Redox Reactions and Electrochemistry — JEE Main Chemistry Notes

1. Chapter at a Glance

This chapter joins two NCERT units: Redox Reactions (Class 11) and Electrochemistry (Class 12). Redox reactions involve a transfer of electrons: one species is oxidised (loses electrons) while another is reduced (gains electrons). Electrochemistry uses these reactions to produce electricity (galvanic cells) or uses electricity to drive non-spontaneous reactions (electrolysis). JEE Main regularly asks about oxidation numbers, cell potential, the Nernst equation, Faraday's laws and conductance.

What JEE asks most

Oxidation numbers, balancing redox equations and equivalent weight (n-factor).
E°_cell from standard reduction potentials; relation with ΔG° and K.
Nernst equation for cell potential at non-standard concentrations.
Faraday's laws of electrolysis; molar conductivity and Kohlrausch's law.

2. Oxidation, Reduction and Oxidation Number

Oxidation: loss of electrons (increase in oxidation number). Reduction: gain of electrons (decrease in oxidation number). Memory aid: OIL RIG — Oxidation Is Loss, Reduction Is Gain.
An oxidising agent is itself reduced; a reducing agent is itself oxidised.
Rules for oxidation number: free elements = 0; monatomic ion = its charge; F is always −1; O is usually −2 (−1 in peroxides, +2 in OF₂); H is +1 with non-metals and −1 in metal hydrides; the sum in a neutral molecule is 0 and in an ion equals its charge.
Disproportionation: the same element is both oxidised and reduced, e.g. 2H₂O₂ → 2H₂O + O₂ (O goes from −1 to −2 and 0).
Comproportionation: two oxidation states of an element combine to give an intermediate one.

Species	Element	Oxidation number
K₂Cr₂O₇	Cr	+6
KMnO₄	Mn	+7
H₂SO₅ (peroxy acid)	S	+6 (one peroxide link)
Fe₃O₄	Fe	+8/3 (average)
CrO₅	Cr	+6 (two peroxide links)

Balancing and n-factor

Balance by the ion–electron (half-reaction) method: balance atoms other than O and H, then O with H₂O, H with H⁺ (acid) or with H₂O/OH⁻ (base), and finally charge with electrons. The n-factor is the number of electrons gained or lost per formula unit. KMnO₄ has n = 5 in acid (Mn⁷⁺ → Mn²⁺), n = 3 in neutral (→ MnO₂) and n = 1 in strongly basic medium (→ MnO₄²⁻). K₂Cr₂O₇ in acid has n = 6.

Equivalents of oxidant = equivalents of reductant (n₁M₁V₁ = n₂M₂V₂)

3. Galvanic Cells and Electrode Potential

In a galvanic (voltaic) cell, a spontaneous redox reaction produces electricity. Oxidation occurs at the anode (negative terminal) and reduction at the cathode (positive terminal). A salt bridge completes the circuit and keeps the solutions electrically neutral.

Daniell cell: Zn | Zn²⁺(aq) || Cu²⁺(aq) | Cu. Electrons move from zinc to copper through the external wire.

Cell notation: anode on the left, cathode on the right, || for the salt bridge.
Standard hydrogen electrode (SHE) is the reference with E° = 0.00 V.
E°_cell = E°_cathode − E°_anode (both as reduction potentials). A positive E°_cell means a spontaneous reaction.
More negative reduction potential = stronger reducing agent (Li is the strongest in water); more positive = stronger oxidising agent (F₂ is the strongest).
A metal with a lower reduction potential displaces one with a higher potential from its salt solution (Zn displaces Cu²⁺).

4. Nernst Equation, ΔG and K

For a cell reaction with n electrons and reaction quotient Q, at 298 K:

E_cell = E°_cell − (0.059/n) log Q

ΔG° = −nFE°_cell (F = 96500 C mol⁻¹). Spontaneous: ΔG < 0, E > 0.
At equilibrium, E_cell = 0, so E°_cell = (0.059/n) log K.
For a single electrode Mⁿ⁺ + ne⁻ → M: E = E° − (0.059/n) log(1/[Mⁿ⁺]).
Concentration cell: the same electrodes at different concentrations; E° = 0, and E = (0.059/n) log(C_cathode/C_anode).

5. Conductance of Electrolytic Solutions

Quantity	Meaning	Unit
Resistance R = ρ(l/A)	Opposition to current	Ω
Conductivity κ = 1/ρ	Conductance of 1 cm cube; κ = G × (l/A) = G × cell constant	S cm⁻¹
Molar conductivity Λ_m	Λ_m = κ × 1000/C (C in mol L⁻¹)	S cm² mol⁻¹

On dilution, κ decreases (fewer ions per volume) but Λ_m increases.
Strong electrolytes: Λ_m = Λ°_m − A√C (Debye–Hückel–Onsager); a nearly straight line against √C.
Weak electrolytes: Λ_m rises steeply at high dilution; Λ°_m cannot be obtained by extrapolation.
Kohlrausch's law: Λ°_m = ν₊λ°₊ + ν₋λ°₋. It gives Λ°_m of weak electrolytes from strong ones, and the degree of dissociation α = Λ_m/Λ°_m.

6. Electrolysis and Batteries

Faraday's first law: w = ZIt = (M/nF) × I × t

Second law: when the same charge passes through different electrolytes, the masses deposited are proportional to their equivalent weights. One faraday (96500 C) is the charge of one mole of electrons.

Electrolysis of aqueous NaCl: H₂ at the cathode (water is reduced more easily than Na⁺), Cl₂ at the anode (due to overpotential of O₂).
Electrolysis of molten NaCl: Na at the cathode, Cl₂ at the anode.
Primary cells (not rechargeable): dry cell (Leclanché), mercury cell (constant 1.35 V).
Secondary cells (rechargeable): lead storage battery — anode Pb, cathode PbO₂, electrolyte H₂SO₄ (about 38%); nickel–cadmium cell.
Fuel cells: H₂–O₂ fuel cell used in the Apollo space programme; product is water.
Corrosion of iron is an electrochemical process; prevented by galvanising (Zn coating) or sacrificial anodes (Mg, Zn).

7. Formula Master Table

Quantity	Formula
Cell potential	E°_cell = E°_cathode − E°_anode
Nernst (298 K)	E = E° − (0.059/n) log Q
Free energy	ΔG° = −nFE°
Equilibrium constant	log K = nE°/0.059
Conductivity	κ = G × (l/A)
Molar conductivity	Λ_m = 1000κ/C
Kohlrausch	Λ°_m = ν₊λ°₊ + ν₋λ°₋
Degree of dissociation	α = Λ_m/Λ°_m
Faraday's law	w = MIt/(nF)

8. Mind Map

Redox & Electrochemistry

Redox

Oxidation number
Balancing, n-factor
Disproportionation

Galvanic cells

E°_cell, SHE
Nernst equation
ΔG° and K

Conductance

κ, Λ_m
Kohlrausch's law

Electrolysis

Faraday's laws
Batteries, fuel cells
Corrosion

9. Tips and Tricks

Shortcuts

E° is an intensive property: never multiply it by the coefficients when you multiply a half-reaction. Add ΔG° values instead when combining half-reactions to get a new half-reaction.
0.059/n with n = 1: each tenfold change in concentration changes E by 0.059 V.
For Faraday problems, first find moles of electrons = It/96500, then divide by n.
"An Ox, Red Cat": Anode = Oxidation, Reduction = Cathode.

Mistakes students make

Writing E°_cell = E°_anode − E°_cathode (wrong order).
Forgetting that in electrolysis the anode is positive, while in a galvanic cell the anode is negative.
Taking the oxidation number of Cr in CrO₅ as +10 — it has peroxide oxygens, so it is +6.

10. Important One-liners

The salt bridge usually contains KCl or KNO₃ in agar-agar gel.
Fluorine is the strongest oxidising agent; lithium is the strongest reducing agent in aqueous solution.
One faraday deposits one gram-equivalent of any substance.
Molar conductivity of an electrolyte increases on dilution.
The lead storage battery's electrolyte becomes less dense on discharge because H₂SO₄ is consumed.
Rusting needs both water and oxygen; salt water speeds it up.

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main. (F = 96500 C mol⁻¹.)

Q1Easy

The oxidation number of chromium in K₂Cr₂O₇ is:

Show solution

2(+1) + 2x + 7(−2) = 0 ⇒ 2x = 12 ⇒ x = +6.

Answer: (B)

Q2Medium

How many moles of KMnO₄ are needed to oxidise 1 mol of Fe²⁺ to Fe³⁺ in acidic medium?

Show solution

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O: each MnO₄⁻ accepts 5 electrons. Each Fe²⁺ gives 1 electron. 1 mol Fe²⁺ gives 1 mol e⁻, which needs 1/5 = 0.2 mol MnO₄⁻.

Answer: (A)

Q3Easy

Given E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V, the standard emf of the Daniell cell is:

0.42 V
−1.10 V
1.10 V
0.76 V

Show solution

Cu is the cathode and Zn the anode: E° = 0.34 − (−0.76) = 1.10 V.

Answer: (C)

Q4Medium

For the Daniell cell (E° = 1.10 V, n = 2), ΔG° is:

−106.2 kJ
−212.3 kJ
+212.3 kJ
−424.6 kJ

Show solution

ΔG° = −nFE° = −2 × 96500 × 1.10 = −212300 J = −212.3 kJ. Negative, so the reaction is spontaneous.

Answer: (B)

Q5Hard

The cell Zn | Zn²⁺(0.1 M) || Cu²⁺(1 M) | Cu has E° = 1.10 V. Its emf at 298 K is about:

1.07 V
1.10 V
1.13 V
1.16 V

Show solution

Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, Q = [Zn²⁺]/[Cu²⁺] = 0.1.

E = 1.10 − (0.059/2) log 0.1 = 1.10 + 0.0295 ≈ 1.13 V. Lower product concentration increases the emf.

Answer: (C)

Q6Numerical

A current of 2 A is passed through CuSO₄ solution for 965 s. If the mass of copper deposited is x × 10⁻³ g (Cu = 63.5), find x.

Show solution

Charge = 2 × 965 = 1930 C = 0.02 F. Cu²⁺ + 2e⁻ → Cu, so moles of Cu = 0.01. Mass = 0.01 × 63.5 = 0.635 g = 635 × 10⁻³ g.

Answer: 635

Q7Hard

Λ°_m values (S cm² mol⁻¹) are: CH₃COONa 91.0, HCl 426.2, NaCl 126.5. Λ°_m of CH₃COOH is:

390.7
643.7
208.7
517.2

Show solution

By Kohlrausch's law: Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl) = 91.0 + 426.2 − 126.5 = 390.7 S cm² mol⁻¹.

Answer: (A)

Q8Numerical

A two-electron cell reaction has E°_cell = 0.295 V at 298 K. Find log K for the reaction (use 2.303RT/F = 0.059 V).

Show solution

log K = nE°/0.059 = (2 × 0.295)/0.059 = 10, so K = 10¹⁰.

Answer: 10

12. Quick Revision Summary

Remember before the exam

Oxidation = loss of e⁻ (ON increases); reduction = gain of e⁻ (ON decreases).
KMnO₄ n-factor: 5 (acid), 3 (neutral), 1 (strong base); K₂Cr₂O₇: 6.
E°_cell = E°_cathode − E°_anode; ΔG° = −nFE°; log K = nE°/0.059.
Nernst: E = E° − (0.059/n) log Q.
Λ_m = 1000κ/C; Kohlrausch for weak electrolytes; α = Λ_m/Λ°_m.
Faraday: w = MIt/(nF); 1 F = 96500 C.

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