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Electrochemistry

Physical Chemistry Weightage: 9 Marks CBSE Unit 2
“Hi there! Electrochemistry is one of the biggest chapters of CBSE Class 12 Chemistry. It carries 9 marks (about 13% of the theory paper) and is a favourite for a 5-mark long answer, a 3-mark numerical and several 1-mark MCQs every year. The good news: it is highly formula-driven, so if you master the Nernst equation, conductivity numericals and Faraday's laws, most of the marks are yours. Read the traps in the red boxes carefully; they are exactly where students lose marks.”
— SCORECHEM ACADEMIC TEAM

1. Galvanic Cell & Cell Notation

A galvanic (voltaic) cell converts the energy of a spontaneous redox reaction into electrical energy. An electrolytic cell does the reverse: electrical energy forces a non-spontaneous reaction.

V Salt bridge (KCl / NH₄NO₃) Zn | Zn²⁺ (1 M) Cu²⁺ (1 M) | Cu ANODE (−) Oxidation CATHODE (+) Reduction Zn → Zn²⁺ + 2e⁻ Cu²⁺ + 2e⁻ → Cu e⁻ flow current
Daniell cell (E°cell = 1.10 V). Electrons flow from Zn (anode, −) to Cu (cathode, +) through the wire; conventional current flows the opposite way. The salt bridge completes the circuit by keeping both half-cells electrically neutral.
Galvanic cell Electrolytic cell
Reaction Spontaneous (ΔG<0\Delta G<0) Non-spontaneous (ΔG>0\Delta G>0)
Anode Negative (−) Positive (+)
Cathode Positive (+) Negative (−)
Energy change Chemical → electrical Electrical → chemical

Golden rule: Anode = oxidation, Cathode = reduction in both cells. Only the signs swap.

Cell notation: anode on the left, cathode on the right; a single line | for a phase boundary, a double line || for the salt bridge. Example: Zn ∣ Zn2+(1 M) ∣∣ Cu2+(1 M) ∣ Cu\text{Zn}\,|\,\text{Zn}^{2+}(1\text{ M})\,||\,\text{Cu}^{2+}(1\text{ M})\,|\,\text{Cu}

Salt bridge: a U-tube of inert electrolyte (KCl, NH₄NO₃, KNO₃) in agar. It (i) completes the circuit and (ii) keeps both half-cells electrically neutral. Its ions must not react with the cell solutions and should have nearly equal ionic mobilities (this is why KCl is used, not NaCl).

⚠️ Board Exam Trap: Direction of current. Electrons flow anode → cathode in the external wire, but conventional current flows cathode → anode. If the question asks for "direction of current", write cathode to anode.

2. Electrode Potential & the E° Series

Ecell∘=Ecathode∘−Eanode∘(both as reduction potentials)E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}\quad(\text{both as reduction potentials})

Reading the series:

⚠️ Board Exam Trap: Do not flip the sign of the anode value twice. Use reduction potentials for both electrodes and then subtract: $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$. For Sn4+/Sn2+ (+0.15 V) with Cr3+/Cr (−0.74 V): $0.15-(-0.74)=+0.89$ V.

3. Nernst Equation, ΔG° and Equilibrium Constant

For aA+bB→cC+dDaA+bB\rightarrow cC+dD with nn electrons transferred, at 298 K:

Ecell=Ecell∘−0.0591nlog⁡Q,Q=[C]c[D]d[A]a[B]bE_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log Q,\qquad Q=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The three linked equations (learn them together):

ΔrG∘=−nFEcell∘ΔrG∘=−RTln⁡KEcell∘=0.0591nlog⁡Kc\Delta_rG^\circ=-nFE^\circ_{cell}\qquad \Delta_rG^\circ=-RT\ln K\qquad E^\circ_{cell}=\frac{0.0591}{n}\log K_c

with F=96500F=96500 C mol⁻¹. Spontaneous ⇒ Ecell∘>0E^\circ_{cell}>0, ΔG∘<0\Delta G^\circ<0, K>1K>1.

Worked example (2023 board): Mg | Mg²⁺(0.1 M) || Cu²⁺(0.01 M) | Cu, E∘=2.71E^\circ=2.71 V, n=2n=2:

E=2.71−0.05912log⁡0.10.01=2.71−0.02955=2.68 VE=2.71-\frac{0.0591}{2}\log\frac{0.1}{0.01}=2.71-0.02955=2.68\text{ V}

⚠️ Board Exam Trap: n is not always 1. Count the electrons in the balanced overall reaction. Also, $E^\circ$ never changes when you multiply a half-reaction by 2, but $\Delta G^\circ$ does (it depends on $n$).

4. Conductance & Kohlrausch's Law

Electrolytic (ionic) conductance uses AC (to avoid electrolysis) and depends on ion concentration, charge, size and temperature.

Quantity Formula SI unit
Resistance R=ρ l/AR=\rho\,l/A ohm (Ω)
Conductance G=1/RG=1/R siemens (S)
Conductivity κ=1/ρ=G×lA=G∗R\kappa=1/\rho=G\times\dfrac{l}{A}=\dfrac{G^*}{R} S m⁻¹ (S cm⁻¹)
Cell constant G∗=l/A=κRG^*=l/A=\kappa R m⁻¹ (cm⁻¹)
Molar conductivity Λm=κ×1000M\Lambda_m=\dfrac{\kappa\times1000}{M} S cm² mol⁻¹

Variation with dilution (very important):

Molar conductivity, Λm √c (concentration) 0 Strong electrolyte (KCl) Λm = Λ°m − A√c (straight line) Weak electrolyte (CH₃COOH) steep rise near c → 0 Λ°m (intercept) Λ°m cannot be got by extrapolation
Strong electrolytes give a straight line, so Λ°m is read as the intercept. Weak electrolytes rise so steeply at low concentration that extrapolation is impossible — hence Kohlrausch’s law is used for them.

Kohlrausch's law of independent migration of ions: at infinite dilution each ion contributes a fixed amount, irrespective of the counter-ion.

Λm∘=ν+λ+∘+ν−λ−∘\Lambda_m^\circ=\nu_+\lambda_+^\circ+\nu_-\lambda_-^\circ

Applications (both are board favourites):

  1. Get Λm∘\Lambda_m^\circ of a weak electrolyte: Λm∘(CH3COOH)=Λ∘(CH3COONa)+Λ∘(HCl)−Λ∘(NaCl)\Lambda^\circ_m(\text{CH}_3\text{COOH})=\Lambda^\circ(\text{CH}_3\text{COONa})+\Lambda^\circ(\text{HCl})-\Lambda^\circ(\text{NaCl}).
  2. Degree of dissociation and dissociation constant of a weak electrolyte:

α=ΛmΛm∘,Ka=cα21−α=cΛm2Λm∘(Λm∘−Λm)\alpha=\frac{\Lambda_m}{\Lambda_m^\circ},\qquad K_a=\frac{c\alpha^2}{1-\alpha}=\frac{c\Lambda_m^2}{\Lambda_m^\circ(\Lambda_m^\circ-\Lambda_m)}

Worked example (2024 board): cell constant from 0.05 M KCl: G∗=κR=1.35×10−2×100=1.35G^*=\kappa R=1.35\times10^{-2}\times100=1.35 cm⁻¹. For AgNO₃ (R = 90 Ω): κ=1.35/90=1.5×10−2\kappa=1.35/90=1.5\times10^{-2} S cm⁻¹ and Λm=1.5×10−2×10000.02=750\Lambda_m=\dfrac{1.5\times10^{-2}\times1000}{0.02}=750 S cm² mol⁻¹.

⚠️ Board Exam Trap: Unit conversion. The factor 1000 in $\Lambda_m=\kappa\times1000/M$ works only if $\kappa$ is in S cm−1 and $M$ is in mol L−1. If $\kappa$ is in S m−1, use $\Lambda_m=\kappa/c$ with $c$ in mol m−3. Also note: H+ and OH− have unusually high ionic conductance (Grotthuss mechanism), so acids and bases conduct best.

5. Electrolysis & Faraday's Laws

Faraday's first law: mass deposited is proportional to the charge passed.

m=Z I t=MnF×Q,Q=I×t (coulomb)m=Z\,I\,t=\frac{M}{nF}\times Q,\qquad Q=I\times t\ (\text{coulomb})

Faraday's second law: for the same charge through different electrolytes in series, masses deposited are proportional to their equivalent weights (M/nM/n).

Useful numbers: 1 F = 96500 C = charge on 1 mol of electrons. Charge needed to deposit 1 mol: Ag⁺ = 1 F, Cu²⁺ = 2 F, Al³⁺ = 3 F.

Which species is discharged? (Products in aqueous solutions)

− + CATHODE (−) ANODE (+) Reduction: Na⁺ vs H₂O H₂O wins → H₂↑ Oxidation: Cl⁻ vs H₂O Cl⁻ wins → Cl₂↑ aq. NaCl (brine) NaOH left in solution Lower E° is reduced first at cathode; lower E° is oxidised first at anode (unless overpotential interferes)
Electrolysis of aqueous NaCl with inert electrodes. Na+ (E° = −2.71 V) cannot compete with water at the cathode, so H2 is released. At the anode, Cl− beats water because of the high overpotential of O2 formation.
Electrolyte Electrodes Cathode product Anode product
Molten NaCl Inert Na Cl₂
Aq. NaCl Inert (Pt) H₂ Cl₂ (overpotential of O₂)
Aq. CuSO₄ Inert (Pt) Cu O₂
Aq. CuSO₄ Cu Cu Cu dissolves (Cu → Cu²⁺)
Dil. H₂SO₄ Inert H₂ O₂
⚠️ Board Exam Trap: Inert vs active electrodes. With aqueous CuCl2 and Pt electrodes, Cu deposits at the cathode and Cl2 at the anode (2026 board). Water should oxidise first by E° (+1.23 V vs +1.36 V), but O2 has a high overpotential, so Cl2 comes out. Say "overpotential" in your answer.

6. Batteries & Fuel Cells

Primary cells (not rechargeable): Dry (Leclanché) cell — Zn anode, graphite rod cathode with MnO₂ + NH₄Cl paste; about 1.5 V. Mercury cell — Zn(Hg) anode, HgO/C cathode, KOH-ZnO paste; constant 1.35 V because the overall reaction Zn(Hg)+HgO→ZnO+Hg\text{Zn(Hg)}+\text{HgO}\rightarrow\text{ZnO}+\text{Hg} involves no ion whose concentration changes. Used in watches and hearing aids.

Secondary cells (rechargeable):

Fuel cells convert the energy of combustion directly into electricity. Reactants are fed continuously.

Hot aq. KOH electrolyte porous C + Pt/Pd porous C + Pt/Pd H₂ in O₂ in ANODE (−) CATHODE (+) 2H₂ + 4OH⁻ → 4H₂O + 4e⁻ O₂ + 2H₂O + 4e⁻ → 4OH⁻ Overall: 2H₂ + O₂ → 2H₂O (E° = 1.23 V)
H2–O2 fuel cell: reactants are supplied continuously (unlike a battery) and the only product is water. Efficiency about 70% versus about 40% for a thermal power plant.

Advantages of the H₂–O₂ fuel cell: high efficiency (~70%), pollution-free (product is water), continuous operation, and the water can be consumed by astronauts.

Primary vs secondary battery (asked in 2025-style questions): a primary battery is used once and cannot be recharged (dry cell, mercury cell); a secondary battery can be recharged by passing current in the reverse direction (lead-acid, Ni-Cd).

7. Corrosion

Corrosion is an electrochemical process: a tiny galvanic cell forms on the metal surface in the presence of moisture and air.

water droplet (with dissolved O₂, CO₂) Iron surface ANODE Fe → Fe²⁺ + 2e⁻ CATHODE O₂ + 4H⁺ + 4e⁻ → 2H₂O e⁻ move through metal Fe²⁺ → (O₂, H₂O) → Fe₂O₃·xH₂O (rust) 2Fe + O₂ + 4H⁺ → 2Fe²⁺ + 2H₂O ; E°cell = 1.67 V
Rusting is an electrochemical process: one spot of the iron acts as anode, another as cathode, and the water film is the electrolyte. Sacrificial coating with Zn or Mg (lower E° than Fe) protects iron even when scratched.

Prevention: paint or oil coating; galvanisation (Zn); electroplating; sacrificial anode (Mg, Zn), cathodic protection; alloying (stainless steel).

⚠️ Board Exam Trap: Best metal to coat iron. Choose the metal with an E° more negative than iron (−0.44 V). Given X (−2.36 V) and Y (−0.14 V), X is better: it corrodes first and sacrifices itself even when the coating is scratched. Y is less reactive than Fe, so it would accelerate rusting of iron once scratched.

8. Formula Sheet & Last-Minute Checklist

Concept Formula
Cell EMF Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}
Nernst E=E∘−0.0591nlog⁡QE=E^\circ-\frac{0.0591}{n}\log Q
Gibbs energy ΔG∘=−nFE∘=−2.303RTlog⁡K\Delta G^\circ=-nFE^\circ=-2.303RT\log K
Equilibrium log⁡K=nE∘0.0591\log K=\frac{nE^\circ}{0.0591}
Conductivity κ=G∗/R\kappa=G^*/R
Molar conductivity Λm=1000κ/M\Lambda_m=1000\kappa/M
Kohlrausch Λm∘=ν+λ+∘+ν−λ−∘\Lambda_m^\circ=\nu_+\lambda_+^\circ+\nu_-\lambda_-^\circ
Degree of dissociation α=Λm/Λm∘\alpha=\Lambda_m/\Lambda_m^\circ
Faraday m=M I tnFm=\dfrac{M\,I\,t}{nF}

Before the exam, check:

  1. Can I write cell notation from a reaction (anode on the left)?
  2. Do I remember to omit solids from Q and to use the balanced n?
  3. Do I know the direction of current, and the difference between primary and secondary cells?
  4. Can I state Kohlrausch's law and one application in two lines?
  5. Do I know why Cl₂ (not O₂) forms in brine and why the mercury cell has constant voltage?