrations

Iodometry (Indirect)

Stronger oxidizing agents oxidize KI to I₂, which is then titrated with Na₂S₂O₃

oxidizing\ agent + KI \rightarrow I_2

I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6

Indicator: Starch (blue-black → colorless)

Iodimetry (Direct)

Direct titration of I₂ with reducing agent (like Na₂S₂O₃)

I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6

Indicator: Starch (added near endpoint)

PYQ — JEE Main 2022

Question: The equivalent weight of KMnO₄ in acidic medium is:

(a) 158 (b) 31.6 (c) 52.67 (d) 79

Solution:

In acidic medium, KMnO₄ is reduced as:

MnO_4^- \rightarrow Mn^{2+}

Mn: +7 → +2, so change = 5 electrons

n-factor = 5

Molar mass of KMnO₄ = 39 + 55 + 4(16) = 158 g/mol

\text{Equivalent weight} = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{158}{5} = 31.6

\text{Equivalent weight} = 31.6 \text{ g/equivalent}

Answer: (b) 31.6

6

Equivalent Weight & n-Factor in Redox

The concept of equivalent weight is essential for solving redox titration problems. The n-factor represents the number of electrons transferred per mole of the substance in a redox reaction.

6.1 Definition and Formula

Important Formula — High PYQ Frequency

\text{Equivalent Weight} = \frac{\text{Molar Mass}}{n\text{-factor}}

Where n-factor = number of electrons lost or gained per molecule

6.2 n-Factor Table for Common Redox Agents

Substance	Half Reaction	n-factor	Eq. Wt
KMnO₄ (acidic)	MnO₄⁻ → Mn²⁺	5	31.6
KMnO₄ (neutral)	MnO₄⁻ → MnO₂	3	52.67
KMnO₄ (alkaline)	MnO₄⁻ → MnO₄²⁻	1	158
K₂Cr₂O₇ (acidic)	Cr₂O₇²⁻ → 2Cr³⁺	6	49
FeSO₄	Fe²⁺ → Fe³⁺	1	151.8
FeC₂O₄	Fe²⁺ → Fe³⁺, C₂O₄²⁻ → 2CO₂	3	143.9/3 ≈ 48
H₂O₂ (as oxidant)	H₂O₂ → 2OH⁻	2	17
H₂O₂ (as reductant)	H₂O₂ → O₂	2	17
Na₂S₂O₃	2S₂O₃²⁻ → S₄O₆²⁻	1	248
SnCl₂	Sn²⁺ → Sn⁴⁺	2	189.6/2 = 94.8
K₂C₂O₄	C₂O₄²⁻ → 2CO₂	2	166.2/2 = 83.1

Key Formula — Law of Equivalence

At the equivalence point in a redox titration:

\[N_1 V_1 = N_2 V_2\]

Where N = Normality = (Number of equivalents)/(Volume in L)

N = \frac{w \times 1000}{\text{Eq. wt} \times V(mL)} = \frac{M \times n}{1}

Normality = Molarity × n-factor

PYQ — JEE Main 2024

Question: 1.6 g of KMnO₄ (acidic medium) is used to oxidize Fe²⁺. The number of moles of Fe²⁺ oxidized is: (Molar mass of KMnO₄ = 158)

(a) 0.005 (b) 0.025 (c) 0.0507 (d) 0.0101

Solution:

Step 1: Find moles of KMnO₄

n_{KMnO_4} = \frac{1.6}{158} = 0.01013 \text{ mol}

Step 2: In acidic medium, each KMnO₄ accepts 5 electrons

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

Step 3: Total electrons accepted = 0.01013 × 5 = 0.05063 mol e⁻

Each Fe²⁺ donates 1 electron: Fe²⁺ → Fe³⁺ + e⁻

So moles of Fe²⁺ = moles of electrons = 0.05063

n_{Fe^{2+}} = 0.01013 \times 5 = 0.0507 \text{ mol}

Answer: (c) 0.0507

Special Case — FeC₂O₄ (Both Fe and C₂O₄ oxidized)

In FeC₂O₄ (ferrous oxalate), BOTH Fe²⁺ and C₂O₄²⁻ get oxidized:

Fe^{2+} \rightarrow Fe^{3+} + e^- \quad (1 \text{ e}^-)

C_2O_4^{2-} \rightarrow 2CO_2 + 2e^- \quad (2 \text{ e}^-)

Total electrons per FeC₂O₄ = 3, so n-factor = 3

Eq. wt of FeC₂O₄ = 143.9/3 = 47.97

7

Electrochemical Cells

An electrochemical cell is a device that converts chemical energy into electrical energy (galvanic/voltaic cell) or vice versa (electrolytic cell). Understanding the working of electrochemical cells is crucial for JEE.

7.1 Galvanic (Voltaic) Cell

Important — JEE Advanced Concept

Galvanic Cell: A device that converts chemical energy of a spontaneous redox reaction into electrical energy.

Daniell Cell (Classic Example): Zn | Zn²⁺ || Cu²⁺ | Cu

Daniell Cell — Working

Anode (Oxidation) — Left

Zn \rightarrow Zn^{2+} + 2e^-

Zn electrode dissolves, loses mass

Negative terminal (source of electrons)

Cathode (Reduction) — Right

Cu^{2+} + 2e^- \rightarrow Cu

Cu electrode gains mass (deposition)

Positive terminal (accepts electrons)

\text{Cell: } Zn(s)|Zn^{2+}(1M)||Cu^{2+}(1M)|Cu(s)

Overall: Zn + Cu²⁺ → Zn²⁺ + Cu

Cell Notation Rules

Anode on the left, cathode on the right
Single vertical line | = phase boundary (metal | solution)
Double line || = salt bridge
Order: Anode | Anode electrolyte || Cathode electrolyte | Cathode
Electrons flow: Anode → External circuit → Cathode
Current flow: Opposite to electron flow

7.2 Salt Bridge and Its Functions

Salt Bridge

A U-tube containing an inert electrolyte gel (KCl, KNO₃, NH₄NO₃) in agar-agar.

Functions:

Completes the electrical circuit (allows ion flow)
Maintains electrical neutrality in each half-cell
Prevents liquid-liquid junction potential

Criteria for salt bridge electrolyte:

Ions should have similar mobility (K⁺, Cl⁻ or NH₄⁺, NO₃⁻)
Should not react with cell solutions
Should not interfere with redox reaction

PYQ — JEE Main 2023

Question: In a galvanic cell, the salt bridge is used to:

(a) Generate electricity

(b) Allow flow of electrons

(c) Maintain electrical neutrality and complete the circuit

(d) Act as the cathode

Solution:

A salt bridge contains inert electrolyte (like KCl, KNO₃)

It does NOT generate electricity (a is wrong)

It does NOT allow electron flow — electrons flow through external wire (b is wrong)

It allows ION flow to maintain electrical neutrality (c is correct)

It completes the internal circuit (c is correct)

\text{Answer: (c) Maintain electrical neutrality and complete the circuit}

Answer: (c)

8

Electrode Potential & EMF of Cell

Electrode potential measures the tendency of an electrode to lose or gain electrons. The EMF (electromotive force) of a cell is the potential difference that drives electrons through the external circuit.

8.1 Electrode Potential

Important Concept

Electrode Potential (E): The potential difference between the electrode and its electrolyte.

Oxidation potential: Tendency to lose electrons (M → Mⁿ⁺ + ne⁻)
Reduction potential: Tendency to gain electrons (Mⁿ⁺ + ne⁻ → M)
By IUPAC convention, reduction potential is used as standard

8.2 Standard Hydrogen Electrode (SHE)

Standard Hydrogen Electrode (SHE)

Construction:

Pt electrode coated with Pt black (platinized platinum)
Immersed in 1M H⁺ solution (1M HCl)
H₂ gas bubbled at 1 bar pressure
Temperature: 298 K (25°C)

2H^+(1M) + 2e^- \rightarrow H_2(1\ bar) \quad E° = 0.00 \text{ V}

SHE is the universal reference electrode with E° = 0 V at all temperatures

SHE can act as both anode and cathode depending on the other electrode

8.3 Standard Electrode Potential and EMF

Most Important Formula — EMF Calculation

E°_{cell} = E°_{cathode} - E°_{anode}

Where both E° values are reduction potentials

If E°cell > 0, the reaction is spontaneous (feasible)

If E°cell < 0, the reaction is non-spontaneous

Nernst Equation (298 K)

E_{cell} = E°_{cell} - \frac{0.0591}{n} \log_{10} Q

Where:

n = number of electrons transferred
Q = reaction quotient = [Products]/[Reactants]
At equilibrium: Ecell = 0, Q = K (equilibrium constant)

E°_{cell} = \frac{0.0591}{n} \log_{10} K

Relationship with ΔG and K

\Delta G° = -nFE°_{cell}

\Delta G° = -RT \ln K = -2.303 RT \log_{10} K

Where F = 96500 C/mol, R = 8.314 J/(mol·K), T = 298 K

PYQ — JEE Main 2024

Question: Calculate the EMF of the cell at 298 K:
Zn|Zn²⁺(0.1M)||Cu²⁺(1M)|Cu
Given: E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V

Solution:

Step 1: Identify anode and cathode

Zn is anode (oxidation), Cu is cathode (reduction)

Step 2: Calculate E°cell

E°_{cell} = E°_{cathode} - E°_{anode} = 0.34 - (-0.76) = +1.10 \text{ V}

Step 3: Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2

Step 4: Q = [Zn²⁺]/[Cu²⁺] = 0.1/1 = 0.1

Step 5: Apply Nernst equation

E_{cell} = 1.10 - \frac{0.0591}{2} \log(0.1)

E_{cell} = 1.10 - \frac{0.0591}{2} \times (-1)

E_{cell} = 1.10 + 0.02955 = 1.1295 \text{ V}

E_{cell} \approx 1.13 \text{ V}

Answer: Ecell ≈ 1.13 V

Common Mistakes in EMF Problems

Using oxidation potential instead of reduction potential (always use reduction potentials)
Forgetting to divide 0.0591 by n (number of electrons)
Writing Q incorrectly — solids and liquids have activity = 1 (not included)
Sign error in E°cell = E°cathode − E°anode (NOT +)
Confusing spontaneity: E°cell > 0 means ΔG < 0 (spontaneous)

9

Applications of Redox Reactions

Redox reactions have wide-ranging applications in industry, biology, and daily life. Understanding these applications helps in answering conceptual JEE questions.

9.1 Industrial Applications

Extraction of metals: Reduction of metal oxides (e.g., Fe₂O₃ + 3CO → 2Fe + 3CO₂)
Bleaching: Chlorine and bleaching powder (CaOCl₂) are oxidizing bleaches
Photography: AgBr → Ag (reduction during development)
Batteries: Dry cell, lead storage battery, Ni-Cd cell
Electroplating: Deposition of metal on another surface
Combustion of fuels: Oxidation of hydrocarbons releases energy

9.2 Biological Applications

Respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy (oxidation of glucose)
Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ (reduction of CO₂)
Biological redox: NAD⁺/NADH, FAD/FADH₂ electron transport chain
Haemoglobin: Fe²⁺ carries oxygen (Fe remains +2, O₂ is just bound)

9.3 Environmental Applications

Water treatment: Chlorination (Cl₂ oxidizes bacteria)
Hydrogen peroxide: Used as antiseptic (oxidizes bacteria)
Ozone layer: O₃ acts as UV shield (redox cycles)
Rust prevention: Galvanization (Zn coating on Fe)

PYQ — JEE Main 2022

Question: In the reaction:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ (photosynthesis)
The substance that undergoes reduction is:

(a) H₂O (b) CO₂ (c) O₂ (d) C₆H₁₂O₆

Solution:

Assign oxidation numbers:

In CO₂: C = +4, O = −2

In H₂O: H = +1, O = −2

In C₆H₁₂O₆: C = 0 (average), H = +1, O = −2

In O₂: O = 0

Changes:

C: +4 → 0 (decrease → REDUCTION, CO₂ is reduced)

O (in H₂O): −2 → 0 (increase → OXIDATION, H₂O is oxidized)

\text{CO}_2 \text{ undergoes reduction}

Answer: (b) CO₂

Common Mistakes in Redox

O.N. of O: Always check for peroxides/superoxides before assigning −2
O.N. of H: Check for metal hydrides (H = −1)
OF₂: O = +2 (not −2) — F is more electronegative
Disproportionation: Same element must be both oxidized and reduced
EMF sign: E°cell = E°cathode − E°anode (always subtraction)
n-factor: Changes with medium (KMnO₄ has different n in acidic/neutral/alkaline)
Balancing basic medium: Don't use H⁺, use OH⁻/H₂O

Previous Year Questions Analysis

JEE Main (Last 10 Years)

✓ Oxidation Number Calculation: 30%
✓ Balancing Redox Equations: 20%
✓ Equivalent Weight & n-Factor: 18%
✓ Disproportionation: 12%
✓ Redox Titration Problems: 10%
✓ Electrochemical Cell/EMF: 10%

JEE Advanced (Last 10 Years)

✓ Balancing in Basic Medium: 25%
✓ Disproportionation/Comproportionation: 20%
✓ Equivalent Weight (complex cases): 20%
✓ Nernst Equation Applications: 15%
✓ Identifying Oxidizing/Reducing Agents: 10%
✓ Electrochemical Cell Problems: 10%

Weightage Analysis

JEE Main: 4-8 marks (1-2 questions)

JEE Advanced: 6-12 marks (1-3 questions)

Difficulty Level: Easy to Moderate

Time Required: 3-4 hours practice

Download Complete PYQ PDF (2015-2024)

Practice Problem Set

Level 1: Basic (JEE Main Standard)

Find the oxidation number of S in H₂SO₄, Na₂S₂O₃, Na₂S₄O₆, and SO₂.
Identify the oxidizing and reducing agent in: 2Al + Fe₂O₃ → 2Fe + Al₂O₃
Calculate the equivalent weight of K₂Cr₂O₇ in acidic medium (M = 294).
Balance: MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O
Is the reaction Cl₂ + 2NaOH → NaCl + NaClO + H₂O a disproportionation? Explain.

Level 2: Intermediate (JEE Main/Advanced)

Balance in basic medium: CrO₄²⁻ + SO₃²⁻ → Cr(OH)₃ + SO₄²⁻
The equivalent weight of FeC₂O₄ when oxidized by KMnO₄ in acidic medium is: (M = 143.9)
Calculate EMF of cell: Cu|Cu²⁺(0.1M)||Ag⁺(0.1M)|Ag. E°(Cu²⁺/Cu) = +0.34 V, E°(Ag⁺/Ag) = +0.80 V
0.5 g of oxalic acid (H₂C₂O₄·2H₂O) is titrated with KMnO₄ in acidic medium. Find moles of KMnO₄ used.
Identify which is NOT a disproportionation: (a) 2H₂O₂→2H₂O+O₂ (b) Cl₂+2NaOH→NaCl+NaClO+H₂O (c) 2Cu⁺→Cu²⁺+Cu (d) 2FeCl₃+SnCl₂→2FeCl₂+SnCl₄

Level 3: Advanced (JEE Advanced/Olympiad)

Balance: As₂S₃ + NO₃⁻ + H⁺ → H₃AsO₄ + S + NO + H₂O (acidic medium)
Calculate the equilibrium constant K for: Zn + Cu²⁺ → Zn²⁺ + Cu. E°cell = 1.10 V
An alloy of Cu and Zn weighing 1.0 g was dissolved in acid and Cu²⁺ was titrated with KI (liberated I₂ titrated with Na₂S₂O₃). Find % composition.
Prove that FeO₄²⁻ (ferrate) cannot exist in acidic medium by calculating E° and comparing with O₂/H₂O.
For the cell: Pt|H₂(1bar)|H⁺(pH=2)||Cu²⁺(0.1M)|Cu, calculate Ecell at 298K. E°(Cu²⁺/Cu) = +0.34 V

Get Solutions PDF with Step-by-Step Working

> FAQ

Frequently Asked Questions

What is a redox reaction?

A redox reaction is a chemical reaction in which both oxidation and reduction occur simultaneously. Oxidation involves loss of electrons or increase in oxidation number, while reduction involves gain of electrons or decrease in oxidation number.

What is oxidation number?

Oxidation number is the charge an atom would have if all bonds to atoms of different elements were ionic. It can be positive, negative, zero, or fractional. It helps in identifying oxidizing and reducing agents and balancing redox equations.

What is the difference between oxidation number and valency?

Valency is the combining capacity of an element and is always positive. Oxidation number can be positive, negative, or zero depending on the electronegativity of the bonded atoms. Valency of N in NH3 and NO2 is 3, but oxidation numbers are -3 and +4 respectively.

What is a disproportionation reaction?

A disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced. Example: 2H2O2 → 2H2O + O2, where oxygen in H2O2 (-1) is both oxidized to O2 (0) and reduced to H2O (-2).

How do you balance redox reactions by the half-reaction method?

Steps: (1) Write the skeletal equation. (2) Identify oxidation and reduction half-reactions. (3) Balance atoms other than O and H. (4) Balance O by adding H2O. (5) Balance H by adding H+ (acidic) or H2O/OH- (basic). (6) Balance charge by adding electrons. (7) Equalize electrons in both half-reactions. (8) Add and simplify.

What is the oxidation number of oxygen in OF2?

In OF2, fluorine is more electronegative than oxygen. Fluorine has oxidation number -1. Since the compound is neutral: O + 2(-1) = 0, so O = +2. This is one of the rare cases where oxygen has a positive oxidation number.

How many questions come from redox reactions in JEE Mains?

On average 1-2 questions (4-8 marks) appear from redox reactions in every JEE Main session. Most questions are from oxidation number calculation, balancing redox reactions, and equivalent weight in redox titrations.

What is the equivalent weight of KMnO4 in acidic medium?

In acidic medium, KMnO4 gains 5 electrons (MnO4- → Mn2+). Equivalent weight = Molar mass / n-factor = 158/5 = 31.6 g/equivalent. In neutral/weakly acidic medium it is 158/3, and in strongly alkaline medium it is 158/1.

What is a standard hydrogen electrode (SHE)?

Standard Hydrogen Electrode is the reference electrode with an arbitrary electrode potential of 0.00 V at all temperatures. It consists of a platinum electrode coated with platinum black, immersed in 1M H+ solution, with H2 gas at 1 bar pressure bubbling through it.

What is the Nernst equation for EMF of a cell?

The Nernst equation is E_cell = E°_cell - (0.0591/n) log Q at 298K, where E°_cell is standard EMF, n is the number of electrons transferred, and Q is the reaction quotient. It relates cell potential to concentration of species involved.

Redox Reactions notes, formulas & PYQs for JEE 2026

📑 Table of Contents

Classical & Electronic Concept of Redox Reactions

1.1 Classical Concept of Oxidation and Reduction

Classical Definitions

⭐ Important for JEE — Oxidizing & Reducing Agents

1.2 Electronic Concept of Redox (Modern)

⭐ Most Repeated PYQ Foundation

💡 Memory Trick — OIL RIG

1.3 Redox Reactions — Both Occur Simultaneously

Complete Redox Reaction Example

📝 PYQ — JEE Main 2023

Oxidation Number — The Most Important Topic

2.1 Definition of Oxidation Number

⭐ Most Important — Highest PYQ Frequency

2.2 Rules for Assigning Oxidation Numbers

🔥 Must-Memorize Rules — 90% of PYQs Based on These

⚠️ Exception Alert (JEE Favorite!)

2.3 Difference Between Oxidation Number and Valency

📝 PYQ — JEE Main 2024

2.4 Fractional Oxidation Numbers

Fractional O.N. Examples

📝 PYQ — JEE Main 2022

🎯 JEE Shortcut — Quick O.N. Calculation

Balancing Redox Equations

3.1 Oxidation Number Method

⭐ Important — Frequently Asked in JEE Main

Worked Example: Balance P₄ + OH⁻ → PH₃ + HPO₃²⁻

3.2 Half-Reaction (Ion-Electron) Method

⭐ Most Repeated Balancing Method in JEE

💡 JEE Pro Tip — Basic Medium Balancing

Worked Example: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (Acidic)

📝 PYQ — JEE Main 2024

⚠️ Common Mistakes

Types of Redox Reactions

4.1 Intermolecular Redox Reactions

4.2 Intramolecular Redox Reactions

4.3 Disproportionation Reactions

🔥 High PYQ Frequency — JEE Favorite!

Common Disproportionation Reactions

💡 How to Check if Disproportionation is Possible

4.4 Comproportionation (Synproportionation)

📝 PYQ — JEE Main 2023

4.5 Auto-oxidation and Auto-reduction

Redox Titrations

5.1 Potassium Permanganate (KMnO₄) Titrations

⭐ Most Repeated Titration in JEE

⚠️ Critical Note — Why HCl and HNO₃ Cannot Be Used

5.2 Potassium Dichromate (K₂Cr₂O₇) Titrations

K₂Cr₂O₇ in Acidic Medium

5.3 Iodometric and Iodimetric Titrations

Iodometry (Indirect)

Iodimetry (Direct)

📝 PYQ — JEE Main 2022

Equivalent Weight & n-Factor in Redox

6.1 Definition and Formula

⭐ Important Formula — High PYQ Frequency

6.2 n-Factor Table for Common Redox Agents

💡 Key Formula — Law of Equivalence

📝 PYQ — JEE Main 2024

🔥 Special Case — FeC₂O₄ (Both Fe and C₂O₄ oxidized)

Electrochemical Cells

7.1 Galvanic (Voltaic) Cell

⭐ Important — JEE Advanced Concept

Daniell Cell — Working

💡 Cell Notation Rules

7.2 Salt Bridge and Its Functions

Salt Bridge

📝 PYQ — JEE Main 2023

Electrode Potential & EMF of Cell

8.1 Electrode Potential

⭐ Important Concept

8.2 Standard Hydrogen Electrode (SHE)

Standard Hydrogen Electrode (SHE)

8.3 Standard Electrode Potential and EMF

⭐ Most Important Formula — EMF Calculation

Nernst Equation (298 K)

💡 Relationship with ΔG and K

📝 PYQ — JEE Main 2024

⚠️ Common Mistakes in EMF Problems

Table of Contents

Important for JEE — Oxidizing & Reducing Agents

Most Repeated PYQ Foundation

Memory Trick — OIL RIG

PYQ — JEE Main 2023

Most Important — Highest PYQ Frequency

Must-Memorize Rules — 90% of PYQs Based on These

Exception Alert (JEE Favorite!)

PYQ — JEE Main 2024

PYQ — JEE Main 2022

JEE Shortcut — Quick O.N. Calculation

Important — Frequently Asked in JEE Main

Most Repeated Balancing Method in JEE

JEE Pro Tip — Basic Medium Balancing

PYQ — JEE Main 2024

Common Mistakes

High PYQ Frequency — JEE Favorite!

How to Check if Disproportionation is Possible

PYQ — JEE Main 2023

Most Repeated Titration in JEE

Critical Note — Why HCl and HNO₃ Cannot Be Used

PYQ — JEE Main 2022

Important Formula — High PYQ Frequency

Key Formula — Law of Equivalence

PYQ — JEE Main 2024

Special Case — FeC₂O₄ (Both Fe and C₂O₄ oxidized)

Important — JEE Advanced Concept

Cell Notation Rules

PYQ — JEE Main 2023

Important Concept

Most Important Formula — EMF Calculation

Relationship with ΔG and K

PYQ — JEE Main 2024

Common Mistakes in EMF Problems

PYQ — JEE Main 2022

Common Mistakes in Redox

Previous Year Questions Analysis

Practice Problem Set

Why Redox Reactions is Important

Must-Remember Formulas & Values

5-Step Exam Strategy

7-Day Redox Reactions Mastery Roadmap