Chapters

Everything NCERT Class 12 Unit 2 (Electrochemistry) says that can be asked, plus what JEE Main adds beyond the book (marked Beyond NCERT). The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Every formula in the chapter on one screen, for the last pass. Details and examples are in the sections below.

Quantity	Formula	Watch
Cell potential	E°cell = E°cathode − E°anode	Both as reduction potentials
Nernst, electrode	E = E° − (0.059/n) log (1/[Mⁿ⁺])	298 K; solids = 1
Nernst, cell	E = E° − (0.059/n) log Q	Same n both sides
Equilibrium	E° = (0.059/n) log K	At equilibrium E = 0
Gibbs energy	ΔG = −nFE; ΔG° = −nFE° = −RT ln K	E intensive, ΔG extensive
Combining half-cells	n₃E°₃ = n₁E°₁ + n₂E°₂	Add nE°, never E°
Temperature coefficient	ΔS = nF(dE/dT); ΔH = −nFE + nFT(dE/dT)	Beyond NCERT
Resistance	R = ρ l/A	ρ in Ω m or Ω cm
Conductance, conductivity	G = 1/R; κ = 1/ρ = G × (l/A)	1 S cm⁻¹ = 100 S m⁻¹
Cell constant	G* = l/A = R κ	Unit m⁻¹ or cm⁻¹
Molar conductivity	Λm = κ × 1000 / M	κ in S cm⁻¹ → S cm² mol⁻¹
Strong electrolyte	Λm = Λ°m − A√c	A same for one charge type
Kohlrausch	Λ°m = ν₊λ°₊ + ν₋λ°₋	Weak from strong by + and −
Weak electrolyte	α = Λm / Λ°m; Kₐ = cα²/(1 − α)
Charge	Q = I t; 1 F = 96 487 ≈ 96 500 C mol⁻¹	t in seconds
Faraday I	m = (M / nF) × I t	n = electrons per ion
Faraday II	m₁/m₂ = E₁/E₂ (equivalent masses)	Same charge

0 · What JEE Main asks from this chapter

170 questions in JEE Main 2019–2026, about 1 per paper. Grouped by type (counts are approximate):

Type	Share	Typical question
Conductance: κ, Λm, cell constant, Kohlrausch, α and Kₐ	~28%	Λ°m of BaSO₄ from BaCl₂, H₂SO₄ and HCl
Nernst: E of a cell or half-cell, pH, concentration ratios	~22%	Zn │ Zn²⁺(0.1 M) ║ Ag⁺(0.01 M) │ Ag → E
E° series: oxidising/reducing power, feasibility, combining E°	~19%	E° of Cu²⁺/Cu⁺ from Cu²⁺/Cu and Cu⁺/Cu
Faraday's laws and products of electrolysis	~15%	AlCl₃, 2 A for 30 min → mass of Al
ΔG°, K and E°	~9%	Disproportionation of Cu⁺ → ln K
Batteries, fuel cells, corrosion	~7%	Change in oxidation state of Pb on charging

Most numericals are rounded to the nearest integer, often as "x × 10⁻²". Read the multiplier before you answer.

1 · Electrochemical cells

Electrochemistry: producing electricity from spontaneous reactions, and using electricity to drive non-spontaneous ones.

	Galvanic (voltaic) cell	Electrolytic cell
Energy	chemical → electrical	electrical → chemical
Reaction	spontaneous	non-spontaneous
Anode (oxidation)	negative	positive
Cathode (reduction)	positive	negative

In both, oxidation is at the anode and reduction at the cathode. Only the signs swap.

The Daniell cell: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), E = 1.1 V with both ions at 1 M. Apply an external voltage Eext against it:

• Eext < 1.1 V: electrons flow Zn → Cu (current Cu → Zn); Zn dissolves, Cu deposits. Galvanic.
• Eext = 1.1 V: no current, no reaction.
• Eext > 1.1 V: electrons flow Cu → Zn; Zn deposits at the zinc electrode, Cu dissolves at the copper electrode. Now an electrolytic cell.

(2020 and 2024 ask exactly these three cases.)

2 · Galvanic cells

• The cell reaction is two half reactions in two half-cells (redox couples): reduction at the copper electrode, oxidation at the zinc.
• Each half-cell is a metal dipped in an electrolyte. The two are joined outside by a wire (through a voltmeter and switch) and inside by a salt bridge. If both electrodes share one electrolyte, no salt bridge is needed.
• Electrode potential: the potential difference between electrode and electrolyte, set by the competition between ions depositing (making the metal +) and atoms dissolving (making it −). With every species at unit concentration it is the standard electrode potential. IUPAC: standard reduction potentials are now simply standard electrode potentials.
• Electrons flow from the negative electrode (anode) to the positive (cathode); current flows the other way.

Cell notation: anode on the left, cathode on the right; a single line │ between a metal and its solution, a double line ║ for the salt bridge. Written this way the emf is positive:

Example (NCERT) · Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) is written Cu(s)│Cu²⁺(aq)║Ag⁺(aq)│Ag(s); E = E(Ag⁺/Ag) − E(Cu²⁺/Cu).

The cell emf is the potential difference when no current is drawn.

Standard hydrogen electrode (SHE): Pt(s)│H₂(g)│H⁺(aq), platinum coated with platinum black, in acid, with pure H₂ bubbled through; p(H₂) = 1 bar, [H⁺] = 1 M. Assigned zero potential at all temperatures for H⁺ + e⁻ → ½H₂. A single half-cell's potential can't be measured; only differences can, so every E° is measured against the SHE as anode.

• Pt│H₂│H⁺ ║ Cu²⁺│Cu gives 0.34 V, so E°(Cu²⁺/Cu) = +0.34 V. Cu²⁺ is reduced more easily than H⁺: Cu doesn't dissolve in HCl (in HNO₃ it is oxidised by nitrate, not by H⁺).
• Pt│H₂│H⁺ ║ Zn²⁺│Zn gives −0.76 V: H⁺ can oxidise Zn.
• Daniell: E° = 0.34 − (−0.76) = 1.10 V.
• Inert electrodes (Pt, Au) only provide a surface and carry electrons: the hydrogen electrode, and the bromine electrode Pt(s)│Br₂(aq)│Br⁻(aq).

Reading the E° table (NCERT Table 2.1):

• E° > 0: the reduced form is more stable than H₂; E° < 0: H₂ is more stable than the reduced form.
• Higher E° = stronger oxidising agent (left side). F₂ (+2.87 V) is the strongest oxidising agent; F⁻ the weakest reducing agent.
• Lower E° = stronger reducing agent (right side). Li (−3.05 V) is the strongest reducing agent in water; Li⁺ the weakest oxidising agent.
• A negative E° means the couple is a stronger reducing agent than H⁺/H₂; positive, weaker.
• A species reduces anything above it in the table. Zn displaces Cu from CuSO₄, so CuSO₄ can't be stored in a zinc pot (NCERT intext 2.2).
• Electrochemical cells also give pH, solubility products, equilibrium constants, thermodynamic data, and are used in potentiometric titrations.

Example (2025) · Strongest reducing agent among Cr, Cl⁻, MnO₄⁻, Mn²⁺ given E°: Cr₂O₇²⁻/Cr³⁺ 1.33, Cl₂/Cl⁻ 1.36, MnO₄⁻/Mn²⁺ 1.51, Cr³⁺/Cr −0.74. The reduced form of the lowest couple: Cr.

Combining half-cells to make a third half-cell · Beyond NCERT. E° isn't additive; ΔG° = −nFE° is. Add n × E°.

Example (2020) · Cu²⁺ + 2e⁻ → Cu (0.34 V) minus Cu⁺ + e⁻ → Cu (0.522 V) gives Cu²⁺ + e⁻ → Cu⁺: E° = 2(0.34) − 1(0.522) = 0.158 V.

Example (2023) · FeO₄²⁻ → Fe³⁺ (+2.2 V, 3e⁻) then Fe³⁺ → Fe²⁺ (+0.70 V, 1e⁻): E°(FeO₄²⁻/Fe²⁺) = (3 × 2.2 + 0.70)/4 = 1.825 V.

Example (2019, 2025) · Fe²⁺ + Ag⁺ → Fe³⁺ + Ag with E°(Ag⁺/Ag) = x, E°(Fe²⁺/Fe) = y, E°(Fe³⁺/Fe) = z. E°(Fe³⁺/Fe²⁺) = 3z − 2y, so E°cell = x + 2y − 3z.

3 · Nernst equation

For Mⁿ⁺(aq) + ne⁻ → M(s), with the solid's concentration taken as 1:

R = 8.314 J K⁻¹ mol⁻¹, F = 96 487 C mol⁻¹. 2.303RT/F = 0.059 V at 298 K (papers often give 0.06).

For a cell reaction aA + bB → cC + dD with n electrons:

• Daniell: E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]). E rises with [Cu²⁺] and falls with [Zn²⁺].
• Use the same n for both electrodes: Ni + 2Ag⁺ → Ni²⁺ + 2Ag has n = 2 and Q = [Ni²⁺]/[Ag⁺]².
• Gases enter Q as partial pressures (bar); pure solids and liquids as 1.

Example (NCERT 2.1) · Mg + 2Ag⁺(0.0001 M) → Mg²⁺(0.130 M) + 2Ag, E° = 3.17 V. E = 3.17 − (0.059/2) log(0.130/10⁻⁸) = 3.17 − 0.21 = 2.96 V.

Example (2021) · Zn│Zn²⁺(0.1 M)║Ag⁺(0.01 M)│Ag: E° = 0.80 + 0.76 = 1.56 V. Q = 0.1/(0.01)² = 10³. E = 1.56 − 0.0295 × 3 = 1.47 V.

Hydrogen electrode and pH · H⁺ + e⁻ → ½H₂ at 1 bar: E = −0.059 pH.

Example (2024) · Hydrogen electrode at pH 3: E = −0.059 × 3 = −0.177 V. To make its emf zero in pure water ([H⁺] = 10⁻⁷), p(H₂) must equal [H⁺]² = 10⁻¹⁴ bar (2024).

Example (2022) · Pt│H₂│H⁺║Cu²⁺(0.01 M)│Cu gives 0.576 V (E° 0.34, factor 0.06). H₂ + Cu²⁺ → 2H⁺ + Cu, n = 2: 0.576 = 0.34 − 0.03 log([H⁺]²/0.01) → log[H⁺]² = −9.87 → pH ≈ 5.

Couples with H⁺ in them · Beyond NCERT. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O carries [H⁺]⁸ in Q, so the oxidising power falls as pH rises.

Example (2021) · [H⁺] from 1 M to 10⁻⁴ M, other concentrations fixed: ΔE = (0.059/5) × 8 × 4 = 0.3776 V.

Concentration cell · Beyond NCERT. Same electrode, two concentrations: E° = 0, and the cell is spontaneous (E > 0, ΔG < 0) only when the cathode side is the more concentrated.

4 · E°, K and ΔG°

Equilibrium constant. A closed Daniell cell runs down: [Zn²⁺] rises, [Cu²⁺] falls, the voltmeter drops to zero. At equilibrium E = 0 and Q = K:

So equilibrium constants too large to measure come from E°. Daniell: log K = 2 × 1.1/0.059 = 37.3 → K = 2 × 10³⁷.

Example (NCERT 2.2) · Cu + 2Ag⁺ → Cu²⁺ + 2Ag, E° = 0.46 V. log K = 0.46 × 2/0.059 = 15.6 → K = 3.92 × 10¹⁵.

Gibbs energy. The maximum (reversible) electrical work a cell does equals the fall in its Gibbs energy:

• E is intensive; ΔG is extensive and depends on n. For Zn + Cu²⁺, ΔG = −2FE; for 2Zn + 2Cu²⁺, ΔG = −4FE with the same E.
• E° > 0 ↔ ΔG° < 0 ↔ K > 1: spontaneous.

Example (NCERT 2.3) · Daniell: ΔG° = −2 × 96 487 × 1.1 = −212.3 kJ mol⁻¹. (NCERT's line "−21227 J mol⁻¹" drops a digit; the kJ figure is right.)

Example (2020) · 2Cu⁺ ⇌ Cu + Cu²⁺ with E°(Cu⁺/Cu) = 0.52, E°(Cu²⁺/Cu⁺) = 0.16, RT/F = 0.025. E° = 0.52 − 0.16 = 0.36 V, n = 1: ln K = 0.36/0.025 = 14.4.

Temperature coefficient · Beyond NCERT

Example (2019) · Zn + Cu²⁺, E° = 2 V, dE/dT = −5 × 10⁻⁴ V K⁻¹ at 300 K, F = 96 000. ΔH° = −384 000 + 2 × 96 000 × 300 × (−5 × 10⁻⁴) = −412.8 kJ mol⁻¹.

5 · Conductance of electrolytic solutions

Quantity	Symbol	SI unit (common unit)
Resistance	R	ohm, Ω = kg m² s⁻³ A⁻²
Resistivity (specific resistance)	ρ	Ω m (Ω cm); 1 Ω m = 100 Ω cm
Conductance	G	siemens S = Ω⁻¹ (mho)
Conductivity (specific conductance)	κ	S m⁻¹ (S cm⁻¹); 1 S cm⁻¹ = 100 S m⁻¹
Cell constant	G* = l/A	m⁻¹ (cm⁻¹)
Molar conductivity	Λm	S m² mol⁻¹ (S cm² mol⁻¹)
Degree of dissociation	α	dimensionless

Resistivity is the resistance of a 1 m long, 1 m² cross-section piece; conductivity is its conductance. IUPAC prefers "resistivity" and "conductivity".

Materials by conductivity (NCERT Table 2.2, S m⁻¹, 298 K):

Class	Examples	κ
Conductors	Na, Cu, Ag, Au, Fe; graphite	1.0–6.2 × 10³ (NCERT's values); graphite 1.2 × 10
Aqueous solutions	pure water; 0.1 M HCl; 0.01 M KCl; 0.01 M NaCl; 0.1 M and 0.01 M acetic acid	3.5 × 10⁻⁵; 3.91; 0.14; 0.12; 0.047, 0.016
Semiconductors	CuO, Si, Ge	1 × 10⁻⁷, 1.5 × 10⁻², 2.0
Insulators	glass, Teflon	10⁻¹⁶, 10⁻¹⁸

• Some non-metals conduct electronically: carbon black, graphite, conducting polymers. Superconductors have zero resistivity: metals and alloys at 0–15 K, and now ceramics and mixed oxides up to about 150 K.
• Electronic (metallic) conductance is due to electrons; depends on the metal's nature and structure and valence electrons per atom; decreases as temperature rises. The metal's composition doesn't change.
• Electrolytic (ionic) conductance is due to ions. Depends on (i) nature of the electrolyte, (ii) size of the ions and their solvation, (iii) nature and viscosity of the solvent, (iv) concentration, (v) temperature: it increases with temperature. Not on the electrode (2024).
• Pure water conducts slightly (H⁺ and OH⁻ about 10⁻⁷ M each; 3.5 × 10⁻⁵ S m⁻¹). Distilled water has the lowest conductance of the waters JEE lists (2020).

Measuring κ:

• DC would change the solution's composition, so use AC (an oscillator, 550–5000 Hz).
• The solution sits in a conductivity cell: two Pt electrodes coated with platinum black, area A, distance l apart.
• Cell constant G* = l/A, found by measuring R with a KCl solution of known κ: G* = R κ. Then for any solution κ = G*/R.
• Wheatstone bridge balanced (no current in the detector, a headphone): R₂ = R₁R₄/R₃. Conductivity meters now read it directly.

Molar conductivity:

With κ in S m⁻¹ and c in mol m⁻³ (= 1000 × molarity), Λm is in S m² mol⁻¹. 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹.

Example (NCERT 2.4) · Cell with 0.1 M KCl (κ = 1.29 S m⁻¹): 100 Ω, so G* = 129 m⁻¹ = 1.29 cm⁻¹. Same cell with 0.02 M KCl: 520 Ω → κ = 129/520 = 0.248 S m⁻¹; Λm = 0.248/20 mol m⁻³ = 124 × 10⁻⁴ S m² mol⁻¹ = 124 S cm² mol⁻¹.

Example (NCERT 2.5) · 0.05 M NaOH, column 1 cm across and 50 cm long, R = 5.55 × 10³ Ω. A = 0.785 cm². ρ = RA/l = 87.1 Ω cm; κ = 0.01148 S cm⁻¹; Λm = 0.01148 × 1000/0.05 = 229.6 S cm² mol⁻¹.

Example (2021) · Cell constant 1.14 cm⁻¹, 0.001 M KCl, 1500 Ω. κ = 1.14/1500 = 7.6 × 10⁻⁴ S cm⁻¹; Λm = 7.6 × 10⁻⁴ × 1000/0.001 = 760 S cm² mol⁻¹.

6 · Conductivity and molar conductivity against concentration

• κ always decreases on dilution, for strong and weak electrolytes alike: fewer ions per unit volume carry current. κ is the conductance of a unit volume (unit length, unit area).
• Λm always increases on dilution. Λm = κV, the conductance of the volume V that holds 1 mol between electrodes a unit distance apart. V grows faster than κ falls.
• Limiting molar conductivity Λ°m: Λm as concentration → 0.

Strong electrolytes: Λm rises slowly and linearly with √c (Debye–Hückel–Onsager):

Plot Λm against √c: a straight line, intercept Λ°m, slope −A. A depends on the solvent, temperature and the charge type: NaCl is 1-1, CaCl₂ 2-1, MgSO₄ 2-2; all of one type share A. Unit of A: S cm² mol⁻³ᐟ² L¹ᐟ² (2024).

Example (NCERT 2.6) · KCl: Λm 148.61, 148.29, 147.81, 147.09 at √c 0.01407, 0.01758, 0.02283, 0.03145. The line gives Λ°m = 150.0 S cm² mol⁻¹ and A = 87.46.

Weak electrolytes (acetic acid): α rises sharply on dilution, so Λm rises steeply at low concentration. Λ°m can't be found by extrapolation (and at such dilution κ is too small to measure), so it comes from Kohlrausch's law.

Kohlrausch's law of independent migration of ions: the limiting molar conductivity of an electrolyte is the sum of the contributions of its cation and anion.

His evidence (298 K): Λ°(KCl) − Λ°(NaCl) = Λ°(KBr) − Λ°(NaBr) = Λ°(KI) − Λ°(NaI) ≈ 23.4 S cm² mol⁻¹, and Λ°(NaBr) − Λ°(NaCl) = Λ°(KBr) − Λ°(KCl) ≈ 1.8.

λ° of ions (S cm² mol⁻¹, 298 K, NCERT Table 2.4):

Cation	λ°	Anion · λ°
H⁺	349.6	OH⁻ · 199.1
Na⁺	50.1	Cl⁻ · 76.3
K⁺	73.5	Br⁻ · 78.1
Ca²⁺	119.0	CH₃COO⁻ · 40.9
Mg²⁺	106.0	SO₄²⁻ · 160.0

H⁺ and OH⁻ are far ahead. Cations: H⁺ > Ca²⁺ > Mg²⁺ > K⁺ > Na⁺ (2025).

Applications:

1. Λ°m of any electrolyte from its ions. CaCl₂ = 119.0 + 2(76.3) = 271.6; MgSO₄ = 106.0 + 160.0 = 266 S cm² mol⁻¹ (NCERT 2.7).
2. Λ°m of a weak electrolyte from strong ones (NCERT 2.8): Λ°(HAc) = Λ°(HCl) + Λ°(NaAc) − Λ°(NaCl) = 425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻¹.
3. α and Kₐ of a weak electrolyte:

The second form (Ostwald's dilution law) is Beyond NCERT; 2024 asked it rearranged: Λm²c − KₐΛ°m² + KₐΛmΛ°m = 0.

Example (NCERT 2.9) · 0.001028 M acetic acid, κ = 4.95 × 10⁻⁵ S cm⁻¹. Λm = 4.95 × 10⁻⁵ × 1000/0.001028 = 48.15. α = 48.15/390.5 = 0.1233. Kₐ = 0.001028 × 0.1233²/0.8767 = 1.78 × 10⁻⁵.

Example (2021) · Λ°m(BaSO₄) = Λ°(BaCl₂) + Λ°(H₂SO₄) − 2Λ°(HCl) = 280 + 860 − 2(426) = 288 S cm² mol⁻¹.

4. Solubility of a sparingly soluble salt · Beyond NCERT. The saturated solution is so dilute that Λm ≈ Λ°m: solubility s (mol L⁻¹) = 1000 κ/Λ°m, then Ksp from s.

7 · Electrolysis

An electrolytic cell: two Cu strips in CuSO₄ with DC applied. Cathode: Cu²⁺ + 2e⁻ → Cu (deposits). Anode: Cu → Cu²⁺ + 2e⁻ (dissolves). This is how impure copper is refined: impure Cu as the anode, pure Cu deposits on the cathode.

Metals made by electrolysis (no good chemical reducing agent): Na and Mg from their fused chlorides; Al from Al₂O₃ with cryolite.

Faraday's laws (1833–34):

1. First law: the amount of reaction at an electrode is proportional to the quantity of electricity passed.
2. Second law: the amounts of different substances liberated by the same quantity of electricity are proportional to their equivalent weights (atomic mass ÷ electrons needed to reduce the cation).

• 1 F is the charge on 1 mol of electrons. Ag⁺ + e⁻ → Ag needs 1 F per mol; Mg²⁺ needs 2 F; Al³⁺ needs 3 F.
• M/(nF) is the electrochemical equivalent Z: the mass deposited by 1 C (2024). Coulometers (a standard cell weighing deposited Ag or Cu) were how charge was once measured.
• Industrial cells use up to 50 000 A, about 0.518 F per second.

Example (NCERT 2.10) · CuSO₄, 1.5 A for 10 min: Q = 900 C. Cu = 63 × 900/(2 × 96 487) = 0.2938 g.

Example (2025) · AlCl₃, 2 A for 30 min: Q = 3600 C = 0.0373 F → Al = 0.0373/3 × 27 = 0.336 g.

Example (2022) · Dilute H₂SO₄, 0.10 A for 2 h: Q = 720 C = 7.46 × 10⁻³ F. H₂ = F/2 = 3.73 × 10⁻³ mol, O₂ = F/4 = 1.87 × 10⁻³ mol. Total 5.60 × 10⁻³ mol × 22.7 L = 127 cm³.

Example (2019) · Recharging the anodic half-cell of a lead-acid battery: PbSO₄ + 2e⁻ → Pb + SO₄²⁻. 0.05 F → 0.025 mol PbSO₄ × 303 = 7.6 g.

Faraday counts from a half-equation: MnO₄⁻ → Mn²⁺ needs 5 F per mol (5 mol → 25 F); Cr₂O₇²⁻ → 2Cr³⁺ needs 6 F; Fe₃O₄ → Fe needs 8/3 F per mol Fe (nearest integer 3); 1 mol H₂O → ½O₂ gives 2 F = 1.93 × 10⁵ C.

Products of electrolysis depend on what is electrolysed and on the electrodes. An inert electrode (Pt, Au) only gives or takes electrons; a reactive one takes part. Some reactions need an extra overpotential because they are slow.

Electrolysed	Cathode	Anode
Molten NaCl	Na	Cl₂
Aqueous NaCl	H₂ (H₂O + e⁻ → ½H₂ + OH⁻; E° 0.00 beats Na⁺'s −2.71)	Cl₂ (O₂ has E° 1.23 < 1.36 but needs overpotential)
Dilute H₂SO₄	H₂	O₂
Concentrated H₂SO₄	H₂	S₂O₈²⁻ (peroxodisulphate; E° 1.96)
AgNO₃, Pt electrodes · Beyond NCERT	Ag	O₂
AgNO₃, Ag electrodes · Beyond NCERT	Ag	Ag dissolves
CuSO₄, Cu electrodes	Cu	Cu dissolves

Net for aqueous NaCl: NaCl + H₂O → Na⁺ + OH⁻ + ½H₂ + ½Cl₂; the solution becomes NaOH. At the cathode the couple with the higher E° wins; at the anode the one with the lower E° wins, unless overpotential intervenes. Nernst potentials replace E° when concentrations matter.

8 · Batteries

A battery is one or more galvanic cells in series. A useful one is light, compact, and its voltage stays steady in use.

Primary batteries (react once, can't be recharged):

Part	Dry (Leclanché) cell	Mercury cell
Anode	Zn container	Zn–Hg amalgam
Cathode	graphite rod in MnO₂ + C	HgO + C paste
Electrolyte	moist NH₄Cl + ZnCl₂ paste	KOH + ZnO paste
Voltage	~1.5 V	~1.35 V

• Dry cell: anode Zn → Zn²⁺ + 2e⁻; cathode MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃. Mn goes from +4 to +3. The NH₃ complexes Zn²⁺ as [Zn(NH₃)₄]²⁺. Used in transistors and clocks.
• Mercury cell: anode Zn(Hg) + 2OH⁻ → ZnO + H₂O + 2e⁻; cathode HgO + H₂O + 2e⁻ → Hg + 2OH⁻; overall Zn(Hg) + HgO → ZnO + Hg. Its voltage stays constant because no ion whose concentration changes appears in the overall reaction. Used in hearing aids and watches.

Secondary batteries (recharged by passing current the opposite way):

• Lead storage battery (cars, invertors): Pb anode; cathode a lead grid packed with PbO₂; electrolyte 38% H₂SO₄.
  • Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻
  • Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O
  • Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
  • Charging reverses it: PbSO₄ on the anode plate → Pb (Pb +2 → 0); on the cathode plate → PbO₂ (+2 → +4) (2025).
• Nickel–cadmium cell: longer life than lead storage, dearer to make. Discharge: Cd + 2Ni(OH)₃ → CdO + 2Ni(OH)₂ + H₂O.

9 · Fuel cells

Galvanic cells that turn the energy of combustion of a fuel (H₂, CH₄, CH₃OH) straight into electricity, with reactants fed in and products removed continuously.

• H₂–O₂ fuel cell: H₂ and O₂ bubbled through porous carbon electrodes into concentrated aqueous NaOH; finely divided Pt or Pd catalysts in the electrodes.
  • Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻
  • Anode: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻
  • Overall: 2H₂ + O₂ → 2H₂O
• Used in the Apollo space programme; the water was drunk by the astronauts.
• Efficiency about 70%, against about 40% for thermal plants. Pollution free; tried in cars.

10 · Corrosion

Corrosion coats metals with oxides or other salts: rusting of iron, tarnishing of silver, the green coat on copper and bronze. It is an electrochemical process. Rusting needs water and air.

• Anode (a spot on the iron): 2Fe → 2Fe²⁺ + 4e⁻, E°(Fe²⁺/Fe) = −0.44 V.
• Electrons travel through the metal to another spot, the cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O, E° = 1.23 V. The H⁺ comes from H₂CO₃ (dissolved CO₂) and other acidic oxides.
• Overall: 2Fe + O₂ + 4H⁺ → 2Fe²⁺ + 2H₂O, E° = 1.67 V.
• Fe²⁺ is oxidised further by air to rust, Fe₂O₃·xH₂O, releasing more H⁺.

Prevention:

• Keep air away: paint, or chemicals such as bisphenol.
• Cover with another metal (Sn, Zn) that is inert or reacts to save the object.
• Sacrificial electrode: a more reactive metal (Mg, Zn) corrodes instead of the iron.

Beyond NCERT (asked 2024): tin is less reactive than iron, so once a tin coat cracks the iron rusts faster; zinc (galvanising) protects even when scratched. Rusting stops above pH 9–10, where there is too little H⁺.

11 · Traps

Trap	What goes wrong	Do this
Adding E°	E°(Cu²⁺/Cu⁺) = 0.34 − 0.522	Add n × E°, then divide by the new n
Multiplying a half-reaction	Doubling E° with the coefficients	E° is intensive: never scale it
Sign of E°cell	Subtracting oxidation potentials	Cathode minus anode, both as reduction
n in Nernst	Different n for the two electrodes	Balance the whole cell; one n
Q upside down	[reactants]/[products]	Products over reactants; solids = 1
Λm units	Missing the 1000, or using it with SI	S cm⁻¹ and mol L⁻¹ → × 1000; S m⁻¹ and mol m⁻³ → none
κ on dilution	Saying κ rises on dilution	κ falls, Λm rises
Weak Λ°m	Extrapolating acetic acid's curve	Kohlrausch: HCl + NaAc − NaCl
Anode sign	Calling the anode negative in electrolysis	Galvanic anode −; electrolytic anode +
Aqueous NaCl	Writing Na at the cathode	H₂ at the cathode; Na only from molten NaCl
Faraday's n	Using the ion's charge when the half-equation differs	Count electrons in the balanced half-equation
Lead battery on charge	Naming anode and cathode products the wrong way	Anode plate PbSO₄ → Pb; cathode plate PbSO₄ → PbO₂
Fuel cell facts	Picking 40% efficiency or Al catalyst	~70%; Pt or Pd catalyst

12 · Numbers worth knowing

• F = 96 487 ≈ 96 500 C mol⁻¹; e = 1.602 × 10⁻¹⁹ C; R = 8.314 J K⁻¹ mol⁻¹.
• 2.303RT/F = 0.059 V at 298 K (0.06 when a paper says so); RT/F = 0.0257 V.
• Faradays per mol: Ag⁺ 1, Cu²⁺ 2, Al³⁺ 3, MnO₄⁻ → Mn²⁺ 5, Cr₂O₇²⁻ → 2Cr³⁺ 6, H₂O → ½O₂ 2.
• Daniell 1.10 V; dry cell ~1.5 V; mercury cell ~1.35 V; corrosion of iron E° 1.67 V; fuel cell ~70% efficient.

Standard reduction potentials (V, 298 K), the ones questions use:

Couple	E°
F₂/F⁻	+2.87
Co³⁺/Co²⁺	+1.81
H₂O₂/H₂O	+1.78
MnO₄⁻/Mn²⁺	+1.51
Au³⁺/Au	+1.40
Cl₂/Cl⁻	+1.36
Cr₂O₇²⁻/Cr³⁺	+1.33
O₂/H₂O	+1.23
Br₂/Br⁻	+1.09
NO₃⁻/NO	+0.97
Ag⁺/Ag	+0.80
Fe³⁺/Fe²⁺	+0.77
I₂/I⁻	+0.54
Cu⁺/Cu	+0.52
Cu²⁺/Cu	+0.34
AgCl/Ag, Cl⁻	+0.22
H⁺/H₂	0.00
Pb²⁺/Pb	−0.13
Sn²⁺/Sn	−0.14
Ni²⁺/Ni	−0.25
Fe²⁺/Fe	−0.44
Cr³⁺/Cr	−0.74
Zn²⁺/Zn	−0.76
H₂O/H₂, OH⁻	−0.83
Al³⁺/Al	−1.66
Mg²⁺/Mg	−2.36
Na⁺/Na	−2.71
Ca²⁺/Ca	−2.87
K⁺/K	−2.93
Li⁺/Li	−3.05

13 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• Electrochemistry makes metals, NaOH, Cl₂, F₂; it can be energy efficient and less polluting. Nerve signals have an electrochemical origin.
• Conducting polymers: MacDiarmid, Heeger and Shirakawa (1977) made polyacetylene conduct by exposing it to iodine vapour; polyaniline, polypyrrole and polythiophene followed. Light, flexible, used in light batteries and bendable transistors. Nobel Prize in Chemistry, 2000.
• The hydrogen economy: burning fossil fuels releases CO₂ and warms the Earth (Maldives at risk of submergence). Hydrogen burns to water only; made by splitting water with solar energy and used in fuel cells, both electrochemical technologies.
• Corrosion costs crores of rupees a year and can cause accidents such as bridge collapses.