Electrochemistry — Class 12

Electrochemical cell with glowing electrodes and electron flow accents

A handwritten notes set for CBSE / NCERT Class 12 board exam prep.

Owner: AshnaAI · Created: Aug 22, 2026
Scope: 5-leaf unit — cells, potential, Nernst, conductance, electrolysis.

NB — this cover introduces the resource only. Teaching begins on Leaf 2.

Electrochemical Cells & Electrode Potential

A **galvanic cell** converts chemical energy into electricity via a spontaneous redox reaction.

Daniell Cell — AN OX, RED CAT

Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)
Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

AN OX — anode is oxidation; RED CAT — reduction at cathode
Salt bridge: anions → anode, cations → cathode
Electrons flow external wire from anode to cathode

Cell Notation & SHE

Zn(s) | Zn²⁺ || Cu²⁺ | Cu(s)

| phase boundary, || salt bridge; left = anode, right = cathode
SHE is the reference: $E ° = 0.00 V$ at 298 K, 1 bar H₂, [H⁺] = 1 M

Quick check: Which electrode is negative in a galvanic cell?

Show answer

The anode — it releases electrons (oxidation), so it accumulates negative charge.

AshnaAI · Aug 22, 2026 · Page 2 of 5

Nernst Equation & Conductance

The Nernst equation gives cell potential under non-standard conditions; conductance links to molar conductivity via the cell constant.

Nernst equation

$E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} lo g Q$

Q = reaction quotient (products/reactants, solids = 1), n = electrons transferred.

Worked example: Daniell cell at 298 K,

[Zn^{2 +}] = 0.1

,

[Cu^{2 +}] = 0.01

,

n = 2

.

E_{cell} = 1.10 - \frac{0.0591}{2} lo g \frac{0.1}{0.01} = 1.10 - 0.0296 = 1.07 V

Thermodynamic links

$Δ G^{\circ} = - n F E_{cell}^{\circ} = - RT ln K_{c}$

Conductance & conductivity

Quantity	Symbol	Relation
Conductivity	κ	$κ = G \times \frac{l}{A}$
Molar cond.	Λₘ	$Λ_{m} = \frac{κ \times 1000}{c}$

Strong electrolytes: Λₘ rises slightly on dilution.
Weak: jumps near infinite dilution; use Kohlrausch: $Λ_{m}^{\circ} = λ_{+}^{\circ} + λ_{-}^{\circ}$

Memory tip: Solids and pure liquids stay as 1 in Q.

I can find E_cell from E°_cell and Q.
I can use κ and cell constant to find Λ_m.

Aug 22, 2026 · AshnaAI · Page 4 of 5

Electrolysis & Corrosion

Link the charge passed to the mass liberated, then see how the same idea explains batteries, fuel cells, and rust.

Faraday’s Laws & Electrolysis

First law: m = Z·Q = Z·I·t, where

Z = \frac{M}{n F}

and

F = 96500

C mol⁻¹.
Second law: for equal charge, masses ∝ chemical equivalent mass (

M / n

).

Molten NaCl: Na at cathode, Cl₂ at anode.
Aqueous NaCl: H₂ at cathode, Cl₂ at anode.
CuSO₄/Cu electrodes: Cu deposits; anode Cu dissolves.

Batteries, Fuel Cells & Rust

Lead-storage (secondary):

Pb + PbO X_{2} + 2 H X_{2} SO X_{4} 2 PbSO X_{4} + 2 H X_{2} O

.
H₂–O₂ fuel cell:

2 H X_{2} + O X_{2} 2 H X_{2} O

; clean, efficient.

NB Rusting needs O₂ + water; Fe oxidises to Fe²⁺, O₂ reduces to OH⁻, giving hydrated Fe₂O₃.

Galvanisation: zinc coat corrodes first (sacrificial).
Cathodic protection: link Fe to Mg or Zn.
Barrier: paint, oil, enamel.

Tip: Remove air or water → no rust cell. 🛡️

State Faraday’s first law.
Why is lead-storage “secondary”?

Electrochemistry Quick Revision & Practice

Board-exam formula sheet, key definitions, and short-answer practice.

Formula Sheet & Definitions

Cell potential:

E_{cell} = E_{cathode} - E_{anode}

Nernst (298 K):

E = E^{\circ} - \frac{0.0591}{n} lo g Q

Free energy:

Δ G^{\circ} = - n F E_{cell}^{\circ} = - RT ln K

Molar conductivity:

Λ_{m} = \frac{1000 κ}{c}

; unit **S cm² mol⁻¹**
Faraday's law:

m = \frac{E I t}{F}

,

F = 96487 C mol^{- 1}

Standard electrode potential — measured at 298 K, 1 bar, 1 M.
Cell constant —

l / A

, unit cm⁻¹.
✗Anode is negative in a galvanic cell; don't reverse signs.

Quick-Revision Table

Topic	One-line Recall
Galvanic cell	Spontaneous redox → electrical energy.
SHE	Reference; $E^{\circ} = 0.00 V$ .
Kohlrausch law	$Λ_{m}^{\circ}$ = sum of limiting ionic conductivities.
Faraday's laws	Mass deposited ∝ charge passed.

Board-Style Questions

Write the cell reaction for $Zn (s) ∣ Zn X^{2 +} (aq) ∣∣ Cu X^{2 +} (aq) ∣ Cu (s)$ .

Show working

Overall:

Zn (s) + Cu X^{2 +} (aq) Zn X^{2 +} (aq) + Cu (s)

Calculate $E_{cell}^{\circ}$ if $E_{Cu X^{2 +} / Cu}^{\circ} = + 0.34 V$ , $E_{Zn X^{2 +} / Zn}^{\circ} = - 0.76 V$ .

Show working

E_{cell}^{\circ} = 0.34 - (- 0.76) = + 1.10 V

A current of 2 A flows for 30 min through $CuSO X_{4}$ . Find charge passed.

Show working

Q = I t = 2 \times 30 \times 60 = 3600 C

Why does conductivity decrease with dilution?

Show working

Fewer ions per unit volume, so conductivity falls though

Λ_{m}

rises.

I can apply the Nernst equation and relate $E^{\circ}$ , $Δ G^{\circ}$ , $K$ .
I can use Faraday's law for mass/charge calculations.