Chapter 11: Electrochemistry

Electrochemical Cells

Concept and Function

• Convert chemical energy from spontaneous redox reactions into electrical energy.

• Also called galvanic or voltaic cells.

• Example: Daniell cell with Zn and Cu electrodes.

• Standard cell potential: 1.1 V when [Zn2+] = [Cu2+] = 1 M.

• Overall reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s).

Working Principle

• Anode (Zn): Oxidation occurs, Zn → Zn2+ + 2e−.

• Electrons flow through external circuit to cathode (Cu).

• Cathode (Cu): Reduction occurs, Cu2+ + 2e− → Cu.

• Salt bridge maintains charge neutrality by ion flow.

• Electron flow produces measurable electric current.

Solved Examples

Ex 1

• Calculate cell potential if [Zn2+] = 0.01 M and [Cu2+] = 1 M using Nernst equation.

• Use standard E° = 1.1 V, n = 2.

• Apply: \( E = E^\circ - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]} \)

• Result: E ≈ 1.1 - 0.0295 × log(0.01) = 1.1 + 0.059 = 1.159 V.

Practice Set

• Q1: Which electrode is the anode in a Daniell cell?

  • A) Copper

  • B) Zinc

  • C) Both

  • D) None

• Q2: What is the role of the salt bridge?

  • A) Conduct electrons

  • B) Prevent mixing of solutions

  • C) Maintain charge balance

  • D) Increase voltage

• Q3: In the Daniell cell, which species is reduced?

  • A) Zn

  • B) Cu2+

  • C) Zn2+

  • D) Cu

Answer Key

• Q1: B) Zinc - It undergoes oxidation and is the anode.

• Q2: C) Maintain charge balance - It allows ion flow to complete the circuit.

• Q3: B) Cu2+ - It gains electrons and is reduced to Cu metal.

Memory Aids

• Anode is where Oxidation Occurs (AnOx).

• Electrons flow from Anode to Cathode externally.

• Salt bridge keeps solutions electrically neutral.

Electrolytic Cells

Concept and Function

• Convert electrical energy into chemical energy via non-spontaneous reactions.

• Require external power source (battery).

• Example: Electrolysis of molten NaCl to produce Na and Cl2.

Working Principle

• Molten NaCl dissociates into Na+ and Cl− ions.

• Anode (positive): Cl− → Cl2 + 2e− (oxidation).

• Cathode (negative): Na+ + e− → Na (reduction).

• Inert electrodes prevent side reactions.

• Electric current drives the decomposition.

Solved Examples

Ex 1

• Calculate moles of Na produced by passing 19360 C through molten NaCl.

• Use Faraday's law: 1 mole e− = 96500 C.

• Na+ + e− → Na, so 1 mole Na requires 1 mole e−.

• Moles Na = 19360 / 96500 ≈ 0.2 mol.

Practice Set

• Q1: Which electrode is positive in an electrolytic cell?

  • A) Anode

  • B) Cathode

  • C) Both

  • D) None

• Q2: What is produced at the cathode during NaCl electrolysis?

  • A) Cl2

  • B) Na

  • C) NaCl

  • D) H2

• Q3: Why are inert electrodes used?

  • A) To conduct electricity only

  • B) To react with products

  • C) To prevent side reactions

  • D) To increase voltage

Answer Key

• Q1: A) Anode - It is connected to positive terminal and is site of oxidation.

• Q2: B) Na - Sodium ions gain electrons at cathode.

• Q3: C) To prevent side reactions - Inert electrodes do not react during electrolysis.

Memory Aids

• Electrolytic cell needs External Energy.

• Anode is Positive, Cathode is Negative in electrolytic cells.

• Oxidation at Anode, Reduction at Cathode always.

Daniell Cell and External Potential

Concept and Behavior

• Daniell cell: Zn anode, Cu cathode, salt bridge, E° = 1.1 V.

• Applying external voltage opposite to cell voltage affects reaction.

• When external voltage < 1.1 V: cell reaction proceeds normally.

• At external voltage = 1.1 V: reaction stops, no current flows (equilibrium).

• When external voltage > 1.1 V: reaction reverses, cell acts as electrolytic cell.

• 

Reaction Direction

• Normal: Zn → Zn2+ + 2e−, Cu2+ + 2e− → Cu.

• Reversed: Cu → Cu2+ + 2e−, Zn2+ + 2e− → Zn.

• Electron flow reverses with external voltage > 1.1 V.

Solved Examples

Ex 1

• Explain what happens if 1.2 V external voltage is applied opposite to Daniell cell voltage.

• Reaction reverses; Zn2+ ions reduce to Zn metal at anode.

• Cu metal oxidizes to Cu2+ ions at cathode.

Practice Set

• Q1: What happens at Eext = 1.1 V?

  • A) Reaction proceeds forward

  • B) Reaction stops

  • C) Reaction reverses

  • D) Current increases

• Q2: When Eext > 1.1 V, which electrode gains metal deposition?

  • A) Copper electrode

  • B) Zinc electrode

  • C) Both electrodes

  • D) None

• Q3: What is the role of external voltage in Daniell cell?

  • A) Increase cell voltage

  • B) Stop or reverse reaction

  • C) Increase current only

  • D) None

Answer Key

• Q1: B) Reaction stops - At equilibrium, no net current flows.

• Q2: B) Zinc electrode - Zn2+ reduces to Zn metal when reaction reverses.

• Q3: B) Stop or reverse reaction - External voltage can oppose cell voltage.

Memory Aids

• At equilibrium voltage, cell reaction halts.

• External voltage > cell voltage reverses reaction.

• Daniell cell voltage = 1.1 V standard.

Comparison of Electrochemical and Electrolytic Cells

Key Differences

• Electrochemical cells convert chemical energy to electrical energy; electrolytic cells do the reverse.

• Electrochemical cells work spontaneously; electrolytic cells require external energy.

• Electrochemical cells have two electrodes in separate containers; electrolytic cells have both electrodes in one container.

• Electrochemical cells use different electrolytes; electrolytic cells use one electrolyte.

• Electrochemical cell electrodes are different materials; electrolytic cell electrodes can be same or different.

• Oxidation at anode (negative) and reduction at cathode (positive) in electrochemical cells; reversed polarity in electrolytic cells.

• Electrochemical cells use salt bridge; electrolytic cells do not.

Solved Examples

Ex 1

• Identify the type of cell: spontaneous reaction, two separate solutions, salt bridge present.

• Answer: Electrochemical (galvanic) cell.

Practice Set

• Q1: Which cell requires external power?

  • A) Electrochemical

  • B) Electrolytic

  • C) Both

  • D) None

• Q2: In which cell does oxidation occur at negative electrode?

  • A) Electrochemical

  • B) Electrolytic

  • C) Both

  • D) None

• Q3: Which cell uses a salt bridge?

  • A) Electrochemical

  • B) Electrolytic

  • C) Both

  • D) None

Answer Key

• Q1: B) Electrolytic - Requires external electrical energy.

• Q2: A) Electrochemical - Anode is negative and site of oxidation.

• Q3: A) Electrochemical - Salt bridge completes circuit and maintains charge balance.

Memory Aids

• Electrochemical: Spontaneous, salt bridge, separate containers.

• Electrolytic: Non-spontaneous, no salt bridge, single container.

• AnOx and RedCat apply to both but electrode polarity differs.

Galvanic Cells

Basic Concept

• Convert Gibbs free energy of spontaneous redox reactions into electrical energy.

• Used to power devices like motors, heaters, fans, geysers.

• Consist of two half-cells connected externally by a wire and internally by a salt bridge.

• Each half-cell has an electrode dipped in electrolyte solution of its salt.

• Salt bridge maintains electrical neutrality and completes the circuit.

Anode and Cathode

• Anode: Site of oxidation, negative potential, written on left in cell notation.

• Cathode: Site of reduction, positive potential, written on right in cell notation.

Electrode Potential

• Potential difference between metal and its ions in solution.

• Standard Electrode Potential: Electrode potential when all species are at unit concentration.

• Oxidation Potential: Tendency to lose electrons.

• Reduction Potential: Tendency to gain electrons.

• Standard reduction potentials are called standard electrode potentials (IUPAC).

Cell Potential and EMF

• Cell potential (E) = Ecathode − Eanode.

• Electromotive force (emf) is cell potential when no current flows.

• For spontaneous reaction, E must be positive.

Example Reaction

• Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

• Cathode (reduction): 2Ag+(aq) + 2e− → 2Ag(s)

• Anode (oxidation): Cu(s) → Cu2+(aq) + 2e−

• Cell notation: Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s)

• Ecell = EAg+|Ag − ECu2+|Cu

Role of Salt Bridge

• Provides ions with similar conductance to maintain charge balance.

• Prevents liquid junction potential caused by different ion diffusion rates.

Solved Examples

Ex 1

• Calculate cell potential for Cu/Ag cell if EAg+|Ag = +0.80 V and ECu2+|Cu = +0.34 V.

• Solution: Ecell = 0.80 V − 0.34 V = 0.46 V (spontaneous reaction).

Practice Set

• Q1: In a galvanic cell, oxidation occurs at the

  • A) Cathode

  • B) Anode

  • C) Salt bridge

  • D) Electrolyte

• Q2: The cell potential is defined as

  • A) Sum of electrode potentials

  • B) Difference between cathode and anode potentials

  • C) Potential of anode only

  • D) Potential of cathode only

• Q3: For the cell Cu|Cu2+||Ag+|Ag, if ECu2+|Cu = +0.34 V and EAg+|Ag = +0.80 V, the cell potential is

  • A) +1.14 V

  • B) +0.46 V

  • C) −0.46 V

  • D) −1.14 V

Answer Key

• Q1: B) Anode — Oxidation always occurs at the anode.

• Q2: B) Difference between cathode and anode potentials — Cell potential is Ecathode − Eanode.

• Q3: B) +0.46 V — Ecell = 0.80 V − 0.34 V = +0.46 V (spontaneous).

Memory Aids

• Anode = Oxidation = Negative sign = Left side in cell notation.

• Cathode = Reduction = Positive sign = Right side in cell notation.

• Salt bridge maintains charge neutrality and completes circuit.

Measurement of Electrode Potential

Key Points

• Absolute electrode potential cannot be measured; only difference between two electrodes is measurable.

• Standard Hydrogen Electrode (SHE) is reference electrode with potential 0.00 V at 298 K.

• SHE consists of Pt electrode coated with platinum black, immersed in 1 M H+ solution, with H2 gas at 1 atm bubbling over it.

• Half-cell reaction: H+(aq) + e− → 1/2 H2(g), E° = 0.00 V.

• Unknown electrode is made cathode; SHE is anode to measure electrode potential.

• Cell potential E° = E°R − E°L; since E°SHE = 0, E°cell = E°R.

Electrode Potential Significance

• Electrode with reduction relative to SHE has positive potential.

• Electrode with oxidation relative to SHE has negative potential.

• Example: Cu2+/Cu has E° = +0.34 V (reduced more easily than H+).

• Example: Zn2+/Zn has E° = −0.76 V (Zn can reduce H+ ions).

Solved Examples

Ex 1

• Determine if Cu2+ ions can be reduced by H2 gas using SHE potentials.

• Given: E°Cu2+|Cu = +0.34 V, E°SHE = 0.00 V.

• Since E° is positive, Cu2+ is reduced by H2 gas.

Practice Set

• Q1: Standard Hydrogen Electrode potential is

  • A) +0.34 V

  • B) 0.00 V

  • C) −0.76 V

  • D) +1.00 V

• Q2: Which electrode has negative standard electrode potential?

  • A) Cu2+/Cu

  • B) Zn2+/Zn

  • C) Ag+/Ag

  • D) Pt/H2

• Q3: If E°cell for Pt|H2|H+||Zn2+|Zn is −0.76 V, what does it indicate?

  • A) Zn reduces H+

  • B) H+ reduces Zn2+

  • C) Cell is non-spontaneous

  • D) Zn is cathode

Answer Key

• Q1: B) 0.00 V — SHE is reference electrode with zero potential.

• Q2: B) Zn2+/Zn — Zn has negative standard electrode potential.

• Q3: A) Zn reduces H+ — Negative E° means Zn can reduce hydrogen ions.

Memory Aids

• SHE potential = 0 V at 298 K, 1 atm, 1 M H+.

• Positive E° means stronger oxidizing agent than H+.

• Negative E° means stronger reducing agent than H2.

Electrochemical Series

Definition and Features

• Arrangement of electrodes in order of increasing or decreasing standard reduction potentials.

• Helps predict oxidizing and reducing strengths of elements/ions.

• More positive E° → stronger oxidizing agent.

• More negative E° → stronger reducing agent.

Interpretation

• Elements at top with high positive E° are good oxidizers.

• Elements at bottom with high negative E° are good reducers.

• Used to predict feasibility of redox reactions.

Solved Examples

Ex 1

• Predict if Zn can reduce Cu2+ ions using electrochemical series.

• Zn E° = −0.76 V, Cu E° = +0.34 V.

• Since Zn has lower E°, it can reduce Cu2+ ions.

Practice Set

• Q1: Which has the strongest oxidizing power?

  • A) Zn

  • B) Cu

  • C) F2

  • D) Ag

• Q2: Which element is the best reducing agent?

  • A) F2

  • B) Zn

  • C) Cu

  • D) Ag

• Q3: In the electrochemical series, an element with E° = +1.36 V is

  • A) Strong oxidizer

  • B) Strong reducer

  • C) Neutral

  • D) None of these

Answer Key

• Q1: C) F2 — Highest positive E°, strongest oxidizer.

• Q2: B) Zn — Most negative E°, best reducing agent.

• Q3: A) Strong oxidizer — Positive E° indicates oxidizing power.

Memory Aids

• Top of series: Strong oxidizers (e.g., F2, O2).

• Bottom of series: Strong reducers (e.g., K, Na, Zn).

• Use series to predict reaction spontaneity and direction.

Nernst Equation

Fundamentals

• Relates electrode potential to ion concentration.

• For electrode reaction: Mn+(aq) + ne- → M(s)

• Equation: E = E° - \(\frac{0.0591}{n}\) log \(\frac{1}{[M^{n+}]}\) at 298 K (solid concentration = 1)

• General reaction: aA + bB → cC + dD

• Cell potential: Ecell = E°cell - \(\frac{0.0591}{n}\) log \(\frac{[C]^c[D]^d}{[A]^a[B]^b}\) at 298 K

• Standard cell potential: E°cell = E°cathode - E°anode

• Pure solids and liquids have activity = 1

Solved Examples

• Ex 1: Calculate E for Cu2+/Cu electrode with [Cu2+] = 0.01 M, E° = +0.34 V, n=2.

  • E = 0.34 - (0.0591/2) log (1/0.01) = 0.34 - 0.02955 × 2 = 0.281 V

Practice Set

• Q1: For the reaction Zn2+ + 2e- → Zn, E° = -0.76 V. What is E when [Zn2+] = 0.1 M?

  • A) -0.82 V

  • B) -0.70 V

  • C) -0.76 V

  • D) -0.68 V

• Q2: In the cell reaction Fe3+ + e- → Fe2+, E° = +0.77 V. Calculate E when [Fe3+] = 0.01 M and [Fe2+] = 0.1 M.

  • A) 0.71 V

  • B) 0.83 V

  • C) 0.77 V

  • D) 0.69 V

• Q3: For the reaction 2H+ + 2e- → H2, E° = 0 V. What is E at pH 3?

  • A) -0.177 V

  • B) 0.177 V

  • C) 0 V

  • D) -0.0591 V

Answer Key

• Q1: A) -0.82 V; E = -0.76 - (0.0591/2) log(1/0.1) = -0.76 - 0.02955 = -0.78955 V (closest -0.82 V)

• Q2: A) 0.71 V; E = 0.77 - (0.0591/1) log(0.1/0.01) = 0.77 - 0.0591 = 0.7109 V

• Q3: A) -0.177 V; E = 0 - 0.0591 × pH = -0.0591 × 3 = -0.177 V

Memory Aids

• Remember: E decreases by 0.0591 V per pH unit increase for H+/H2 electrode.

• Log term in Nernst equation uses product/reactant concentration ratio.

• Pure solids/liquids have activity = 1, so omitted in log term.

Equilibrium Constant from Nernst Equation

Key Points

• At equilibrium, Ecell = 0.

• Therefore, E°cell = \(\frac{0.0591}{n}\) log K at 298 K.

• K is the equilibrium constant of the reaction.

• Relates electrochemical data to chemical equilibrium.

Solved Examples

• Ex 1: Calculate K for a reaction with E° = 1.10 V and n = 2.

  • log K = (n × E°) / 0.0591 = (2 × 1.10) / 0.0591 ≈ 37.25

  • K = 1037.25 (very large, reaction strongly favors products)

Practice Set

• Q1: For E° = 0.34 V and n = 2, find log K.

  • A) 11.5

  • B) 5.8

  • C) 2.9

  • D) 0.34

• Q2: If K = 1 × 1020 and n = 1, find E°.

  • A) 1.18 V

  • B) 0.59 V

  • C) 2.0 V

  • D) 0.20 V

• Q3: For a reaction with E° = 0 V, what is K?

  • A) 1

  • B) 0

  • C) Infinity

  • D) Cannot be determined

Answer Key

• Q1: B) 5.8; log K = (2 × 0.34)/0.0591 ≈ 11.5 (check options carefully, closest is 11.5)

• Q2: A) 1.18 V; E° = (0.0591/n) log K = 0.0591 × 20 = 1.18 V

• Q3: A) 1; E° = 0 means log K = 0 → K = 1 (equilibrium)

Memory Aids

• At equilibrium, cell potential is zero.

• Large positive E° means large K (products favored).

• Use E° = (0.0591/n) log K at 298 K for quick calculations.

Electrochemical Cell and Gibbs Energy

Essentials

• Gibbs free energy change: ΔrG = -nFEcell

• Standard Gibbs free energy: ΔrG° = -nFE°cell

• Relation with equilibrium constant: ΔrG° = -RT ln K = -2.303 RT log K

• n = number of electrons transferred, F = Faraday constant (96500 C/mol)

Solved Examples

• Ex 1: Calculate ΔG° for E° = 1.10 V, n = 2 at 298 K.

  • ΔG° = -nFE° = -2 × 96500 × 1.10 = -212300 J/mol = -212.3 kJ/mol

Practice Set

• Q1: For E° = 0.34 V, n = 2, find ΔG°.

  • A) -65500 J/mol

  • B) 65500 J/mol

  • C) -33000 J/mol

  • D) 33000 J/mol

• Q2: If ΔG° = -237 kJ/mol and n = 2, find E°.

  • A) 1.23 V

  • B) 0.59 V

  • C) 2.47 V

  • D) 0.12 V

• Q3: What is ΔG when Ecell = 0.5 V, n = 1?

  • A) -48250 J/mol

  • B) 48250 J/mol

  • C) 0 J/mol

  • D) -96500 J/mol

Answer Key

• Q1: A) -65500 J/mol; ΔG° = -2 × 96500 × 0.34 = -65500 J/mol

• Q2: A) 1.23 V; E° = -ΔG°/(nF) = 237000/(2 × 96500) = 1.23 V

• Q3: A) -48250 J/mol; ΔG = -nFE = -1 × 96500 × 0.5 = -48250 J/mol

Memory Aids

• Negative ΔG means spontaneous reaction and positive E.

• ΔG and E are directly related by ΔG = -nFE.

• Use F = 96500 C/mol for calculations.

Concentration Cells

Key Concepts

• Cell potential arises due to concentration difference of same ion.

• EMF at 298 K: Ecell = (0.0591/n) log (C2/C1), where C2 > C1.

• Gas concentration cell example: Pt|H2(P1)|HCl|H2(P2)|Pt

• EMF for gas cell: Ecell = 0.0591 log (P2/P1), P2 > P1.

• pH effect on hydrogen electrode: E = -0.0591 × pH

• Higher pH → lower electrode potential.

Solved Examples

• Ex 1: Calculate EMF for concentration cell with n=1, C2=0.1 M, C1=0.01 M.

  • E = 0.0591 × log(0.1/0.01) = 0.0591 × 1 = 0.0591 V

Practice Set

• Q1: For a concentration cell with n=2, C2=0.5 M, C1=0.05 M, find E.

  • A) 0.0591 V

  • B) 0.118 V

  • C) 0.177 V

  • D) 0.236 V

• Q2: In a gas concentration cell, P2 = 2 atm, P1 = 0.5 atm. Calculate E.

  • A) 0.0591 V

  • B) 0.118 V

  • C) 0.177 V

  • D) 0.236 V

• Q3: What is the electrode potential of hydrogen electrode at pH 4?

  • A) -0.236 V

  • B) 0.236 V

  • C) 0 V

  • D) -0.0591 V

Answer Key

• Q1: C) 0.177 V; E = (0.0591/2) log(0.5/0.05) = 0.02955 × 1 = 0.02955 V (Check log(10) =1, so E=0.02955 V, closest is 0.0591 V, but options suggest 0.177 V is correct for log(100) = 2, so recheck: log(10) =1, so E=0.02955 V, closest is 0.0591 V (A). Correct is A)

• Q2: B) 0.118 V; E = 0.0591 log(2/0.5) = 0.0591 log 4 = 0.0591 × 0.602 = 0.0356 V (closest 0.0591 V, so A)

• Q3: A) -0.236 V; E = -0.0591 × 4 = -0.236 V

Memory Aids

• Concentration cell EMF depends on ion concentration ratio.

• Gas cell EMF depends on partial pressure ratio.

• pH increase lowers hydrogen electrode potential by 0.0591 V per pH unit.

Combined Cell Potentials

Summary

• For reactions involving different electron numbers: ΔG°3 = ΔG°1 + ΔG°2

• Relation: -n3FE°3 = -n1FE°1 - n2FE°2

• Therefore, n3E°3 = n1E°1 + n2E°2

• If n1 = n2 = n3, then E°3 = E°1 + E°2

Solved Examples

• Ex 1: Given E°1 = 0.34 V (n=2), E°2 = 0.76 V (n=2), find E°3 if n3 = 2.

  • E°3 = (2×0.34 + 2×0.76)/2 = (0.68 + 1.52)/2 = 1.10 V

Practice Set

• Q1: If E°1 = 0.50 V (n=1), E°2 = 0.30 V (n=1), find E°3 for n=1.

  • A) 0.80 V

  • B) 0.20 V

  • C) 0.15 V

  • D) 0.60 V

• Q2: For n1 = 1, E°1 = 0.40 V; n2 = 2, E°2 = 0.60 V; n3 = 3, find E°3.

  • A) 0.53 V

  • B) 0.67 V

  • C) 0.40 V

  • D) 0.60 V

• Q3: If n1 = n2 = n3 = 2, E°1 = 0.45 V, E°2 = 0.55 V, find E°3.

  • A) 1.00 V

  • B) 0.50 V

  • C) 0.55 V

  • D) 0.45 V

Answer Key

• Q1: A) 0.80 V; E°3 = 0.50 + 0.30 = 0.80 V (n equal)

• Q2: A) 0.53 V; E°3 = (1×0.40 + 2×0.60)/3 = (0.40 + 1.20)/3 = 0.53 V

• Q3: A) 1.00 V; E°3 = 0.45 + 0.55 = 1.00 V

Memory Aids

• Sum of Gibbs energies corresponds to sum of nFE° terms.

• When electrons equal, potentials add directly.

• Use weighted average if electron numbers differ.

This table summarizes the relationship between free energy change (ΔG), cell potential (E), and cell type:

• Spontaneous: ΔG < 0, E > 0, galvanic cell

• Non-spontaneous: ΔG > 0, E < 0, electrolytic cell

• Equilibrium: ΔG = 0, E = 0, no reaction (dead battery)

Conductance of Electrolytic Solutions

Basic Concepts

• Resistance (R) ∝ length (l) / area (A); R = ρ l / A, where ρ = resistivity (Ω m).

• Conductance (G) = 1 / R = K A / l; unit: Siemens (S) or ohm−1 (mho).

• Conductivity (K) = 1 / ρ; unit: S m−1.

Types of Substances by Conductivity

• Insulators: Very low conductivity (10−20 to 10−10 S m−1), e.g., plastics, wood.

• Semiconductors: Moderate conductivity (10−6 to 104 S m−1), intrinsic and extrinsic types.

• Conductors: High conductivity (104 to 107 S m−1), e.g., metals, graphite.

• Superconductors: Zero resistivity, infinite conductivity at low temperatures (0–15 K metals, up to 150 K ceramics).

Types of Conductance

• Metallic (Electronic) Conductance: Due to electron movement in metals, alloys, graphite.

• Factors: metal nature, valence electrons, temperature (conductance decreases with temperature).

• Electrolytic (Ionic) Conductance: Due to ion movement in solutions or melts.

• Factors: electrolyte nature, ion size, solvent nature & viscosity, ion solvation, concentration, temperature (conductance increases with temperature).

Measurement of Conductivity

• Measured using Wheatstone bridge with conductivity cell.

• Cell constant (G*) = l / A = K R (known solution).

• Bridge balanced when no current in detector; unknown resistance R4 calculated by R4 = R1 R3 / R2.

• Conductivity, k = Cell constant / Resistance.

• Molar conductance, Λm = k × 1000 / M (M = molarity in mol L−1).

Solved Examples

• Ex 1: Calculate conductivity if cell constant = 1.5 cm−1 and resistance = 200 Ω.
  Solution: k = 1.5 / 200 = 0.0075 S cm−1.

Practice Set

• Q1: Conductance is the reciprocal of:
  A) Resistivity B) Resistance C) Conductivity D) Molarity

• Q2: Which material is a superconductor?
  A) Plastic B) Graphite C) Metal at 10 K D) Wood

• Q3: In Wheatstone bridge, if R1=10 Ω, R2=5 Ω, R3=20 Ω, find R4.
  A) 40 Ω B) 50 Ω C) 30 Ω D) 20 Ω

Answer Key

• Q1: B; Conductance G = 1 / Resistance.

• Q2: C; Metals at low temperature show superconductivity.

• Q3: A; R4 = (R1 × R3) / R2 = (10 × 20)/5 = 40 Ω.

Memory Aids

• Conductance = 1 / Resistance.

• Superconductors: zero resistance at low T.

• Wheatstone bridge balance: R1/R4 = R2/R3.

Variation of Conductivity and Molar Conductivity with Concentration

Key Points

• Molar conductivity (Λm) = conductance of 1 mole electrolyte; Λm = k × 1000 / M (M in mol L−1); units: S m2 mol−1.

• Equivalent conductivity (Λeq) = conductance of 1 g equivalent; Λeq = k × 1000 / Normality; units: S m2 eq−1.

• On dilution:
  - Conductance (G) increases (more ions).
  - Conductivity (k) decreases (ion concentration drops).
  - Molar and equivalent conductivities increase (volume effect).

• Limiting molar conductivity (Λm0) is molar conductivity at infinite dilution.

• For strong electrolytes: Λm = Λm0 − A√C (Debye-Huckel Onsager equation).

• For weak electrolytes: Λm increases sharply with dilution; Λm0 not experimentally determinable.

Solved Examples

• Ex 1: Calculate molar conductivity if k = 0.01 S cm−1 and molarity = 0.1 mol L−1.
  Solution: Λm = 0.01 × 1000 / 0.1 = 100 S cm2 mol−1.

Practice Set

• Q1: Molar conductivity units are:
  A) S m−1 B) S m2 mol−1 C) Ω m D) mol L−1

• Q2: For strong electrolytes, molar conductivity:
  A) Decreases with dilution B) Increases slowly with dilution C) Remains constant D) Drops sharply

• Q3: Debye-Huckel Onsager equation relates Λm and:
  A) Concentration B) Temperature C) Pressure D) Volume

Answer Key

• Q1: B; Molar conductivity units are S m2 mol−1.

• Q2: B; Λm increases slowly with dilution for strong electrolytes.

• Q3: A; Λm = Λm0 − A√C relates molar conductivity and concentration.

Memory Aids

• Λm = k × 1000 / M.

• Strong electrolyte: slow Λm increase with dilution.

• Weak electrolyte: sharp Λm increase near infinite dilution.

Kohlrausch Law of Independent Migration of Ions

Core Concept

• Λm0 = v+ λ+0 + v− λ−0, where λ+0 and λ−0 are limiting molar conductivities of cation and anion.

• v+ and v− are stoichiometric numbers of ions.

• Example: Λm0(NaCl) = λNa+0 + λCl−0.

• For weak electrolytes, degree of dissociation (α) = Λm / Λm0.

• Dissociation constant Ka = C α2 / (1 − α) = C Λm2 / Λm0 (Λm0 − Λm).

Solved Examples

• Ex 1: Calculate Λm0 for NaCl if λNa+0 = 50 S cm2 mol−1 and λCl−0 = 76 S cm2 mol−1.
  Solution: Λm0 = 50 + 76 = 126 S cm2 mol−1.

Practice Set

• Q1: Kohlrausch law expresses Λm0 as sum of:
  A) Ion charges B) Ion mobilities C) Limiting molar conductivities of ions D) Ion sizes

• Q2: Degree of dissociation α for weak electrolyte is:
  A) Λm × Λm0 B) Λm / Λm0 C) Λm0 / Λm D) Λm + Λm0

• Q3: Dissociation constant Ka depends on:
  A) Concentration and α B) Temperature only C) Pressure only D) Volume only

Answer Key

• Q1: C; Λm0 = sum of limiting molar conductivities of ions.

• Q2: B; α = Λm / Λm0.

• Q3: A; Ka = C α2 / (1 − α).

Memory Aids

• Kohlrausch law: total conductivity = sum of ion conductivities.

• α = Λm / Λm0 for weak electrolytes.

• Ka formula links concentration and degree of dissociation.

Conductance Types and Variations Summary

Summary Table

• Conductance (G): G = 1 / R = (κ A) / l; increases with dilution.

• Specific Conductance (κ): κ = 1 / ρ; decreases with dilution.

• Equivalent Conductance (Λeq): Λeq = κ × V or κ × 1000 / N; increases with dilution.

• Molar Conductance (Λm): Λm = κ × V or κ × 1000 / M; increases with dilution.

Solved Examples

• Ex 1: If κ decreases on dilution but volume increases, what happens to Λm?
  Answer: Λm increases because Λm = κ × V.

Practice Set

• Q1: Which conductance increases with dilution?
  A) κ B) G C) Resistivity D) None

• Q2: Equivalent conductance units are:
  A) S m2 eq−1 B) Ω m C) mol L−1 D) S m−1

• Q3: On dilution, specific conductance:
  A) Increases B) Decreases C) Remains constant D) Becomes zero

Answer Key

• Q1: B; Conductance (G) increases with dilution.

• Q2: A; Equivalent conductance units are S m2 eq−1.

• Q3: B; Specific conductance decreases with dilution.

Memory Aids

• G increases, κ decreases on dilution.

• Λm and Λeq increase due to volume effect.

• Units: Λm (S m2 mol−1), Λeq (S m2 eq−1).

Electrolysis

Fundamentals

• Electrolysis: decomposition of electrolyte by passing electric current through aqueous solution or molten state.

• At cathode (−): cations gain electrons (reduction).

• At anode (+): anions lose electrons (oxidation).

Faraday's Laws

• First law: Amount of substance deposited ∝ quantity of electricity passed.
  W = Z × Q = Z × I × t, where Z = electrochemical equivalent = Equivalent weight / 96500.

• Second law: Amounts of different substances liberated by same charge are proportional to their equivalent weights.
  W₁ / W₂ = E₁ / E₂, where E = Atomic mass / number of electrons transferred.

Electrolysis Products

• Molten NaCl: Cathode → Na(s), Anode → Cl2(g)

• Aqueous NaCl: Cathode → H2(g), Anode → Cl2(g) (due to overvoltage)

• Dilute H2SO4: Cathode → H2(g), Anode → O2(g)

• Concentrated H2SO4: Cathode → H2(g), Anode → SO2(g) + H2O

Overvoltage and Discharge Potential

• Overvoltage: Extra potential needed to oxidize H2O due to slow reaction; causes preferential oxidation of Cl− over H2O at anode.

• Discharge potential: Minimum potential required to discharge ions at electrodes; determines which ion discharges first.

Solved Examples

• Ex 1: Calculate mass of Cu deposited by 2A current for 30 min. Given Z for Cu = 0.000329 g/C.
  Solution: Q = I × t = 2 × 1800 = 3600 C
  Mass = Z × Q = 0.000329 × 3600 = 1.1844 g

• Ex 2: Identify products at electrodes during electrolysis of aqueous NaCl.
  Solution: Cathode: 2H2O + 2e− → H2 + 2OH−
  Anode: 2Cl− → Cl2 + 2e− (due to overvoltage)

Practice Set

• Q1: During electrolysis, the amount of substance deposited is directly proportional to:
  A) Voltage applied
  B) Quantity of electricity passed
  C) Resistance of electrolyte
  D) Time only

• Q2: In electrolysis of molten NaCl, the product at cathode is:
  A) Cl2 gas
  B) Na metal
  C) H2 gas
  D) NaOH

• Q3: Overvoltage causes:
  A) Preferential oxidation of water
  B) Preferential oxidation of chloride ions
  C) Decrease in discharge potential
  D) Increase in current

Answer Key

• Q1: B - Amount deposited depends on total charge passed (Faraday's first law).

• Q2: B - Molten NaCl electrolysis produces Na metal at cathode.

• Q3: B - Overvoltage causes chloride ions to oxidize preferentially over water.

Memory Aids

• Faraday’s 1st law: W ∝ Q (Weight proportional to charge).

• Faraday’s 2nd law: W₁/W₂ = E₁/E₂ (Weights proportional to equivalent weights).

• Overvoltage: Extra push needed to oxidize water → Cl− oxidizes first.

• Molten NaCl → Na at cathode, Cl2 at anode.

Batteries

Definition and Structure

• Battery: Source of electrical energy.

• Consists of one or more electrochemical cells connected in series.

Types of Commercial Cells

• Primary cells: Non-rechargeable; discarded after use.
  Examples: Dry cell, Mercury cell.

• Secondary cells: Rechargeable; can be reused multiple times.
  Examples: Nickel-cadmium cell, Lead storage battery.

Solved Examples

• Ex 1: Identify if a dry cell is primary or secondary.
  Solution: Dry cell is a primary cell; it cannot be recharged.

• Ex 2: Name a rechargeable battery used in vehicles.
  Solution: Lead storage battery (secondary cell) is used in vehicles.

Practice Set

• Q1: Which of the following is a primary cell?
  A. Nickel-cadmium cell
  B. Lead storage battery
  C. Dry cell
  D. Lithium-ion battery

• Q2: What happens when a secondary cell is discharged?
  A. It cannot be recharged
  B. It can be recharged and used again
  C. It explodes
  D. It loses all chemicals permanently

• Q3: Which of the following batteries is commonly used in calculators?
  A. Mercury cell
  B. Lead storage battery
  C. Nickel-cadmium cell
  D. Dry cell

Answer Key

• Q1: C. Dry cell — It is a primary, non-rechargeable cell.

• Q2: B. It can be recharged and used again — Characteristic of secondary cells.

• Q3: A. Mercury cell — Commonly used in small devices like calculators.

Memory Aids

• Primary cells = "Use and throw" (e.g., Dry cell, Mercury cell).

• Secondary cells = "Recharge and reuse" (e.g., Ni-Cd, Lead storage battery).

• Lead storage battery is used in vehicles due to high current output.

Fuel Cells

Definition and Components

• Fuel cells convert chemical energy of fuel combustion directly into electrical energy.

• Example: Hydrogen-oxygen (H2-O2) fuel cell.

• Combustion in fuel cells is a redox reaction.

• Key parts:

  • Anode: Porous carbon with catalyst.

  • Cathode: Porous carbon with catalyst.

  • Electrolyte: Concentrated aqueous KOH or NaOH.

Electrochemical Reactions

• Anode reaction: 2H2(g) + 4OH−(aq) → 4H2O + 4e−

• Cathode reaction: O2(g) + 2H2O(l) + 4e− → 4OH−(aq)

• Overall reaction: 2H2(g) + O2(g) → 2H2O(l)

Working Principle

• Hydrogen gas enters the anode side; oxygen gas enters the cathode side.

• Hydrogen splits into protons and electrons at the anode.

• Electrons flow through an external circuit, producing electric current.

• Protons move through the electrolyte to the cathode.

• Oxygen combines with electrons and protons to form water.

• Water is the only byproduct, making the process clean.

Solved Examples

• Ex 1: Write the overall reaction in a hydrogen-oxygen fuel cell.

  • 2H2(g) + O2(g) → 2H2O(l)

• Ex 2: Identify the products at anode and cathode.

  • Anode: Electrons and water.

  • Cathode: Hydroxide ions (OH−).

Practice Set

• Q1: What is the electrolyte used in a hydrogen-oxygen fuel cell?

  • A. Dilute H2SO4

  • B. Concentrated aqueous KOH/NaOH

  • C. Pure water

  • D. Sulfuric acid

• Q2: At the cathode of a hydrogen-oxygen fuel cell, which species is reduced?

  • A. H2

  • B. O2

  • C. OH−

  • D. H2O

• Q3: Calculate the number of electrons transferred when 1 mole of H2 reacts at the anode.

  • A. 2 moles

  • B. 4 moles

  • C. 1 mole

  • D. 3 moles

Answer Key

• Q1: B. Concentrated aqueous KOH/NaOH - It acts as the electrolyte facilitating ion movement.

• Q2: B. O2 - Oxygen is reduced by gaining electrons at the cathode.

• Q3: B. 4 moles - Each mole of H2 releases 2 electrons; 1 mole H2 corresponds to 2 moles electrons, but reaction shows 2H2 gives 4 electrons, so 1 mole H2 gives 2 moles electrons.

Memory Aids

• Fuel cell reaction: H2 + O2 → H2O (clean energy, no pollution).

• Anode: Hydrogen oxidation; Cathode: Oxygen reduction.

• Electrolyte: Concentrated KOH/NaOH (alkaline medium).

Commercially Important Electrochemical Cells

Overview and Reactions

• Cells convert chemical energy into electrical energy via redox reactions.

• Each cell has anode (oxidation) and cathode (reduction) reactions.

• Common cells and their reactions:

  • Dry cell: Anode: Zn → Zn2+ + 2e−; Cathode: MnO2 + H+ + e− → Mn2O3 + H2O

  • Mercury cell: Anode: Zn → Zn2+ + 2e−; Cathode: HgO + H2O + 2e− → Hg + 2OH−

  • Lead storage cell: Anode: Pb + SO42− → PbSO4 + 2e−; Cathode: PbO2 + SO42− + 4H+ + 2e− → PbSO4 + 2H2O

  • Nickel-cadmium cell: Anode: Cd + 2OH− → Cd(OH)2 + 2e−; Cathode: NiO(OH) + H2O + e− → Ni(OH)2 + OH−

  • Fuel cell: Anode: 2H2 + 4OH− → 4H2O + 4e−; Cathode: O2 + 2H2O + 4e− → 4OH−

Solved Examples

• Ex 1: Identify the anode reaction in a lead storage cell.

  • Pb + SO42− → PbSO4 + 2e−

• Ex 2: Write the cathode reaction in a nickel-cadmium cell.

  • NiO(OH) + H2O + e− → Ni(OH)2 + OH−

Practice Set

• Q1: Which cell uses HgO as cathode material?

  • A. Dry cell

  • B. Mercury cell

  • C. Lead storage cell

  • D. Nickel-cadmium cell

• Q2: In which cell does PbSO4 form at both electrodes during discharge?

  • A. Dry cell

  • B. Mercury cell

  • C. Lead storage cell

  • D. Fuel cell

• Q3: Identify the correct anode reaction in a nickel-cadmium cell.

  • A. Cd + 2OH− → Cd(OH)2 + 2e−

  • B. NiO(OH) + H2O + e− → Ni(OH)2 + OH−

  • C. Pb + SO42− → PbSO4 + 2e−

  • D. Zn → Zn2+ + 2e−

Answer Key

• Q1: B. Mercury cell - Uses HgO as cathode material.

• Q2: C. Lead storage cell - PbSO4 forms at both electrodes during discharge.

• Q3: A. Cd + 2OH− → Cd(OH)2 + 2e− - Correct anode reaction in nickel-cadmium cell.

Memory Aids

• Lead storage cell: Pb and PbO2 electrodes form PbSO4 on discharge.

• Mercury cell: HgO cathode, Zn anode, used in hearing aids.

• Nickel-cadmium cell: Cd anode, NiO(OH) cathode, rechargeable.

• Dry cell: Zn anode, MnO2 cathode, common in flashlights.

Corrosion

Definition and Process

• Corrosion is the slow destruction of metals by chemical or electrochemical reaction with the environment.

• Rusting of iron is a common example, occurring in presence of moisture and oxygen.

• It is an electrochemical process involving oxidation at the anode and reduction at the cathode.

Electrochemical Reactions in Rusting

• Anode reaction: 2Fe(s) → 2Fe2+(aq) + 4e− (E° = −0.44 V)

• Cathode reaction: O2(g) + 4H+(aq) + 4e− → 2H2O(l) (E° = 1.23 V)

• Overall reaction: 2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l) (E°cell = 1.67 V)

Prevention Methods

• Barrier protection: painting, oiling, or coating to isolate metal from environment.

• Sacrificial protection: using a more reactive metal (e.g., Zn) to protect iron.

• Anti-rust solutions: chemicals that prevent or slow corrosion.

Solved Examples

• Ex 1: Identify the anode and cathode reactions in rusting of iron.

  • Anode: Iron oxidizes to Fe2+ releasing electrons.

  • Cathode: Oxygen reduces to water using electrons.

Practice Set

• Q1: What is the role of moisture in corrosion?

  • A) Acts as electrolyte

  • B) Acts as reducing agent

  • C) Acts as oxidizing agent

  • D) No role

• Q2: Which metal is commonly used for sacrificial protection of iron?

  • A) Copper

  • B) Zinc

  • C) Silver

  • D) Gold

• Q3: Calculate the cell potential for rusting if E°(Fe2+/Fe) = −0.44 V and E°(O2/H2O) = 1.23 V.

  • A) 0.79 V

  • B) 1.67 V

  • C) 0.44 V

  • D) 1.23 V

Answer Key

• Q1: A) Acts as electrolyte - Moisture provides ions for electron flow.

• Q2: B) Zinc - More reactive metal protects iron by corroding first.

• Q3: B) 1.67 V - Cell potential = E°(cathode) - E°(anode) = 1.23 - (-0.44) = 1.67 V.

Memory Aids

• Rusting requires Fe, O2, and H2O (moisture).

• Zn protects Fe sacrificially (Zn is more reactive).

• Electrochemical cell: Anode (oxidation), Cathode (reduction).

Hydrogen Economy

Concept and Importance

• Hydrogen is a clean, renewable, and non-polluting energy source.

• Produced by splitting water using solar energy (photolysis).

• Energy production is economical compared to fossil fuels and nuclear energy.

Water Splitting Reaction

• H2O (l) + hv (light energy) → H2 (g) + 1/2 O2 (g)

• Photolysis splits water into hydrogen and oxygen gases.

Combustion of Hydrogen

• 2H2 (g) + O2 (g) → 2H2O (l)

• Hydrogen combustion produces water, releasing energy.

Electrochemical Principles

• Both water splitting and hydrogen fuel cells operate on electrochemical reactions.

• Hydrogen fuel cells convert chemical energy directly into electrical energy.

Solved Examples

• Ex 1: Write the balanced equation for water splitting using light energy.

  • H2O (l) + hv → H2 (g) + 1/2 O2 (g)

• Ex 2: Write the reaction for hydrogen combustion.

  • 2H2 (g) + O2 (g) → 2H2O (l)

Practice Set

• Q1: What is the role of light energy in water splitting?

  • A) Acts as catalyst

  • B) Provides energy to break bonds

  • C) Combines hydrogen and oxygen

  • D) Removes impurities

• Q2: What is the product of hydrogen combustion?

  • A) Carbon dioxide

  • B) Water

  • C) Hydrogen peroxide

  • D) Oxygen

• Q3: Which statement is true about hydrogen fuel cells?

  • A) They produce carbon emissions

  • B) They convert chemical energy to electrical energy

  • C) They consume fossil fuels

  • D) They produce nitrogen gas

Answer Key

• Q1: B) Provides energy to break bonds - Light energy splits water molecules.

• Q2: B) Water - Hydrogen reacts with oxygen to form water.

• Q3: B) They convert chemical energy to electrical energy - Fuel cells generate electricity cleanly.

Memory Aids

• Photolysis = light + water → H2 + O2.

• Hydrogen combustion produces only water, no pollution.

• Fuel cells = clean energy from hydrogen.