“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.
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)
Non-spontaneous (ΔG>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
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
The potential of a single electrode cannot be measured, so it is measured against the Standard Hydrogen Electrode (SHE), assigned E∘=0.00 V (Pt, H₂ at 1 bar, 1 M H⁺).
Standard conditions: 1 M concentration, 1 bar pressure, 298 K.
All tabulated values are reduction potentials.
Ecell∘=Ecathode∘−Eanode∘(both as reduction potentials)
Reading the series:
More positiveE∘ = stronger oxidising agent (e.g. F₂, +2.87 V).
More negativeE∘ = stronger reducing agent (e.g. Li, −3.05 V).
A metal displaces another from its salt solution if its E∘ is lower (more negative). Metals above H (negative E∘) liberate H₂ from dilute acids; Cu, Ag, Au (positive E∘) do not.
⚠️ 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+dD with n electrons transferred, at 298 K:
Ecell=Ecell∘−n0.0591logQ,Q=[A]a[B]b[C]c[D]d
Pure solids and liquids (Cu, Zn, H₂O) are omitted from Q. Gases enter as partial pressure.
For a single electrode Mn++ne−→M: E=E∘−n0.0591log[Mn+]1.
At equilibrium, Ecell=0 (a "dead" battery) and Q=Kc.
with F=96500 C mol⁻¹. Spontaneous ⇒ Ecell∘>0, ΔG∘<0, K>1.
Worked example (2023 board): Mg | Mg²⁺(0.1 M) || Cu²⁺(0.01 M) | Cu, E∘=2.71 V, n=2:
E=2.71−20.0591log0.010.1=2.71−0.02955=2.68 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=ρl/A
ohm (Ω)
Conductance
G=1/R
siemens (S)
Conductivity
κ=1/ρ=G×Al=RG∗
S m⁻¹ (S cm⁻¹)
Cell constant
G∗=l/A=κR
m⁻¹ (cm⁻¹)
Molar conductivity
Λm=Mκ×1000
S cm² mol⁻¹
Variation with dilution (very important):
Conductivity (κ) decreases on dilution (fewer ions per unit volume).
Molar conductivity (Λm) increases on dilution (more dissociation for weak electrolytes; less inter-ionic attraction for strong ones).
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.
Strong electrolytes: Debye-Hückel-Onsager equation Λm=Λm∘−Ac. Extrapolate the line to c=0 to get Λm∘.
Weak electrolytes: the curve is too steep to extrapolate, so use Kohlrausch's law.
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−∘
Applications (both are board favourites):
Get Λm∘ of a weak electrolyte: Λm∘(CH3COOH)=Λ∘(CH3COONa)+Λ∘(HCl)−Λ∘(NaCl).
Degree of dissociation and dissociation constant of a weak electrolyte:
α=Λm∘Λm,Ka=1−αcα2=Λm∘(Λm∘−Λm)cΛm2
Worked example (2024 board): cell constant from 0.05 M KCl: G∗=κR=1.35×10−2×100=1.35 cm⁻¹. For AgNO₃ (R = 90 Ω): κ=1.35/90=1.5×10−2 S cm⁻¹ and Λm=0.021.5×10−2×1000=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=ZIt=nFM×Q,Q=I×t(coulomb)
Faraday's second law: for the same charge through different electrolytes in series, masses deposited are proportional to their equivalent weights (M/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: the species with the higher reduction potential is reduced first. Cu²⁺ (+0.34 V) beats H₂O; Na⁺, K⁺, Ca²⁺ (very negative E°) lose to water, so H₂ is evolved.
Anode (inert electrode): the species that is easier to oxidise (lower E∘) goes first, but O₂ evolution from water needs extra voltage (overpotential), so Cl⁻ is oxidised in brine.
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 involves no ion whose concentration changes. Used in watches and hearing aids.
Discharge: Pb+PbO2+2H2SO4→2PbSO4+2H2O (about 2 V per cell; acid density falls).
Charging reverses this.
Ni-Cd cell: longer life (~1.4 V), but costlier.
Fuel cells convert the energy of combustion directly into electricity. Reactants are fed continuously.
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.
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.
Anode:Fe→Fe2++2e− (E∘=−0.44 V)
Cathode:O2+4H++4e−→2H2O (E∘=+1.23 V)
Then Fe²⁺ is oxidised by air and forms rust, Fe2O3⋅xH2O.
⚠️ 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∘
Nernst
E=E∘−n0.0591logQ
Gibbs energy
ΔG∘=−nFE∘=−2.303RTlogK
Equilibrium
logK=0.0591nE∘
Conductivity
κ=G∗/R
Molar conductivity
Λm=1000κ/M
Kohlrausch
Λm∘=ν+λ+∘+ν−λ−∘
Degree of dissociation
α=Λm/Λm∘
Faraday
m=nFMIt
Before the exam, check:
Can I write cell notation from a reaction (anode on the left)?
Do I remember to omit solids from Q and to use the balanced n?
Do I know the direction of current, and the difference between primary and secondary cells?
Can I state Kohlrausch's law and one application in two lines?
Do I know why Cl₂ (not O₂) forms in brine and why the mercury cell has constant voltage?