Electrochemistry
Learn Electrochemistry with notes, examples, and practice questions.
Chapter Overview
Electrochemistry links chemical reactions to electric current — galvanic cells produce electricity; electrolytic cells use electricity to drive reactions. Nernst equation quantifies cell potential.
Cells
Galvanic
Spontaneous; anode oxidation (−), cathode reduction (+)
EMF
E°_cell = E°_cathode − E°_anode
Nernst
E = E° − (0.0591/n) log Q (at 298 K)
ΔG°
ΔG° = −nFE°
Electrolysis
Faraday 1st
m = ZIt = ZQ
Faraday 2nd
Same charge deposits equivalents proportional to equivalent mass
Conductivity
κ = 1/ρ; molar conductivity Λ_m = κ/c
Solved Examples
Step-by-step solutions — read each step before checking the final answer.
Cell EMF
Zn|Zn²⁺||Cu²⁺|Cu. E°(Zn²⁺/Zn)=−0.76, E°(Cu²⁺/Cu)=+0.34.
- 1E°=0.34−(−0.76)=1.10V
Answer
1.10 V
Faraday
Charge to deposit 1 mol Ag⁺ (96500 C/mol).
- 11 F = 96500 C
Answer
96500 C (1 Faraday)
Key Points to Remember
- ✓ SHE: E°=0
- ✓ Salt bridge maintains neutrality
- ✓ Kohlrausch law for weak electrolytes
Exam Tips
- • Anode = oxidation always
- • Balance redox before cell diagram
- • Use log Q in Nernst
Formula Cheat Sheet
Quick reference — NCERT board exam formulas
Electrochemical Cells
- Galvanic cell:Spontaneous ; anode (−) oxidation, cathode (+) reduction
- EMF:E°_cell = E°_cathode − E°_anode
- Nernst (298 K):E = E° − (0.0591/n) log Q
- Gibbs:ΔG° = −nFE°
Electrolysis & Conductivity
- Faraday 1st:m = ZIt = ZQ
- Faraday constant:F ≈ 96500 C/mol
- Conductivity:κ = 1/ρ
- Molar conductivity:Λ_m = κ/c
Practice MCQs — Electrochemistry
Test your understanding with topic-wise multiple choice questions. Explanations appear after each answer.
Anode is where: