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CHEM 05Chemistry3–4 Q/year★★★★ High Priority
Electrochemistry & Chemical Kinetics
NCERT Class 12 Ch. 2–4. Nernst equation, Faraday's laws, rate laws, Arrhenius — formula-heavy, scoring.
3–4
Questions/Year
12–16
Marks at Stake
Electrochemistry & Kinetics
Electrochemistry: master the cell notation and Nernst equation. Kinetics: know rate laws and half-life formulas cold.
Galvanic Cells & EMF

Galvanic cell: Chemical energy → Electrical energy. Anode (oxidation) on left; Cathode (reduction) on right. Salt bridge maintains electrical neutrality.

Cell notation: Anode | Anode solution || Cathode solution | Cathode. E.g., Zn | ZnSO₄ || CuSO₄ | Cu. E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 1.10 V.

Standard electrode potential: SHE (Standard Hydrogen Electrode) = 0.00 V by definition. Measured at 25°C, 1 M, 1 atm.

Nernst equation: E_cell = E°_cell - (RT/nF) ln Q = E°_cell - (0.0592/n) log Q at 25°C. At equilibrium, E_cell = 0 and Q = Kc. Therefore: log Kc = nE°_cell / 0.0592.

Electrolysis & Faraday's Laws

Electrolytic cell: Electrical energy → Chemical change. External power source forces non-spontaneous reactions. Cathode = reduction (cations deposited). Anode = oxidation.

Faraday's First Law: Mass deposited (m) ∝ charge passed (Q). m = (M × Q) / (n × F). Where M = molar mass, n = valency, F = Faraday = 96,500 C/mol.

Faraday's Second Law: Same charge → masses deposited are proportional to their equivalent weights.

Kohlrausch's law: Λ°_m(electrolyte) = sum of Λ° of individual ions at infinite dilution.

Rate Laws & Order

Rate = k[A]^m [B]^n where m, n = partial orders (experimentally determined, not from stoichiometry).

Zero order: Rate = k. t½ = [A]₀/2k. [A] = [A]₀ - kt. Rate independent of concentration.

First order: Rate = k[A]. t½ = 0.693/k. ln[A] = ln[A]₀ - kt. t½ is CONSTANT (concentration-independent). Radioactive decay, H₂O₂ decomposition.

Second order: Rate = k[A]². t½ = 1/(k[A]₀). t½ depends on initial concentration.

Arrhenius Equation

k = A e^(-Ea/RT) where A = frequency factor (pre-exponential factor), Ea = activation energy, R = 8.314 J/mol·K.

ln k = ln A - Ea/RT. Plot of ln k vs 1/T: slope = -Ea/R.

Effect of temperature: For every 10°C rise, rate approximately doubles (rule of thumb). Van't Hoff factor: temperature coefficient μ ≈ 2–3.

Catalyst: Provides alternate pathway with lower Ea. Does not change ΔH, ΔG, or equilibrium position — only changes rate by lowering Ea. Increases both forward and reverse rate equally.

Electrochemistry & Kinetics Facts
Key EMF Values
F (Faraday): 96,500 C/mol
R: 8.314 J/mol·K
At 25°C: 2.303RT/F = 0.0592
Nernst: E = E° - (0.0592/n)log Q
SHE: 0.00 V (reference)
n = number of electrons transferred. For Zn-Cu cell, n=2; E°=1.10V
Half-Life Formulas
Zero order: t½ = [A]₀/2k
First order: t½ = 0.693/k (constant!)
Second order: t½ = 1/k[A]₀
Radioactive: first order (t½ constant)
ln 2 = 0.693
Only first-order has concentration-independent half-life — key distinguishing feature
Order Determination
Rate = k[A]^m[B]^n
Overall order = m + n
Units of k (0 order): mol/L/s
Units of k (1st order): s⁻¹
Units of k (2nd order): L/mol/s
Units of k can identify reaction order — this appears directly in NEET questions
Conductance Terms
Conductance G = 1/R (Siemens, S)
Specific conductance κ = l/(R×A)
Molar conductance Λ_m = κ × 1000/M
Kohlrausch: Λ° = sum of ionic Λ°
Ostwald dilution law: α = √(Ka/C)
Molar conductance increases on dilution for all electrolytes
Worked Examples
EasyThe half-life of a first-order reaction is 30 minutes. What fraction remains after 90 minutes?
90 minutes = 3 half-lives. After each half-life, half the amount remains. After 1 t½: 1/2. After 2 t½: 1/4. After 3 t½: 1/8.
Answer: 1/8 remains (12.5%)
MediumFor the Zn-Cu galvanic cell (E°Zn²⁺/Zn = -0.76V; E°Cu²⁺/Cu = +0.34V), calculate E°cell.
E°cell = E°cathode - E°anode = E°Cu - E°Zn = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V. Zn is anode (more negative potential, easier to oxidise). Cu is cathode (more positive, easier to reduce).
Answer: E°cell = +1.10 V
HardThe rate constant units are L mol⁻¹ s⁻¹. What is the order of the reaction?
Rate = k[A]^n. Rate units: mol L⁻¹ s⁻¹. [A]^n units: mol^n L^-n. k = rate/[A]^n → k units = (mol L⁻¹ s⁻¹)/(mol^n L^-n) = mol^(1-n) L^(n-1) s⁻¹. For second order: k units = L mol⁻¹ s⁻¹. This matches the given units.
Answer: Second order (n=2)
Mistake DNA
❌ Confusing anode and cathode in galvanic vs electrolytic cells
In BOTH cells: Anode = oxidation, Cathode = reduction. But the signs differ: Galvanic cell — anode is negative (electrode releasing electrons to external circuit). Electrolytic cell — anode is connected to positive terminal.
Fix: Anode = Oxidation (always). An ox, red cat — Anode oxidises, Cathode reduces.
❌ Using stoichiometric coefficients as reaction order
Reaction order is determined EXPERIMENTALLY, not from the balanced equation (except for elementary reactions). For A + B → C, the rate might be rate = k[A]²[B]⁰ — you cannot tell from stoichiometry alone.
Fix: Only elementary single-step reactions have order = stoichiometry. Complex reactions don't.
Chapter Intelligence
PYQ Frequency
Nernst equation: 1 Q/year
Faraday's law calculations: 1 Q/year
Rate law/half-life: 1–2 Q/year
Arrhenius equation: 1 Q/year
2026 Prediction
High: Half-life calculation (first order)
Expected: Units of rate constant → identify order
Watch: Kohlrausch law application
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