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State Board · Class 12 · Chemistry

Lesson 5: Electrochemistry

7 concepts · 56 practice questions with answers · Free to read

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Electrical conductance

From daily life

Current passes easily through a copper wire but not through sugar solution — conductance measures that ease.

Simple definition

Conductance G = 1/R (unit S). Conductivity κ = G × G*, where G* = l/A is the cell constant.

ConductanceElectrical conductance: R, G, κR (Ω)resistanceG = 1/R(S = Ω⁻¹)κ = G × G*(S cm⁻¹)G* = l/A = cell constant (cm⁻¹)Example: R = 100 Ω, G* = 1.29 cm⁻¹G = 1/100 = 0.01 S ; κ = 0.01 × 1.29 = 0.0129 S cm⁻¹Metals: electrons carry, T↑ ⇒ G↓Electrolytes: ions carry, T↑ ⇒ G↑κ: conductance of 1 cm³ of solution1 S = 1 Ω⁻¹; the cell constant is found with a known KCl solution

In the figure : R, G and κ.

Rule 1: G and κ — G = 1/R; κ = G × (l/A); unit of κ is S cm⁻¹ (SI: S m⁻¹).

  • Easy : R = 100 Ω ⇒ G = 0.01 S.
  • Medium : G* = 1.29 cm⁻¹ ⇒ κ = 0.0129 S cm⁻¹.
  • Exam-type : Double R ⇒ half G.

Rule 2: Metals and electrolytes — Metals: electrons carry current, T↑ ⇒ conductance ↓. Electrolytes: ions carry current, T↑ ⇒ ↑.

  • Easy : Copper is a metallic conductor.
  • Medium : NaCl solution is an ionic conductor.
  • Exam-type : Heating raises the conductance of a solution.

Rule 3: Cell constant — Use KCl of known κ: G* = κ / G.

  • Easy : Unit of G* is cm⁻¹.
  • Medium : Larger l ⇒ larger G*.
  • Exam-type : Larger A ⇒ smaller G*.

Calculation: R = 100 Ω, G* = 1.29 cm⁻¹

  • Step 1: G = 1/100 = 0.01 S.
  • Step 2: κ = G × G* = 0.01 × 1.29.
  • Step 3: κ = 0.0129 S cm⁻¹.

CET level (MCQ)

  • CET 1) Unit of κ? 1) S cm⁻¹ 2) S cm 3) Ω cm⁻¹ 4) S — Answer: 1).
  • CET 2) Conductance rises with temperature in? 1) electrolyte 2) copper 3) silver 4) aluminium — Answer: 1).

Real-life use: Water purity, batteries, TDS meters.

Remember

G = 1/R; κ = G × G*; electrolytes: T↑ ⇒ G↑.

Common mistake

Treating G and κ as the same unit.

CET trap

G is in S; κ is in S cm⁻¹.

Shortcut

Memory clue: "G = 1/R; κ = G × G*".

Spot the difference

QuantityUnit
RΩ
GS
κS cm⁻¹
G*cm⁻¹

Practice questions

Easy
  1. 1Relation of G and R.
  2. 2Define κ.
  3. 3What is cell constant?
Medium
  1. 4R = 100 Ω, G* = 1.29 ⇒ κ.
  2. 5Effect of T on metals and solutions.
  3. 6SI unit of κ.
Challenging
  1. 7How is G* found?
  2. 8Doubling l ⇒ G*?
Show answers (try them yourself first)
  1. 1G = 1/R.
  2. 2Conductance of 1 cm³.
  3. 3l/A.
  4. 40.0129 S cm⁻¹.
  5. 5Metal ↓, solution ↑.
  6. 6S m⁻¹.
  7. 7Using KCl solution.
  8. 8Doubles.

Molar conductivity and Kohlrausch's law

From daily life

With a thinner crowd each person moves more freely — ions in a dilute solution move more easily.

Simple definition

Λm = κ × 1000 / c (S cm² mol⁻¹). Kohlrausch: Λ° = λ°(+) + λ°(–): independent ion contributions at infinite dilution.

ΛmΛm versus √c: strong and weak electrolytes√cΛmstrong (KCl)weak (CH₃COOH)Λ°Finding Λ°strong: extrapolateweak: KohlrauschΛm = κ × 1000 / c (S cm² mol⁻¹)Λ°(CH₃COOH) = 426.1 + 91.0 – 126.5 = 390.60.01 M, κ = 1.63×10⁻⁴ ⇒ Λm = 16.3; α = 0.0417; Ka = 1.82×10⁻⁵

In the figure : Λm versus √c.

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Molar conductivity and Kohlrausch's law

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Rule 1: Λm and dilution — Λm rises on dilution (more ionisation, fewer interactions); strong electrolyte: Λm = Λ° – A√c.

  • Easy : KCl gives a straight line.
  • Medium : CH₃COOH gives a steep curve.
  • Exam-type : Λ° is the y-intercept.

Rule 2: Kohlrausch's law — Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) – Λ°(NaCl) = 91.0 + 426.1 – 126.5 = 390.6.

  • Easy : Finds Λ° of weak electrolytes.
  • Medium : Unit S cm² mol⁻¹.
  • Exam-type : Ions move independently.

Rule 3: α and Ka — α = Λm/Λ°; Ka = cα²/(1 – α); 0.01 M, κ = 1.63×10⁻⁴ ⇒ Λm = 16.3, α = 0.0417, Ka = 1.82×10⁻⁵.

  • Easy : Λm = 1.63×10⁻⁴ × 1000 / 0.01.
  • Medium : α = 16.3/390.6.
  • Exam-type : Ka ≈ 10⁻⁵ for acetic acid.

Calculation: 0.01 M acetic acid, κ = 1.63×10⁻⁴ S cm⁻¹

  • Step 1: Λm = κ × 1000 / c = 16.3.
  • Step 2: α = 16.3 / 390.6 = 0.0417.
  • Step 3: Ka = 0.01 × 0.0417² / (1 – 0.0417) = 1.82×10⁻⁵.

CET level (MCQ)

  • CET 1) On dilution Λm? 1) rises 2) falls 3) constant 4) zero — Answer: 1).
  • CET 2) Unit of Λm? 1) S cm² mol⁻¹ 2) S cm⁻¹ 3) S 4) Ω — Answer: 1).

Real-life use: Measuring Ka of weak acids, drug solubility.

Remember

Λm = κ×1000/c; Λ° = λ₊ + λ₋; α = Λm/Λ°.

Common mistake

Using mol cm⁻³ instead of mol L⁻¹.

CET trap

Multiply by 1000 and keep c in mol L⁻¹.

Shortcut

Memory clue: "κ times thousand over c".

Spot the difference

AspectStrongWeak
Λm plotstraight linesteep curve
Finding Λ°extrapolationKohlrausch

Practice questions

Easy
  1. 1Define Λm.
  2. 2Why does Λm rise on dilution?
  3. 3State Kohlrausch's law.
Medium
  1. 4Find Λ°(CH₃COOH).
  2. 5Find α.
  3. 6Find Ka.
Challenging
  1. 7Strong vs weak graph.
  2. 8Unit of Λm.
Show answers (try them yourself first)
  1. 1κ×1000/c.
  2. 2Ionisation increases.
  3. 3Independent ion contributions.
  4. 4390.6.
  5. 50.0417.
  6. 61.82×10⁻⁵.
  7. 7Line vs curve.
  8. 8S cm² mol⁻¹.

Electrolysis and Faraday's laws

From daily life

Gold plating on a temple dome is done by electrolysis.

Simple definition

Electrolysis drives a reaction with electrical energy. Faraday: m = (M/nF) × Q = (M/nF) × I t; F = 96500 C mol⁻¹.

ElectrolysisElectrolysis and Faraday+ –cathode (–)anode (+)Q = I × tm = (M/nF) × I tF = 96500 C mol⁻¹5 A × 30 min, CuSO₄Q = 9000 C ⇒ 2.96 g Cu9 g Al ⇒ 96500 C10 A ⇒ 2.68 hElectrolyteCathodeAnodeNaCl (molten)NaCl₂NaCl (aq)H₂Cl₂CuSO₄ (Pt)CuO₂CuSO₄ (Cu)CuCu dissolves

In the figure : Electrolysis and products.

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Electrolysis and Faraday's laws

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Rule 1: Faraday's laws — 1) m ∝ Q. 2) For the same Q, m ∝ M/n (equivalent mass).

  • Easy : 5 A × 30 min = 9000 C.
  • Medium : Cu²⁺: n = 2 ⇒ m = 2.96 g.
  • Exam-type : 1 mol e⁻ = 96500 C.

Rule 2: Products — Molten NaCl: Na, Cl₂; NaCl(aq): H₂, Cl₂; CuSO₄ (Pt): Cu, O₂; CuSO₄ (Cu electrodes): Cu deposits, anode dissolves.

  • Easy : Water is reduced in preference to Na⁺.
  • Medium : Refining: impure Cu as anode.
  • Exam-type : Plating: object is the cathode.

Rule 3: Uses — Extraction (Na, Al), purification (Cu), electroplating, Cl₂ and NaOH production.

  • Easy : 9 g Al ⇒ 96500 C.
  • Medium : 10 A ⇒ 9648.5 s = 2.68 h.
  • Exam-type : Al³⁺: n = 3.

Calculation: 5 A, 30 min, CuSO₄ (Cu = 63.5)

  • Step 1: Q = I t = 5 × 1800 = 9000 C.
  • Step 2: m = (63.5 / (2 × 96500)) × 9000.
  • Step 3: m = 2.96 g Cu.

CET level (MCQ)

  • CET 1) Charge of 1 mol of electrons? 1) 96500 C 2) 9650 C 3) 1 C 4) 8.314 C — Answer: 1).
  • CET 2) At the cathode of aqueous NaCl? 1) H₂ 2) Na 3) Cl₂ 4) O₂ — Answer: 1).

Real-life use: Electroplating, aluminium extraction, chlor-alkali industry.

Remember

m = (M/nF) I t; reduction at cathode, oxidation at anode.

Common mistake

Leaving t in minutes.

CET trap

Convert t to seconds (30 min = 1800 s).

Shortcut

Memory clue: "I times t is Q".

Spot the difference

ElectrolyteCathodeAnode
NaCl (l)NaCl₂
NaCl (aq)H₂Cl₂
CuSO₄ (Pt)CuO₂

Practice questions

Easy
  1. 1Faraday's first law.
  2. 2Second law.
  3. 35 A × 30 min ⇒ how much Cu?
Medium
  1. 4Time for 9 g Al at 10 A?
  2. 5Products of NaCl(aq) electrolysis.
  3. 6How is Cu refined?
Challenging
  1. 7Where is the object in electroplating?
  2. 8What is F?
Show answers (try them yourself first)
  1. 1m ∝ Q.
  2. 2m ∝ M/n.
  3. 32.96 g.
  4. 42.68 h.
  5. 5H₂, Cl₂.
  6. 6Impure Cu as anode.
  7. 7Cathode.
  8. 896500 C mol⁻¹.

Galvanic cells

From daily life

A phone battery is a galvanic cell: a chemical reaction supplies electrical energy.

Simple definition

A galvanic cell turns a spontaneous redox reaction into electrical energy. Anode = oxidation (–); cathode = reduction (+). E°cell = E°cathode – E°anode.

DaniellDaniell cell: Zn | Zn²⁺ ‖ Cu²⁺ | Cusalt bridgeV = 1.10Zn: anode (–)Cu: cathode (+)Zn → Zn²⁺ + 2e⁻Cu²⁺ + 2e⁻ → CuE°cell = E°cathode – E°anode = 0.34 – (–0.76) = 1.10 VElectrons flow Zn ⇒ Cu externally; ions move through the salt bridgeAnode oxidation · cathode reduction (AN OX, RED CAT)

In the figure : Daniell cell.

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Galvanic cells

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Rule 1: Daniell cell — Zn | Zn²⁺ ‖ Cu²⁺ | Cu; anode Zn → Zn²⁺ + 2e⁻; cathode Cu²⁺ + 2e⁻ → Cu; E°cell = 0.34 + 0.76 = 1.10 V.

  • Easy : The Zn plate wears away.
  • Medium : The Cu plate thickens.
  • Exam-type : Electrons flow Zn ⇒ Cu.

Rule 2: Salt bridge — A KCl/KNO₃ tube completes the circuit, keeps charge balanced and prevents mixing.

  • Easy : Anions move toward the anode.
  • Medium : Cations move toward the cathode.
  • Exam-type : No bridge ⇒ cell stops.

Rule 3: Cell notation — Anode | anode ion ‖ cathode ion | cathode; ‖ is the salt bridge.

  • Easy : Zn | Zn²⁺(1 M) ‖ Cu²⁺(1 M) | Cu.
  • Medium : Oxidation on the left.
  • Exam-type : Reduction on the right.

Calculation: E°cell of the Daniell cell

  • Step 1: E°cathode (Cu²⁺/Cu) = +0.34 V.
  • Step 2: E°anode (Zn²⁺/Zn) = –0.76 V.
  • Step 3: E°cell = 0.34 – (–0.76) = 1.10 V.

CET level (MCQ)

  • CET 1) At the anode? 1) oxidation 2) reduction 3) both 4) none — Answer: 1).
  • CET 2) E°cell of the Daniell cell? 1) 1.10 V 2) 0.42 V 3) 0.76 V 4) 0.34 V — Answer: 1).

Real-life use: Batteries, pacemakers, cars.

Remember

Anode oxidation, cathode reduction; E°cell = E°c – E°a.

Common mistake

Taking the anode as positive in a galvanic cell.

CET trap

Galvanic: anode –; electrolytic: anode +.

Shortcut

Memory clue: "AN OX, RED CAT".

Spot the difference

AspectGalvanicElectrolytic
Energychemical ⇒ electricalelectrical ⇒ chemical
Anode–+
Cathode+–

Practice questions

Easy
  1. 1What is a galvanic cell?
  2. 2Reactions of the Daniell cell.
  3. 3Functions of salt bridge.
Medium
  1. 4How is E°cell found?
  2. 5Write the cell notation.
  3. 6Galvanic vs electrolytic.
Challenging
  1. 7Direction of e⁻ flow.
  2. 8Why is Zn the anode?
Show answers (try them yourself first)
  1. 1Electricity from a spontaneous reaction.
  2. 2Zn → Zn²⁺, Cu²⁺ → Cu.
  3. 3Completes circuit, neutralises charge.
  4. 4E°c – E°a.
  5. 5Zn | Zn²⁺ ‖ Cu²⁺ | Cu.
  6. 6Energy and sign.
  7. 7Zn ⇒ Cu.
  8. 8Lower E° ⇒ easier oxidation.

Electrode potentials and the electrochemical series

From daily life

Heights are measured from a zero level; electrode potentials are measured from the SHE, set to zero.

Simple definition

Standard electrode potential E°: measured against the SHE (1 M H⁺, 1 bar H₂, Pt) at 298 K, 1 M. ΔG° = –nFE°.

E°E° (V): SHE = 0, 298 KLi⁺/Li-3.05K⁺/K-2.93Mg²⁺/Mg-2.37Al³⁺/Al-1.66Zn²⁺/Zn-0.76Fe²⁺/Fe-0.44H⁺/H₂0.00Cu²⁺/Cu+0.34Ag⁺/Ag+0.80Cl₂/Cl⁻+1.36F₂/F⁻+2.87negative ⇒ strong reducing agent; positive ⇒ strong oxidising agentΔG° = –nFE° = –2 × 96500 × 1.10 = –212.3 kJ

In the figure : E° ladder.

Rule 1: The series — Lower E° ⇒ stronger reducing agent; higher E° ⇒ stronger oxidising agent; Li –3.05 ... F₂ +2.87.

  • Easy : Zn (–0.76) reduces Cu²⁺.
  • Medium : Cu (+0.34) cannot reduce Zn²⁺.
  • Exam-type : F₂ is the strongest oxidising agent.

Rule 2: ΔG° and E° — ΔG° = –nFE°; Daniell: –2 × 96500 × 1.10 = –212.3 kJ.

  • Easy : Positive E° ⇒ negative ΔG° ⇒ spontaneous.
  • Medium : Negative E° ⇒ not spontaneous.
  • Exam-type : n = electrons transferred.

Rule 3: Applications — Predicts reactions and displacement; acids liberate H₂ only from metals with E° below 0.

  • Easy : Zn + HCl ⇒ H₂.
  • Medium : Cu + HCl ⇒ no reaction.
  • Exam-type : Fe + CuSO₄ ⇒ Cu.

Calculation: ΔG° for the Daniell cell

  • Step 1: n = 2, E° = 1.10 V.
  • Step 2: ΔG° = –2 × 96500 × 1.10 = –212300 J.
  • Step 3: ≈ –212.3 kJ mol⁻¹ (spontaneous).

CET level (MCQ)

  • CET 1) Strongest reducing agent? 1) Li 2) Zn 3) Cu 4) Ag — Answer: 1).
  • CET 2) E° of SHE? 1) 0 V 2) 1 V 3) –1 V 4) 0.34 V — Answer: 1).

Real-life use: Corrosion, battery design, displacement reactions.

Remember

Low E° ⇒ reducing agent; ΔG° = –nFE°.

Common mistake

Multiplying E° by stoichiometric coefficients.

CET trap

E° is intensive — it does not change with coefficients; ΔG° does.

Shortcut

Memory clue: "low E° gives electrons".

Spot the difference

Half-reactionE° (V)
Li⁺/Li–3.05
Zn²⁺/Zn–0.76
H⁺/H₂0.00
Cu²⁺/Cu+0.34
F₂/F⁻+2.87

Practice questions

Easy
  1. 1What is SHE?
  2. 2Define E°.
  3. 3ΔG° of the Daniell cell.
Medium
  1. 4Which metals give H₂?
  2. 5Why does Zn + CuSO₄ occur?
  3. 6Strongest oxidising agent.
Challenging
  1. 7Relation of E° and ΔG°.
  2. 8Meaning of n.
Show answers (try them yourself first)
  1. 1Standard hydrogen electrode.
  2. 2Potential versus SHE.
  3. 3–212.3 kJ.
  4. 4Those with E° < 0.
  5. 5E°cell = 1.10 V is positive.
  6. 6F₂.
  7. 7ΔG° = –nFE°.
  8. 8Electrons transferred.

The Nernst equation

From daily life

A battery runs down as reactants fall and products rise, so the voltage drops.

Simple definition

Nernst: E = E° – (0.0592/n) log Q at 298 K; at equilibrium E = 0 and Q = K.

NernstNernst equation: E versus log Qlog QE (V)E° (Q = 1)Q↓ ⇒ E↑Q = K ⇒ E = 0(equilibrium, dead cell)E = E° – (0.0592/n) log Q  (298 K)Daniell [Zn²⁺] 0.1, [Cu²⁺] 0.01 ⇒ E = 1.10 – 0.0296 log 10 = 1.07 Vlog K = nE°/0.0592 = 2 × 1.10 / 0.0592 = 37.16Cu²⁺ 0.001 M half-cell: 0.34 – 0.0296 log(1/0.001) = 0.251 V

In the figure : E versus log Q.

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The Nernst equation

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Rule 1: Cell potential — Q = [products]/[reactants]; lower Q ⇒ higher E; Daniell: Q = [Zn²⁺]/[Cu²⁺].

  • Easy : [Zn²⁺] 0.1, [Cu²⁺] 0.01 ⇒ Q = 10.
  • Medium : E = 1.10 – 0.0296 × 1 = 1.07 V.
  • Exam-type : Q = 1 ⇒ E = E°.

Rule 2: Half-cell potential — Cu²⁺/Cu: E = 0.34 – (0.0592/2) log(1/[Cu²⁺]); [Cu²⁺] = 0.001 M ⇒ 0.251 V.

  • Easy : Solid metal activity = 1.
  • Medium : Dilute ⇒ lower E.
  • Exam-type : log(1/0.001) = 3.

Rule 3: Equilibrium constant — At E = 0: log K = nE°/0.0592; Daniell: 2 × 1.10/0.0592 = 37.16 ⇒ K ≈ 1.4×10³⁷.

  • Easy : Large K ⇒ nearly complete reaction.
  • Medium : ΔG° = –RT ln K.
  • Exam-type : E° and K are linked.

Calculation: [Zn²⁺] = 0.1 M, [Cu²⁺] = 0.01 M

  • Step 1: Q = 0.1/0.01 = 10 ⇒ log Q = 1.
  • Step 2: E = 1.10 – (0.0592/2) × 1.
  • Step 3: E = 1.07 V.

CET level (MCQ)

  • CET 1) At equilibrium E = ? 1) 0 2) E° 3) 1 4) –E° — Answer: 1).
  • CET 2) When Q = 1? 1) E = E° 2) E = 0 3) E = 1 4) E negative — Answer: 1).

Real-life use: pH meters, battery testing.

Remember

E = E° – (0.0592/n) log Q; E = 0 at equilibrium.

Common mistake

Including solid metals in Q.

CET trap

Solids and pure liquids are left out of Q.

Shortcut

Memory clue: "E equals E° minus 0.0592 over n times log Q".

Spot the difference

CaseE
Q = 1E°
Q < 1E > E°
Q > 1E < E°
Q = K0

Practice questions

Easy
  1. 1Write the Nernst equation.
  2. 2Daniell: [Zn²⁺] 0.1, [Cu²⁺] 0.01 ⇒ E.
  3. 3E of Cu²⁺ 0.001 M half-cell.
Medium
  1. 4Find log K.
  2. 5Why E = 0 at equilibrium?
  3. 6Why does a battery die?
Challenging
  1. 7What is left out of Q?
  2. 8Relation of ΔG and E.
Show answers (try them yourself first)
  1. 1E = E° – 0.0592/n log Q.
  2. 21.07 V.
  3. 30.251 V.
  4. 437.16.
  5. 5Q = K ⇒ ΔG = 0.
  6. 6Q rises ⇒ E falls.
  7. 7Solids and pure liquids.
  8. 8ΔG = –nFE.

Batteries, fuel cells and corrosion

From daily life

A torch, a car and a laptop each use a different battery.

Simple definition

Primary cells are not rechargeable; secondary cells are. A fuel cell gives continuous electricity from a fuel. Corrosion is electrochemical oxidation of a metal.

BatteriesBatteries, fuel cells and corrosionTypeVFeatureLeclanché1.5Zn case, MnO₂, NH₄ClMercury cell1.35constant voltageLead-acid2.0 / cell12 V = 6 cellsNi–Cd1.4rechargeableLi-ion≈ 3.7light, high energyPb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O (discharge)1 mol Pb ⇒ 2F = 53.6 Ah; H₂SO₄ ≈ 38 %H₂–O₂ fuel cell: anode 2H₂ + 4OH⁻ → 4H₂O + 4e⁻cathode O₂ + 2H₂O + 4e⁻ → 4OH⁻; only water is formedCorrosion: Fe anode ⇒ Fe²⁺; prevention: galvanising, sacrificial Mg

In the figure : Comparison of batteries.

Rule 1: Lead-acid battery — Discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O; 2 V per cell; six cells = 12 V; ≈ 38 % H₂SO₄.

  • Easy : 1 mol Pb ⇒ 2 F = 53.6 Ah.
  • Medium : Charging reverses the reaction.
  • Exam-type : Falling density ⇒ discharging.

Rule 2: Other cells — Leclanché 1.5 V (Zn, MnO₂); mercury cell 1.35 V; Ni–Cd 1.4 V; Li-ion ≈ 3.7 V.

  • Easy : Mercury cell: steady voltage.
  • Medium : Ni–Cd: rechargeable.
  • Exam-type : Li-ion: light.

Rule 3: Fuel cell and corrosion — H₂–O₂: anode 2H₂ + 4OH⁻ → 4H₂O + 4e⁻; cathode O₂ + 2H₂O + 4e⁻ → 4OH⁻; only water; corrosion Fe → Fe²⁺ + 2e⁻; prevention by galvanising or a sacrificial Mg anode.

  • Easy : H₂–O₂ cells in spacecraft.
  • Medium : Rust needs air and water.
  • Exam-type : A zinc coat protects iron.

Calculation: charge from 1 mol Pb on discharge

  • Step 1: Pb → Pb²⁺ + 2e⁻ ⇒ n = 2.
  • Step 2: Q = 2 × 96500 = 193000 C.
  • Step 3: ÷ 3600 = 53.6 Ah.

CET level (MCQ)

  • CET 1) Product of an H₂–O₂ fuel cell? 1) water 2) CO₂ 3) NO₂ 4) H₂S — Answer: 1).
  • CET 2) Secondary cell? 1) lead-acid 2) Leclanché 3) mercury cell 4) none — Answer: 1).

Real-life use: Cars, phones, spacecraft, protecting bridges.

Remember

Primary: not rechargeable; secondary: rechargeable; fuel cell: continuous; corrosion: anodic oxidation.

Common mistake

Calling Li-ion a primary cell.

CET trap

Li-ion and Ni–Cd are rechargeable (secondary).

Shortcut

Memory clue: "lead 2 V, six cells = 12 V".

Spot the difference

CellV
Leclanché1.5
Mercury1.35
Lead-acid2.0
Ni–Cd1.4

Practice questions

Easy
  1. 1Primary vs secondary cell.
  2. 2Reactions of lead-acid battery.
  3. 3Fuel-cell reactions.
Medium
  1. 4How does rusting occur?
  2. 5Ways to prevent corrosion.
  3. 61 mol Pb ⇒ Ah.
Challenging
  1. 7Feature of mercury cell.
  2. 8Advantage of a fuel cell.
Show answers (try them yourself first)
  1. 1Rechargeable or not.
  2. 2Pb + PbO₂ + 2H₂SO₄ ⇒ 2PbSO₄.
  3. 3H₂ at anode, O₂ at cathode.
  4. 4Fe anode, O₂ cathode.
  5. 5Galvanising, Mg.
  6. 653.6 Ah.
  7. 7Steady voltage.
  8. 8Clean and continuous.

Quick revision of the whole lesson

ConceptKey formula / fact
ConductanceG = 1/R ; κ = G × G*
Molar conductivityΛm = κ × 1000 / c
KohlrauschΛ° = λ₊ + λ₋
Faradaym = (M/nF) I t
Cell potentialE°cell = E°cathode – E°anode
EnergyΔG° = –nFE°
NernstE = E° – (0.0592/n) log Q
Equilibriumlog K = nE°/0.0592
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🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Conductance.
  • Molar conductivity.
  • Faraday's laws.
  • Galvanic cells.
  • E° series.
  • Nernst.
  • Batteries, corrosion.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.