Electrochemistry (Class 12)
Why This Chapter Matters
Electrochemistry gives 5-7 marks in CBSE Class 12. EMF of cells, Nernst equation, conductance, and electrolysis are all board topics.
Core Concepts
1. Electrochemical Cells
Galvanic cell (Voltaic): Chemical energy → Electrical energy (spontaneous).
Anode: oxidation (negative in galvanic). Cathode: reduction (positive in galvanic).
Daniell cell: Zn|Zn²⁺||Cu²⁺|Cu.
Zn → Zn²⁺ + 2e⁻ (anode). Cu²⁺ + 2e⁻ → Cu (cathode).
E°cell = E°cathode - E°anode = +0.34 - (-0.76) = +1.10 V.
Salt bridge: Maintains electrical neutrality and ion flow without mixing solutions.
2. Standard Electrode Potential
Measured relative to Standard Hydrogen Electrode (SHE) = 0.00 V.
More positive E° = better oxidising agent (gets reduced more easily).
More negative E° = better reducing agent (gets oxidised more easily).
Standard cell potential: E°cell = E°cathode - E°anode.
Spontaneous if E°cell > 0 ↔ ΔG° < 0 ↔ K > 1.
Relation: ΔG° = -nFE°cell = -RT lnK.
F = Faraday = 96500 C/mol.
3. Nernst Equation
At non-standard conditions: E = E° - (RT/nF)lnQ.
At 25°C: E = E° - (0.0591/n)logQ.
4. Conductance
Resistance R = ρL/A. Conductance G = 1/R = κ·A/L.
Specific conductance κ (= conductivity): G×L/A. Unit: S/m or S/cm.
Molar conductance Λm = κ×1000/M (M = molarity). Unit: S·cm²/mol.
At infinite dilution (Λ°m):
Strong electrolytes: extrapolate linear graph of Λm vs √c.
Weak electrolytes: Kohlrausch's law: Λ°m = Σλ°(ions).
Λm increases with dilution for all electrolytes.
For strong electrolytes: Λm = Λ°m - b√c (Debye-Hückel-Onsager equation).
Degree of dissociation: α = Λm/Λ°m (for weak electrolytes).
5. Faraday's Laws of Electrolysis
1st Law: Mass deposited ∝ charge passed. m = ZQ = ZIt.
2nd Law: Mass deposited ∝ equivalent weight (at same charge).
m = (M/nF)×Q = (E/F)×Q. E = equivalent weight = M/n.
Electrodes: Cathode = reduction (+ions deposit). Anode = oxidation (metal dissolves or water oxidised).
Board Examples
Q1: Calculate E_cell for Daniell cell when [Zn²⁺]=0.1M, [Cu²⁺]=0.01M at 25°C.
E°cell = 1.10V. Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10.
E = 1.10 - (0.0591/2)log10 = 1.10 - 0.02955 = 1.07V.
Q2: How many grams of Ag deposited by 1A current for 1 hour from AgNO₃ solution?
Q = It = 1×3600 = 3600C. m = (108/96500)×3600 = 4.03g.