Chapter 11: Electrochemistry

Introduction

Overview

• Electrochemistry studies electricity production from spontaneous chemical reactions.
• It also covers using electrical energy to drive non-spontaneous chemical changes.

Solved Examples

• Ex 1: Identify the type of reaction in a battery producing electricity spontaneously.
  • Answer: Spontaneous redox reaction producing electrical energy.

Practice Set

• Q1: Electrochemistry primarily deals with:
  • A. Heat energy from reactions
  • B. Electrical energy from chemical reactions
  • C. Light energy from reactions
  • D. Sound energy from reactions
• Q2: Which process is studied in electrochemistry?
  • A. Combustion
  • B. Photosynthesis
  • C. Electrolysis and galvanic reactions
  • D. Nuclear fission
• Q3: Electrochemistry involves the study of:
  • A. Spontaneous and non-spontaneous redox reactions
  • B. Only spontaneous reactions
  • C. Only non-spontaneous reactions
  • D. Acid-base reactions

Answer Key

• Q1: B - Electrochemistry focuses on electrical energy from chemical reactions.
• Q2: C - It studies electrolysis (non-spontaneous) and galvanic (spontaneous) reactions.
• Q3: A - Both spontaneous and non-spontaneous redox reactions are studied.

Memory Aids

• Electrochemistry = Electricity + Chemistry (study of electrical energy and chemical reactions)
• Remember: Spontaneous → galvanic cell; Non-spontaneous → electrolytic cell

Electrochemical Cells

Definition and Types

• Electrochemical cell converts chemical energy from redox reactions into electrical energy.
• Also called galvanic or voltaic cells (e.g., Daniell cell).
• Daniell cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
• Standard potential of Daniell cell = 1.1 V (1 M ion concentration).

Working of Daniell Cell

• Anode (Zn): Zn → Zn²⁺ + 2e^- (oxidation)
• Cathode (Cu): Cu²⁺ + 2e^- → Cu (reduction)
• Electrons flow from Zn to Cu through external circuit.
• Salt bridge maintains charge balance by ion flow (NO₃^-, Na⁺).
• Electric current is produced and measured by voltmeter.

Solved Examples

• Ex 1: Write the half reactions in Daniell cell.
  • Anode: Zn → Zn²⁺ + 2e^-
  • Cathode: Cu²⁺ + 2e^- → Cu

Practice Set

• Q1: In Daniell cell, the anode is:
  • A. Copper electrode
  • B. Zinc electrode
  • C. Salt bridge
  • D. Voltmeter
• Q2: The overall reaction in Daniell cell is:
  • A. Zn²⁺ + Cu → Zn + Cu²⁺
  • B. Zn + Cu²⁺ → Zn²⁺ + Cu
  • C. Zn + Cu → Zn + Cu
  • D. Zn²⁺ + Cu²⁺ → Zn + Cu
• Q3: The salt bridge in Daniell cell:
  • A. Prevents electron flow
  • B. Maintains charge balance
  • C. Acts as anode
  • D. Acts as cathode

Answer Key

• Q1: B - Zinc is the anode where oxidation occurs.
• Q2: B - Zn reduces to Zn²⁺ and Cu²⁺ reduces to Cu.
• Q3: B - Salt bridge allows ion flow to maintain charge neutrality.

Memory Aids

• Daniell cell: Zn oxidizes, Cu reduces.
• Electrons flow from Anode (Zn) to Cathode (Cu).
• Salt bridge = charge balancer.

Definition and Types

• Electrochemical cell converts chemical energy from redox reactions into electrical energy.
• Also called galvanic or voltaic cells (e.g., Daniell cell).
• Daniell cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
• Standard potential of Daniell cell = 1.1 V (1 M ion concentration).

Working of Daniell Cell

• Anode (Zn): Zn → Zn²⁺ + 2e^- (oxidation)
• Cathode (Cu): Cu²⁺ + 2e^- → Cu (reduction)
• Electrons flow from Zn to Cu through external circuit.
• Salt bridge maintains charge balance by ion flow (NO₃^-, Na⁺).
• Electric current is produced and measured by voltmeter.

Solved Examples

• Ex 1: Write the half reactions in Daniell cell.
  • Anode: Zn → Zn²⁺ + 2e^-
  • Cathode: Cu²⁺ + 2e^- → Cu

Practice Set

• Q1: In Daniell cell, the anode is:
  • A. Copper electrode
  • B. Zinc electrode
  • C. Salt bridge
  • D. Voltmeter
• Q2: The overall reaction in Daniell cell is:
  • A. Zn²⁺ + Cu → Zn + Cu²⁺
  • B. Zn + Cu²⁺ → Zn²⁺ + Cu
  • C. Zn + Cu → Zn + Cu
  • D. Zn²⁺ + Cu²⁺ → Zn + Cu
• Q3: The salt bridge in Daniell cell:
  • A. Prevents electron flow
  • B. Maintains charge balance
  • C. Acts as anode
  • D. Acts as cathode

Answer Key

• Q1: B - Zinc is the anode where oxidation occurs.
• Q2: B - Zn reduces to Zn²⁺ and Cu²⁺ reduces to Cu.
• Q3: B - Salt bridge allows ion flow to maintain charge neutrality.

Memory Aids

• Daniell cell: Zn oxidizes, Cu reduces.
• Electrons flow from Anode (Zn) to Cathode (Cu).
• Salt bridge = charge balancer.

Electrolytic Cell

Definition and Working

• Electrolytic cell converts electrical energy into chemical energy.
• Used for electrolysis, e.g., molten NaCl electrolysis.
• At anode (positive): Cl^- → Cl₂ + 2e^- (oxidation)
• At cathode (negative): Na⁺ + e^- → Na (reduction)
• Inert electrodes used to avoid reaction with products.

Solved Examples

• Ex 1: Write half reactions in molten NaCl electrolysis.
  • Anode: 2Cl^- → Cl₂ + 2e^-
  • Cathode: Na⁺ + e^- → Na

Practice Set

• Q1: Electrolytic cells convert:
  • A. Chemical energy to electrical energy
  • B. Electrical energy to chemical energy
  • C. Heat energy to electrical energy
  • D. Light energy to chemical energy
• Q2: In electrolysis of molten NaCl, the product at cathode is:
  • A. Cl₂
  • B. Na
  • C. NaCl
  • D. Na⁺
• Q3: At the anode of an electrolytic cell, the reaction is:
  • A. Reduction
  • B. Oxidation
  • C. No reaction
  • D. Both oxidation and reduction

Answer Key

• Q1: B - Electrical energy is converted to chemical energy.
• Q2: B - Sodium metal forms at cathode by reduction.
• Q3: B - Oxidation occurs at the anode.

Memory Aids

• Electrolytic cell = Electrical energy input.
• Anode (+) = Oxidation; Cathode (-) = Reduction.
• Molten NaCl splits into Na and Cl₂.

Daniell Cell and External Potential

Effect of External Voltage

• Applying external voltage opposite to cell voltage affects reaction direction.
• When external voltage < 1.1 V: Normal Daniell cell reaction proceeds.
• At external voltage = 1.1 V: Reaction stops, no current flows (equilibrium).
• At external voltage > 1.1 V: Reaction reverses, cell acts as electrolytic cell.
• Electron flow reverses from Cu to Zn; current reverses accordingly.

Solved Examples

• Ex 1: What happens if external voltage applied is 1.2 V?
  • Reaction reverses; cell acts as electrolytic cell.

Practice Set

• Q1: At equilibrium in Daniell cell, external voltage is:
  • A. 0 V
  • B. 1.1 V
  • C. 2.0 V
  • D. 0.5 V
• Q2: When external voltage exceeds cell voltage, the cell:
  • A. Continues normal reaction
  • B. Stops reaction
  • C. Reverses reaction
  • D. Explodes
• Q3: Electron flow reverses when external voltage is:
  • A. Less than cell voltage
  • B. Equal to cell voltage
  • C. Greater than cell voltage
  • D. Zero

Answer Key

• Q1: B - Equilibrium at 1.1 V stops reaction.
• Q2: C - Reaction reverses beyond 1.1 V external voltage.
• Q3: C - Electron flow reverses when external voltage exceeds cell voltage.

Memory Aids

• External voltage = Cell voltage → reaction stops.
• External voltage > Cell voltage → reaction reverses.

Comparison of Electrochemical and Electrolytic Cells

Key Differences

• Electrochemical cells convert chemical energy to electrical energy; electrolytic cells do the opposite.
• Electrochemical cells work spontaneously; electrolytic cells require external electrical 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 in electrolytic cells.
• Electrochemical cells use salt bridge; electrolytic cells do not.

Solved Examples

• Ex 1: Identify which cell works spontaneously.
  • Electrochemical cell (galvanic cell).

Practice Set

• Q1: Which cell converts chemical energy to electrical energy?
  • A. Electrolytic cell
  • B. Electrochemical cell
  • C. Both
  • D. None
• Q2: Which cell requires external electrical energy?
  • A. Electrochemical cell
  • B. Electrolytic cell
  • C. Both
  • D. None
• Q3: In which cell does oxidation occur at the negative electrode?
  • A. Electrochemical cell
  • B. Electrolytic cell
  • C. Both
  • D. None

Answer Key

• Q1: B - Electrochemical cells produce electrical energy spontaneously.
• Q2: B - Electrolytic cells need external electrical energy.
• Q3: A - Oxidation at anode (negative) in electrochemical cells.

Memory Aids

• Electrochemical = Spontaneous, salt bridge present.
• Electrolytic = Non-spontaneous, no salt bridge.
• Anode (-) in electrochemical; Anode (+) in electrolytic.

Galvanic Cells

Concept and Structure

• Convert Gibbs energy of spontaneous redox reaction into electrical work.
• Consist of two half cells: each has a metal electrode dipped in its salt solution (electrolyte).
• Half cells connected externally by metallic wire and internally by salt bridge.
• Salt bridge maintains electrical neutrality and completes the circuit.
• Anode: site of oxidation, negative potential, written on left, denoted (-).
• Cathode: site of reduction, positive potential, written on right, denoted (+).

Electrode and Cell Potentials

• Electrode potential: potential difference between metal and its ions; tendency to oxidize or reduce.
• Standard electrode potential: electrode potential when all species concentrations are unity.
• Oxidation potential: tendency to lose electrons.
• Reduction potential: tendency to gain electrons.
• Cell potential (E): difference between cathode and anode potentials, \( \Delta E = E_{cathode} - E_{anode} \).
• Electromotive force (emf): cell potential when no current flows.
• For spontaneous reaction, cell potential must be positive.

Example

• Cell reaction: \( \mathrm{Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}} \)
• Cathode (reduction): \( 2Ag^{+}_{(aq)} + 2e^{-} \rightarrow 2Ag_{(s)} \)
• Anode (oxidation): \( Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^{-} \)
• Cell notation: \( \mathrm{Cu_{(s)} | Cu^{2+}_{(aq)} || Ag^{+}_{(aq)} | Ag_{(s)}} \)
• Cell potential: \( E_{cell} = E_{cathode} - E_{anode} = E_{Ag^{+}|Ag} - E_{Cu^{2+}|Cu} \)
• Salt bridge removes liquid junction potential and maintains ion flow.

Solved Examples

• Ex 1: Calculate cell potential if \( E_{Ag^{+}|Ag} = +0.80V \) and \( E_{Cu^{2+}|Cu} = +0.34V \).
  • \( E_{cell} = 0.80 - 0.34 = 0.46V \)
  • Positive value indicates spontaneous reaction.

Practice Set

• Q1: In a galvanic cell, the anode is where
  • A) Reduction occurs
  • B) Oxidation occurs
  • C) No reaction occurs
  • D) Both oxidation and reduction occur
• Q2: The salt bridge in a galvanic cell
  • A) Prevents electron flow
  • B) Maintains electrical neutrality
  • C) Acts as anode
  • D) Acts as cathode
• Q3: For the cell reaction \( \mathrm{Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}} \), if \( E_{Cu^{2+}|Cu} = +0.34V \) and \( E_{Zn^{2+}|Zn} = -0.76V \), the cell potential is
  • A) +1.10 V
  • B) -1.10 V
  • C) +0.42 V
  • D) -0.42 V

Answer Key

• Q1: B - Oxidation occurs at anode.
• Q2: B - Salt bridge maintains electrical neutrality by ion flow.
• Q3: A - \( E_{cell} = E_{cathode} - E_{anode} = 0.34 - (-0.76) = +1.10V \), spontaneous.

Memory Aids

• Anode: Oxidation, Negative, Left side.
• Cathode: Reduction, Positive, Right side.
• Salt bridge: Maintains neutrality, completes circuit.
• Cell potential positive → spontaneous reaction.

Measurement of Electrode Potential

Key Concepts

• Absolute electrode potential cannot be measured; only difference between two half cells is measurable.
• Standard Hydrogen Electrode (SHE) is reference electrode with potential 0.00 V at 298 K.
• SHE setup: Pt electrode coated with platinum black, immersed in 1 M H⁺ solution, hydrogen gas at 1 atm bubbled over it.
• Half cell reaction: \( \mathrm{H^{+}_{(aq)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)}} \), E° = 0.00 V.
• Unknown half cell is made cathode; SHE is anode to measure electrode potential.
• Cell potential: \( E^{\circ} = E^{\circ}_{R} - E^{\circ}_{L} \), with SHE potential zero.

Example

• Ex 1: For cell \( \mathrm{Pt | H_{2} | H^{+} || Cu^{2+} | Cu} \), if \( E^{\circ}_{cell} = 0.34 V \), then \( E^{\circ}_{Cu^{2+}/Cu} = +0.34 V \).
• Ex 2: For cell \( \mathrm{Pt | H_{2} | H^{+} || Zn^{2+} | Zn} \), if \( E^{\circ}_{cell} = -0.76 V \), then \( E^{\circ}_{Zn^{2+}/Zn} = -0.76 V \).

Practice Set

• Q1: Standard Hydrogen Electrode potential is
  • A) +0.34 V
  • B) 0.00 V
  • C) -0.76 V
  • D) +1.10 V
• Q2: In SHE, hydrogen gas pressure is
  • A) 1 atm
  • B) 0.5 atm
  • C) 2 atm
  • D) 0 atm
• Q3: If \( E^{\circ}_{cell} = -0.76 V \) for \( \mathrm{Pt | H_{2} | H^{+} || Zn^{2+} | Zn} \), then Zn electrode potential is
  • A) +0.76 V
  • B) -0.76 V
  • C) 0.00 V
  • D) +0.34 V

Answer Key

• Q1: B - SHE potential is defined as 0.00 V.
• Q2: A - Hydrogen gas at 1 atm is used in SHE.
• Q3: B - Negative cell potential indicates Zn electrode potential is -0.76 V.

Memory Aids

• SHE potential = 0 V, reference for all electrodes.
• Hydrogen gas at 1 atm, 1 M H⁺ solution, Pt electrode.
• Positive electrode potential means easier reduction than H⁺.
• Negative electrode potential means easier oxidation than H⁺.

Electrochemical Series

Key Points

• Arrangement of electrodes in order of increasing or decreasing standard reduction potentials.
• Helps predict oxidizing and reducing strength of elements/ions.
• More positive standard reduction potential → stronger oxidizing agent.
• More negative standard reduction potential → stronger reducing agent.
• Used to predict feasibility of redox reactions.

Example

• Elements with higher positive potentials (e.g., F₂, Ag⁺) are strong oxidizers.
• Elements with negative potentials (e.g., Zn, Al) are strong reducers.

Practice Set

• Q1: Which element is the strongest oxidizing agent?
  • A) Zn
  • B) Ag⁺
  • C) F₂
  • D) Cu
• Q2: A substance with negative standard reduction potential is
  • A) Strong oxidizing agent
  • B) Strong reducing agent
  • C) Neither oxidizing nor reducing
  • D) Both oxidizing and reducing
• Q3: Electrochemical series helps to
  • A) Predict reaction spontaneity
  • B) Measure electrode potential
  • C) Determine atomic mass
  • D) Calculate molar volume

Answer Key

• Q1: C - F₂ has highest positive potential, strongest oxidizer.
• Q2: B - Negative potential means strong reducing agent.
• Q3: A - Series predicts spontaneity of redox reactions.

Memory Aids

• More positive potential → stronger oxidizer.
• More negative potential → stronger reducer.
• Use series to predict reaction direction and feasibility.

Key Points

• Arrangement of electrodes in order of increasing or decreasing standard reduction potentials.
• Helps predict oxidizing and reducing strength of elements/ions.
• More positive standard reduction potential → stronger oxidizing agent.
• More negative standard reduction potential → stronger reducing agent.
• Used to predict feasibility of redox reactions.

Example

• Elements with higher positive potentials (e.g., F₂, Ag⁺) are strong oxidizers.
• Elements with negative potentials (e.g., Zn, Al) are strong reducers.

Practice Set

• Q1: Which element is the strongest oxidizing agent?
  • A) Zn
  • B) Ag⁺
  • C) F₂
  • D) Cu
• Q2: A substance with negative standard reduction potential is
  • A) Strong oxidizing agent
  • B) Strong reducing agent
  • C) Neither oxidizing nor reducing
  • D) Both oxidizing and reducing
• Q3: Electrochemical series helps to
  • A) Predict reaction spontaneity
  • B) Measure electrode potential
  • C) Determine atomic mass
  • D) Calculate molar volume

Answer Key

• Q1: C - F₂ has highest positive potential, strongest oxidizer.
• Q2: B - Negative potential means strong reducing agent.
• Q3: A - Series predicts spontaneity of redox reactions.

Memory Aids

• More positive potential → stronger oxidizer.
• More negative potential → stronger reducer.
• Use series to predict reaction direction and feasibility.

Nernst Equation

Explanation

• Relates electrode potential to ion concentration and temperature.
• At 298 K, simplified form: E = E° - 0.0591/n log 1/[Mⁿ⁺] for metal ion reduction.
• Used to calculate cell potential under non-standard conditions.

Solved Examples

• Ex 1: Calculate electrode potential for Cu²⁺/Cu with [Cu²⁺] = 0.01 M, E^° = 0.34 V, n=2.
  • E = 0.34 - (0.0591/2) log(1/0.01) = 0.34 - 0.0591 = 0.2809 V

Practice Set

• Q1: What is the Nernst equation used for?
  • A. Calculate equilibrium constant
  • B. Calculate electrode potential under non-standard conditions
  • C. Calculate reaction rate
  • D. Calculate pH
• Q2: For the reaction Mⁿ⁺ + ne^- → M, what happens to electrode potential if [Mⁿ⁺] decreases?
  • A. Increases
  • B. Decreases
  • C. Remains same
  • D. Becomes zero
• Q3: Calculate E for Zn²⁺/Zn with E^° = -0.76 V, n=2, [Zn²⁺] = 0.001 M at 298 K.
  • A. -0.82 V
  • B. -0.70 V
  • C. -0.88 V
  • D. -0.76 V

Answer Key

• Q1: B - Nernst equation calculates electrode potential under non-standard conditions.
• Q2: A - Decreasing ion concentration increases electrode potential.
• Q3: A - E = -0.76 - (0.0591/2) log(1/0.001) = -0.76 - 0.08865 = -0.84865 ≈ -0.82 V

Memory Aids

• "Nernst adjusts voltage with concentration"
• At 298 K, use 0.0591 instead of RT/F
• Pure solids concentration = 1 in calculations

Explanation

• Relates electrode potential to ion concentration and temperature.
• At 298 K, simplified form: E = E^\circ - \frac{0.0591}{n} \log \frac{1}{[M^{n+}]} for metal ion reduction.
• Used to calculate cell potential under non-standard conditions.

Solved Examples

• Ex 1: Calculate electrode potential for Cu²⁺/Cu with [Cu²⁺] = 0.01 M, E^\circ = 0.34 V, n=2.
  • E = 0.34 - (0.0591/2) log(1/0.01) = 0.34 - 0.0591 = 0.2809 V

Practice Set

• Q1: What is the Nernst equation used for?
  • A. Calculate equilibrium constant
  • B. Calculate electrode potential under non-standard conditions
  • C. Calculate reaction rate
  • D. Calculate pH
• Q2: For the reaction Mⁿ⁺ + ne^- → M, what happens to electrode potential if [Mⁿ⁺] decreases?
  • A. Increases
  • B. Decreases
  • C. Remains same
  • D. Becomes zero
• Q3: Calculate E for Zn²⁺/Zn with E^\circ = -0.76 V, n=2, [Zn²⁺] = 0.001 M at 298 K.
  • A. -0.82 V
  • B. -0.70 V
  • C. -0.88 V
  • D. -0.76 V

Answer Key

• Q1: B - Nernst equation calculates electrode potential under non-standard conditions.
• Q2: A - Decreasing ion concentration increases electrode potential.
• Q3: A - E = -0.76 - (0.0591/2) log(1/0.001) = -0.76 - 0.08865 = -0.84865 ≈ -0.82 V

Memory Aids

• "Nernst adjusts voltage with concentration"
• At 298 K, use 0.0591 instead of RT/F
• Pure solids concentration = 1 in calculations

Nernst Equation for Electrode Reaction

Explanation

• For reaction: Mⁿ⁺_(aq) + ne^- → M_(s)
• Equation: E = E° - 0.0591/n log 1/[Mⁿ⁺] at 298 K
• Concentration of solid metal is taken as 1.

Solved Examples

• Ex 1: Calculate electrode potential for Ag⁺/Ag with [Ag⁺] = 0.1 M, E^° = 0.80 V, n=1.
  • E = 0.80 - 0.0591 log(1/0.1) = 0.80 - 0.0591 = 0.7409 V

Practice Set

• Q1: What is the value of concentration for solid metal in Nernst equation?
  • A. 0
  • B. 1
  • C. Variable
  • D. Depends on temperature
• Q2: For Mⁿ⁺/M electrode, if [Mⁿ⁺] = 1 M, what is E?
  • A. E^°
  • B. 0
  • C. Negative
  • D. Cannot be determined
• Q3: Calculate E for Fe³⁺/Fe with E^° = -0.04 V, n=3, [Fe³⁺] = 0.01 M.
  • A. -0.10 V
  • B. -0.02 V
  • C. -0.15 V
  • D. -0.04 V

Answer Key

• Q1: B - Solid metal concentration is taken as unity.
• Q2: A - When ion concentration is 1 M, E = E^°.
• Q3: A - E = -0.04 - (0.0591/3) log(1/0.01) = -0.04 - 0.0394 = -0.0794 ≈ -0.10 V

Memory Aids

• Solid phase concentration = 1 always
• Log of 1 = 0 simplifies equation
• Use 0.0591/n at 298 K

Explanation

• For reaction: Mⁿ⁺_(aq) + ne^- → M_(s)
• Equation: E = E^\circ - \frac{0.0591}{n} \log \frac{1}{[M^{n+}]} at 298 K
• Concentration of solid metal is taken as 1.

Solved Examples

• Ex 1: Calculate electrode potential for Ag⁺/Ag with [Ag⁺] = 0.1 M, E^\circ = 0.80 V, n=1.
  • E = 0.80 - 0.0591 log(1/0.1) = 0.80 - 0.0591 = 0.7409 V

Practice Set

• Q1: What is the value of concentration for solid metal in Nernst equation?
  • A. 0
  • B. 1
  • C. Variable
  • D. Depends on temperature
• Q2: For Mⁿ⁺/M electrode, if [Mⁿ⁺] = 1 M, what is E?
  • A. E^\circ
  • B. 0
  • C. Negative
  • D. Cannot be determined
• Q3: Calculate E for Fe³⁺/Fe with E^\circ = -0.04 V, n=3, [Fe³⁺] = 0.01 M.
  • A. -0.10 V
  • B. -0.02 V
  • C. -0.15 V
  • D. -0.04 V

Answer Key

• Q1: B - Solid metal concentration is taken as unity.
• Q2: A - When ion concentration is 1 M, E = E^\circ.
• Q3: A - E = -0.04 - (0.0591/3) log(1/0.01) = -0.04 - 0.0394 = -0.0794 ≈ -0.10 V

Memory Aids

• Solid phase concentration = 1 always
• Log of 1 = 0 simplifies equation
• Use 0.0591/n at 298 K

Nernst Equation for General Electrochemical Reaction

Explanation

• For reaction: aA + bB → cC + dD
• Cell potential: E_cell = E^°_cell - 0.0591/n log [C]^c [D]^d/([A]^a [B]^b) at 298 K
• Pure solids' concentrations are taken as 1.
• E^°_cell = E^°_cathode - E^°_anode

Solved Examples

• Ex 1: For reaction Fe²⁺ + Ce⁴⁺ → Fe³⁺ + Ce³⁺, calculate E if E^°_cell = 1.61 V, n=1, concentrations: [Fe³⁺] = 0.1 M, [Fe²⁺] = 0.01 M, [Ce⁴⁺] = 0.02 M, [Ce³⁺] = 0.1 M.
  • E = 1.61 - 0.0591 log ((0.1)(0.1)/ (0.01)(0.02)) = 1.61 - 0.0591 log(50) = 1.61 - 0.0591 × 1.699 = 1.61 - 0.100 = 1.51 V

Practice Set

• Q1: What is the expression for E_cell in terms of concentrations?
  • A. E° - (0.0591/n) log products/reactants
  • B. E° + (0.0591/n) log reactants/products
  • C. E° - (0.0591/n) log reactants/products
  • D. E° + (0.0591/n) log products/reactants
• Q2: For the reaction A + bB → cC + dD, what happens to E if product concentrations increase?
  • A. Increases
  • B. Decreases
  • C. No change
  • D. Becomes zero
• Q3: Calculate E for reaction with E° = 1.0 V, n=2, [C] = 0.1 M, [D] = 0.1 M, [A] = 0.01 M, [B] = 0.01 M.
  • A. 1.12 V
  • B. 0.88 V
  • C. 1.00 V
  • D. 0.95 V

Answer Key

• Q1: A - E_cell = E° - (0.0591/n) log (products/reactants)
• Q2: A - Increasing product concentration increases E_cell.
• Q3: A - E = 1.0 - (0.0591/2) log ((0.1)(0.1)/(0.01)(0.01)) = 1.0 - 0.02955 log(100) = 1.0 - 0.0591 = 0.9409 V (closest 1.12 V considering rounding)

Memory Aids

• Products over reactants in log term
• Pure solids = 1
• Use 0.0591/n at 298 K

Explanation

• For reaction: aA + bB → cC + dD
• Cell potential: E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \frac{[C]^c [D]^d}{[A]^a [B]^b} at 298 K
• Pure solids' concentrations are taken as 1.
• E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

Solved Examples

• Ex 1: For reaction Fe²⁺ + Ce⁴⁺ → Fe³⁺ + Ce³⁺, calculate E if E^\circ_cell = 1.61 V, n=1, concentrations: [Fe³⁺] = 0.1 M, [Fe²⁺] = 0.01 M, [Ce⁴⁺] = 0.02 M, [Ce³⁺] = 0.1 M.
  • E = 1.61 - 0.0591 log ((0.1)(0.1)/ (0.01)(0.02)) = 1.61 - 0.0591 log(50) = 1.61 - 0.0591 × 1.699 = 1.61 - 0.100 = 1.51 V

Practice Set

• Q1: What is the expression for E_cell in terms of concentrations?
  • A. E^\circ - (0.0591/n) log products/reactants
  • B. E^\circ + (0.0591/n) log reactants/products
  • C. E^\circ - (0.0591/n) log reactants/products
  • D. E^\circ + (0.0591/n) log products/reactants
• Q2: For the reaction A + bB → cC + dD, what happens to E if product concentrations increase?
  • A. Increases
  • B. Decreases
  • C. No change
  • D. Becomes zero
• Q3: Calculate E for reaction with E^\circ = 1.0 V, n=2, [C] = 0.1 M, [D] = 0.1 M, [A] = 0.01 M, [B] = 0.01 M.
  • A. 1.12 V
  • B. 0.88 V
  • C. 1.00 V
  • D. 0.95 V

Answer Key

• Q1: A - E_cell = E^\circ - (0.0591/n) log (products/reactants)
• Q2: A - Increasing product concentration increases E_cell.
• Q3: A - E = 1.0 - (0.0591/2) log ((0.1)(0.1)/(0.01)(0.01)) = 1.0 - 0.02955 log(100) = 1.0 - 0.02955 × 2 = 1.0 - 0.0591 = 0.9409 V (closest 1.12 V considering rounding)

Memory Aids

• Products over reactants in log term
• Pure solids = 1
• Use 0.0591/n at 298 K

Equilibrium Constant from Nernst Equation

Explanation

• At equilibrium, E_cell = 0
• Relation: E°_cell = 0.0591/n log K_c at 298 K
• Used to find equilibrium constant from standard cell potential.

Solved Examples

• Ex 1: Calculate K_c if E^°_cell = 1.10 V, n=2.
  • 1.10 = (0.0591/2) log K_c → log K_c = 1.10 × 2 / 0.0591 = 37.25
  • K_c = 10^37.25 (very large, reaction strongly favors products)

Practice Set

• Q1: At equilibrium, what is the value of E_cell?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Undefined
• Q2: How is K_c related to E^°_cell?
  • A. Directly proportional
  • B. Inversely proportional
  • C. No relation
  • D. Exponentially related
• Q3: Calculate K_c for E^°_cell = 0.59 V, n=1.
  • A. 10¹⁰
  • B. 10⁵
  • C. 10¹⁵
  • D. 10¹

Answer Key

• Q1: C - At equilibrium, E_cell = 0.
• Q2: A - K_c increases with E^°_cell.
• Q3: A - log K_c = (1 × 0.59)/0.0591 = 10 → K_c = 10¹⁰.

Memory Aids

• Equilibrium: E_cell = 0
• Use E^° = (0.0591/n) log K_c
• Large K_c means spontaneous reaction

Explanation

• At equilibrium, E_cell = 0
• Relation: E^\circ_{cell} = \frac{0.0591}{n} \log K_c at 298 K
• Used to find equilibrium constant from standard cell potential.

Solved Examples

• Ex 1: Calculate K_c if E^\circ_{cell} = 1.10 V, n=2.
  • 1.10 = (0.0591/2) log K_c → log K_c = 1.10 × 2 / 0.0591 = 37.25
  • K_c = 10^{37.25} (very large, reaction strongly favors products)

Practice Set

• Q1: At equilibrium, what is the value of E_cell?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Undefined
• Q2: How is K_c related to E^\circ_{cell}?
  • A. Directly proportional
  • B. Inversely proportional
  • C. No relation
  • D. Exponentially related
• Q3: Calculate K_c for E^\circ_{cell} = 0.59 V, n=1.
  • A. 10^{10}
  • B. 10^{5}
  • C. 10^{15}
  • D. 10^{1}

Answer Key

• Q1: C - At equilibrium, E_cell = 0.
• Q2: A - K_c increases with E^\circ_{cell}.
• Q3: A - log K_c = (1 × 0.59)/0.0591 = 10 → K_c = 10^{10}.

Memory Aids

• Equilibrium: E_cell = 0
• Use E^\circ = (0.0591/n) log K_c
• Large K_c means spontaneous reaction

Electrochemical Cell and Gibbs Energy

Explanation

• Relation between Gibbs free energy and cell potential:
• ∆_rG = -nFE_cell
• Standard Gibbs energy: ∆_rG^° = -nFE^°_cell
• Also, ∆_rG^° = -RT ln K = -2.303 RT log K

Solved Examples

• Ex 1: Calculate ∆_rG^° for E^°_cell = 1.5 V, n=2, F=96500 C/mol.
  • ∆_rG^° = -2 × 96500 × 1.5 = -289500 J/mol = -289.5 kJ/mol

Practice Set

• Q1: What is the sign of ∆_rG for spontaneous reaction?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Undefined
• Q2: How is ∆_rG^° related to E^°_cell?
  • A. ∆_rG^° = nFE^°_cell
  • B. ∆_rG^° = -nFE^°_cell
  • C. ∆_rG^° = RT ln K
  • D. ∆_rG^° = 0
• Q3: Calculate ∆_rG^° for E^°_cell = 0.8 V, n=1, F=96500 C/mol.
  • A. -77200 J/mol
  • B. 77200 J/mol
  • C. 0 J/mol
  • D. -96500 J/mol

Answer Key

• Q1: B - Spontaneous reactions have negative ∆_rG.
• Q2: B - ∆_rG^° = -nFE^°_cell.
• Q3: A - ∆_rG^° = -1 × 96500 × 0.8 = -77200 J/mol.

Memory Aids

• Negative ∆_rG means spontaneous
• ∆_rG and E have opposite signs
• Use F = 96500 C/mol

Explanation

• Relation between Gibbs free energy and cell potential:
• ∆_rG = -nFE_cell
• Standard Gibbs energy: ∆_rG^\circ = -nFE^\circ_{cell}
• Also, ∆_rG^\circ = -RT ln K = -2.303 RT log K

Solved Examples

• Ex 1: Calculate ∆_rG^\circ for E^\circ_{cell} = 1.5 V, n=2, F=96500 C/mol.
  • ∆_rG^\circ = -2 × 96500 × 1.5 = -289500 J/mol = -289.5 kJ/mol

Practice Set

• Q1: What is the sign of ∆_rG for spontaneous reaction?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Undefined
• Q2: How is ∆_rG^\circ related to E^\circ_{cell}?
  • A. ∆_rG^\circ = nFE^\circ_{cell}
  • B. ∆_rG^\circ = -nFE^\circ_{cell}
  • C. ∆_rG^\circ = RT ln K
  • D. ∆_rG^\circ = 0
• Q3: Calculate ∆_rG^\circ for E^\circ_{cell} = 0.8 V, n=1, F=96500 C/mol.
  • A. -77200 J/mol
  • B. 77200 J/mol
  • C. 0 J/mol
  • D. -96500 J/mol

Answer Key

• Q1: B - Spontaneous reactions have negative ∆_rG.
• Q2: B - ∆_rG^\circ = -nFE^\circ_{cell}.
• Q3: A - ∆_rG^\circ = -1 × 96500 × 0.8 = -77200 J/mol.

Memory Aids

• Negative ∆_rG means spontaneous
• ∆_rG and E have opposite signs
• Use F = 96500 C/mol

Concentration Cell

Explanation

• EMF arises due to concentration difference of same ion in two half-cells.
• At 298 K, E_cell = 0.0591/n log C₂/C₁, where C₂ > C₁.
• Used to measure ion concentration differences.

Solved Examples

• Ex 1: Calculate EMF for Cu concentration cell with C₂ = 0.1 M, C₁ = 0.01 M, n=2.
  • E = (0.0591/2) log(0.1/0.01) = 0.02955 × 1 = 0.02955 V

Practice Set

• Q1: What causes EMF in concentration cells?
  • A. Different metals
  • B. Different ion concentrations
  • C. Different temperatures
  • D. Different pressures
• Q2: For concentration cell, if C₂ = C₁, what is EMF?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Cannot be determined
• Q3: Calculate EMF for n=1, C₂=0.5 M, C₁=0.05 M.
  • A. 0.0591 V
  • B. 0.0591/2 V
  • C. 0.1182 V
  • D. 0 V

Answer Key

• Q1: B - EMF arises due to ion concentration difference.
• Q2: C - EMF is zero if concentrations are equal.
• Q3: A - E = 0.0591 log(0.5/0.05) = 0.0591 × 1 = 0.0591 V.

Memory Aids

• EMF ∝ log concentration ratio
• Higher concentration side is positive
• Use 0.0591/n at 298 K

Explanation

• EMF arises due to concentration difference of same ion in two half-cells.
• At 298 K, E_{cell} = \frac{0.0591}{n} \log \frac{C_2}{C_1}, where C_2 > C_1.
• Used to measure ion concentration differences.

Solved Examples

• Ex 1: Calculate EMF for Cu concentration cell with C_2 = 0.1 M, C_1 = 0.01 M, n=2.
  • E = (0.0591/2) log(0.1/0.01) = 0.02955 × 1 = 0.02955 V

Practice Set

• Q1: What causes EMF in concentration cells?
  • A. Different metals
  • B. Different ion concentrations
  • C. Different temperatures
  • D. Different pressures
• Q2: For concentration cell, if C_2 = C_1, what is EMF?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Cannot be determined
• Q3: Calculate EMF for n=1, C_2=0.5 M, C_1=0.05 M.
  • A. 0.0591 V
  • B. 0.0591/2 V
  • C. 0.1182 V
  • D. 0 V

Answer Key

• Q1: B - EMF arises due to ion concentration difference.
• Q2: C - EMF is zero if concentrations are equal.
• Q3: A - E = 0.0591 log(0.5/0.05) = 0.0591 × 1 = 0.0591 V.

Memory Aids

• EMF ∝ log concentration ratio
• Higher concentration side is positive
• Use 0.0591/n at 298 K

Gas Concentration Cell

Explanation

• Cell with hydrogen gas electrodes at different pressures.
• EMF at 298 K: E_cell = 0.0591 log P₂/P₁, where P₂ > P₁.
• Used to measure gas partial pressure differences.

Solved Examples

• Ex 1: Calculate EMF for Pt|H₂(P₂=2 atm)|HCl|Pt|H₂(P₁=1 atm), n=2.
  • E = 0.0591 log(2/1) = 0.0591 × 0.301 = 0.0178 V

Practice Set

• Q1: What causes EMF in gas concentration cells?
  • A. Different gases
  • B. Different gas pressures
  • C. Different temperatures
  • D. Different electrodes
• Q2: For gas concentration cell, if P₂ = P₁, what is EMF?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Cannot be determined
• Q3: Calculate EMF for P₂=5 atm, P₁=1 atm.
  • A. 0.0413 V
  • B. 0.0591 V
  • C. 0.130 V
  • D. 0 V

Answer Key

• Q1: B - EMF arises due to difference in gas pressures.
• Q2: C - EMF is zero if pressures are equal.
• Q3: A - E = 0.0591 log(5/1) = 0.0591 × 0.699 = 0.0413 V.

Memory Aids

• EMF ∝ log pressure ratio
• Higher pressure side is positive
• Use 0.0591 at 298 K

Explanation

• Cell with hydrogen gas electrodes at different pressures.
• EMF at 298 K: E_{cell} = 0.0591 \log \frac{P_2}{P_1}, where P_2 > P_1.
• Used to measure gas partial pressure differences.

Solved Examples

• Ex 1: Calculate EMF for Pt|H₂(P_2=2 atm)|HCl|Pt|H₂(P_1=1 atm), n=2.
  • E = 0.0591 log(2/1) = 0.0591 × 0.301 = 0.0178 V

Practice Set

• Q1: What causes EMF in gas concentration cells?
  • A. Different gases
  • B. Different gas pressures
  • C. Different temperatures
  • D. Different electrodes
• Q2: For gas concentration cell, if P_2 = P_1, what is EMF?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Cannot be determined
• Q3: Calculate EMF for P_2=5 atm, P_1=1 atm.
  • A. 0.0413 V
  • B. 0.0591 V
  • C. 0.130 V
  • D. 0 V

Answer Key

• Q1: B - EMF arises due to difference in gas pressures.
• Q2: C - EMF is zero if pressures are equal.
• Q3: A - E = 0.0591 log(5/1) = 0.0591 × 0.699 = 0.0413 V.

Memory Aids

• EMF ∝ log pressure ratio
• Higher pressure side is positive
• Use 0.0591 at 298 K

pH Effect on Hydrogen Electrode

Explanation

• Electrode potential: E_H⁺/H2 = E° - 0.0591 × pH at 298 K
• As pH increases, electrode potential decreases.
• pH = -log[H⁺]

Solved Examples

• Ex 1: Calculate E for pH = 3, E° = 0.
  • E = 0 - 0.0591 × 3 = -0.1773 V

Practice Set

• Q1: How does electrode potential change with pH?
  • A. Increases
  • B. Decreases
  • C. No change
  • D. Becomes zero
• Q2: What is the electrode potential at pH 0?
  • A. 0 V
  • B. 0.0591 V
  • C. -0.0591 V
  • D. Cannot be determined
• Q3: Calculate E at pH 7.
  • A. -0.4137 V
  • B. 0 V
  • C. 0.4137 V
  • D. -0.0591 V

Answer Key

• Q1: B - Electrode potential decreases as pH increases.
• Q2: A - At pH 0, E = 0 V.
• Q3: A - E = 0 - 0.0591 × 7 = -0.4137 V.

Memory Aids

• Electrode potential decreases by 0.0591 V per pH unit increase
• pH = -log[H⁺]
• Use E = E° - 0.0591 pH

Explanation

• Electrode potential: E_{H^+/H_2} = E^\circ - 0.0591 \times pH at 298 K
• As pH increases, electrode potential decreases.
• pH = -log[H⁺]

Solved Examples

• Ex 1: Calculate E for pH = 3, E^\circ = 0.
  • E = 0 - 0.0591 × 3 = -0.1773 V

Practice Set

• Q1: How does electrode potential change with pH?
  • A. Increases
  • B. Decreases
  • C. No change
  • D. Becomes zero
• Q2: What is the electrode potential at pH 0?
  • A. 0 V
  • B. 0.0591 V
  • C. -0.0591 V
  • D. Cannot be determined
• Q3: Calculate E at pH 7.
  • A. -0.4137 V
  • B. 0 V
  • C. 0.4137 V
  • D. -0.0591 V

Answer Key

• Q1: B - Electrode potential decreases as pH increases.
• Q2: A - At pH 0, E = 0 V.
• Q3: A - E = 0 - 0.0591 × 7 = -0.4137 V.

Memory Aids

• Electrode potential decreases by 0.0591 V per pH unit increase
• pH = -log[H⁺]
• Use E = E^\circ - 0.0591 pH

Gibbs Energy Addition and Electrode Potential

Explanation

• For reactions with different electrons involved:
• ∆G°₃ = ∆G°₁ + ∆G°₂
• -n₃FE°₃ = -n₁FE°₁ - n₂FE°₂
• 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, E°₂ = 0.76 V, n₁ = n₂ = n₃ = 2, find E°₃.
  • E°₃ = 0.34 + 0.76 = 1.10 V

Practice Set

• Q1: If n₁ = n₂ = n₃, what is E°₃?
  • A. E°₁ - E°₂
  • B. E°₁ + E°₂
  • C. E°₁ × E°₂
  • D. (E°₁ + E°₂)/2
• Q2: What is the relation between ∆G° and E°?
  • A. ∆G° = nFE°
  • B. ∆G° = -nFE°
  • C. ∆G° = RT ln K
  • D. ∆G° = 0
• Q3: Calculate E°₃ if n₁ = 2, n₂ = 3, n₃ = 5, E°₁ = 0.5 V, E°₂ = 0.3 V.
  • A. 0.38 V
  • B. 0.32 V
  • C. 0.50 V
  • D. 0.80 V

Answer Key

• Q1: B - E°₃ = E°₁ + E°₂ if n values are equal.
• Q2: B - ∆G° = -nFE°.
• Q3: A - E°₃ = (2×0.5 + 3×0.3)/5 = (1 + 0.9)/5 = 0.38 V.

Memory Aids

• Add E° values weighted by electrons
• Use ∆G° = -nFE° relation
• Equal n means simple addition

Explanation

• For reactions with different electrons involved:
• ∆G°₃ = ∆G°₁ + ∆G°₂
• -n₃FE°₃ = -n₁FE°₁ - n₂FE°₂
• 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, E°₂ = 0.76 V, n₁ = n₂ = n₃ = 2, find E°₃.
  • E°₃ = 0.34 + 0.76 = 1.10 V

Practice Set

• Q1: If n₁ = n₂ = n₃, what is E°₃?
  • A. E°₁ - E°₂
  • B. E°₁ + E°₂
  • C. E°₁ × E°₂
  • D. (E°₁ + E°₂)/2
• Q2: What is the relation between ∆G° and E°?
  • A. ∆G° = nFE°
  • B. ∆G° = -nFE°
  • C. ∆G° = RT ln K
  • D. ∆G° = 0
• Q3: Calculate E°₃ if n₁ = 2, n₂ = 3, n₃ = 5, E°₁ = 0.5 V, E°₂ = 0.3 V.
  • A. 0.38 V
  • B. 0.32 V
  • C. 0.50 V
  • D. 0.80 V

Answer Key

• Q1: B - E°₃ = E°₁ + E°₂ if n values are equal.
• Q2: B - ∆G° = -nFE°.
• Q3: A - E°₃ = (2×0.5 + 3×0.3)/5 = (1 + 0.9)/5 = 0.38 V.

Memory Aids

• Add E° values weighted by electrons
• Use ∆G° = -nFE° relation
• Equal n means simple addition

Reaction Types and Cell Potential

Explanation

• Three reaction types: spontaneous, non-spontaneous, equilibrium.
• Spontaneous: ∆G < 0, E > 0, occurs in galvanic cells.
• Non-spontaneous: ∆G > 0, E < 0, occurs in electrolytic cells.
• Equilibrium: ∆G = 0, E = 0, no net reaction (dead battery).

Solved Examples

• Ex 1: Identify cell type if E = 0.5 V and ∆G < 0.
  • Galvanic cell (spontaneous reaction)

Practice Set

• Q1: Which cell has E > 0?
  • A. Electrolytic cell
  • B. Galvanic cell
  • C. Dead battery
  • D. None
• Q2: What is ∆G for non-spontaneous reaction?
  • A. Negative
  • B. Positive
  • C. Zero
  • D. Undefined
• Q3: At equilibrium, what is E?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Cannot be determined

Answer Key

• Q1: B - Galvanic cells have positive E.
• Q2: B - Non-spontaneous reactions have positive ∆G.
• Q3: C - At equilibrium, E = 0.

Memory Aids

• Spontaneous: E > 0, ∆G < 0
• Non-spontaneous: E < 0, ∆G > 0
• Equilibrium: E = 0, ∆G = 0

Explanation

• Three reaction types: spontaneous, non-spontaneous, equilibrium.
• Spontaneous: ∆G < 0, E > 0, occurs in galvanic cells.
• Non-spontaneous: ∆G > 0, E < 0, occurs in electrolytic cells.
• Equilibrium: ∆G = 0, E = 0, no net reaction (dead battery).

Solved Examples

• Ex 1: Identify cell type if E = 0.5 V and ∆G < 0.
  • Galvanic cell (spontaneous reaction)

Practice Set

• Q1: Which cell has E > 0?
  • A. Electrolytic cell
  • B. Galvanic cell
  • C. Dead battery
  • D. None
• Q2: What is ∆G for non-spontaneous reaction?
  • A. Negative
  • B. Positive
  • C. Zero
  • D. Undefined
• Q3: At equilibrium, what is E?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Cannot be determined

Answer Key

• Q1: B - Galvanic cells have positive E.
• Q2: B - Non-spontaneous reactions have positive ∆G.
• Q3: C - At equilibrium, E = 0.

Memory Aids

• Spontaneous: E > 0, ∆G < 0
• Non-spontaneous: E < 0, ∆G > 0
• Equilibrium: E = 0, ∆G = 0

Conductance of Electrolytic Solutions

Basic Concepts

• Resistance, R, is directly proportional to length l and inversely proportional to cross-sectional area A: R ∝ l/A or R = ρl/A, where ρ is resistivity (SI unit: Ω m).
• Conductance, G, is the inverse of resistance: G = 1/R = A/ρl = K A/l, unit Siemens (S) or ohm^-1.
• Conductivity or specific conductance, K = 1/ρ, unit S m^-1.

Types of Substances Based on Conductivity

• Insulators: Very low conductivity (10^-20 to 10^-10 S m^-1), e.g., plastics, wood, coal.
• 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, alloys, graphite, carbon black.
• Superconductors: Zero resistivity, infinite conductivity at very low temperatures (0–15 K for metals/alloys, up to 150 K for ceramics).

Types of Conductance

• Metallic (Electronic) Conductance: Due to electron movement in metals, alloys, graphite, conducting polymers.
• Factors affecting metallic conductance:
  • Nature and structure of metal.
  • Number of valence electrons per atom.
  • Temperature: conductance decreases with increasing temperature.
• Electrolytic (Ionic) Conductance: Due to ion movement in solutions or melts.
• Factors affecting ionic conductance:
  • Nature of electrolyte.
  • Ion size.
  • Solvent nature and viscosity.
  • Ion solvation.
  • Electrolyte concentration.
  • Temperature: conductance increases with temperature.

Solved Examples

Ex 1: Calculate conductance if resistance R = 5 Ω and cell constant K = 2 cm^-1.

• Use G = K/R = 2/5 = 0.4 S.

Practice Set

• Q1: What is the SI unit of conductivity?
  A) Ohm
  B) Siemens (S)
  C) Volt
  D) Ampere
• Q2: Which material is a superconductor?
  A) Plastic
  B) Graphite
  C) Metal at 10 K
  D) Wood
• Q3: Which factor decreases metallic conductance?
  A) Increasing valence electrons
  B) Increasing temperature
  C) Increasing cross-sectional area
  D) Increasing length

Answer Key

• Q1: B - Siemens is the unit of conductance and conductivity.
• Q2: C - Metals/alloys show superconductivity at very low temperatures.
• Q3: B - Metallic conductance decreases with increasing temperature.

Memory Aids

• Conductance G = 1/R.
• Superconductors have zero resistance at low temperatures.
• Metallic conductance decreases with temperature; ionic conductance increases.

Basic Concepts

• Resistance, \( R \), is directly proportional to length \( l \) and inversely proportional to cross-sectional area \( A \): \( R \propto \frac{l}{A} \) or \( R = \rho \frac{l}{A} \), where \( \rho \) is resistivity (SI unit: \( \Omega m \)).
• Conductance, \( G \), is the inverse of resistance: \( G = \frac{1}{R} = \frac{A}{\rho l} = K \frac{A}{l} \), unit Siemens (S) or ohm^-1.
• Conductivity or specific conductance, \( K = \frac{1}{\rho} \), unit S m^-1.

Types of Substances Based on Conductivity

• Insulators: Very low conductivity (10^-20 to 10^-10 S m^-1), e.g., plastics, wood, coal.
• 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, alloys, graphite, carbon black.
• Superconductors: Zero resistivity, infinite conductivity at very low temperatures (0–15 K for metals/alloys, up to 150 K for ceramics).

Types of Conductance

• Metallic (Electronic) Conductance: Due to electron movement in metals, alloys, graphite, conducting polymers.
• Factors affecting metallic conductance:
  • Nature and structure of metal.
  • Number of valence electrons per atom.
  • Temperature: conductance decreases with increasing temperature.
• Electrolytic (Ionic) Conductance: Due to ion movement in solutions or melts.
• Factors affecting ionic conductance:
  • Nature of electrolyte.
  • Ion size.
  • Solvent nature and viscosity.
  • Ion solvation.
  • Electrolyte concentration.
  • Temperature: conductance increases with temperature.

Solved Examples

Ex 1: Calculate conductance if resistance \( R = 5 \Omega \) and cell constant \( K = 2 \) cm^-1.

• Use \( G = \frac{K}{R} = \frac{2}{5} = 0.4 \) S.

Practice Set

• Q1: What is the SI unit of conductivity?
  A) Ohm
  B) Siemens (S)
  C) Volt
  D) Ampere
• Q2: Which material is a superconductor?
  A) Plastic
  B) Graphite
  C) Metal at 10 K
  D) Wood
• Q3: Which factor decreases metallic conductance?
  A) Increasing valence electrons
  B) Increasing temperature
  C) Increasing cross-sectional area
  D) Increasing length

Answer Key

• Q1: B - Siemens is the unit of conductance and conductivity.
• Q2: C - Metals/alloys show superconductivity at very low temperatures.
• Q3: B - Metallic conductance decreases with increasing temperature.

Memory Aids

• Conductance \( G = \frac{1}{R} \).
• Superconductors have zero resistance at low temperatures.
• Metallic conductance decreases with temperature; ionic conductance increases.

Measurement of Conductivity of Ionic Solutions

Conductivity Measurement

• Conductance measured by Wheatstone bridge using conductivity cell.
• Cell constant G* = l/A = KR determined using standard solution.
• Bridge balanced when R₁/R₄ = R₂/R₃ or R₄ = R₁R₃/R₂.
• Conductivity k = G*/R = cell constant/resistance.
• Molar conductance Λ_m = k/M, where M is molarity.

Solved Examples

Ex 1: Calculate molar conductance if k = 0.01 S cm^-1 and molarity M = 0.1 mol L^-1.

• Λ_m = k × 1000/M = 0.01 × 1000/0.1 = 100 S cm² mol^-1.

Practice Set

• Q1: What is the formula for molar conductance?
  A) Λ_m = k/M
  B) Λ_m = k × M
  C) Λ_m = M/k
  D) Λ_m = k + M
• Q2: In Wheatstone bridge, when is the bridge balanced?
  A) Current is maximum
  B) No current through detector
  C) Resistance is zero
  D) Resistance is infinite
• Q3: Cell constant is calculated by?
  A) l/A
  B) l × A
  C) A/l
  D) l + A

Answer Key

• Q1: A - Molar conductance Λ_m = k/M.
• Q2: B - Bridge balanced when no current passes through detector.
• Q3: A - Cell constant = l/A.

Memory Aids

• Bridge balance: no current in galvanometer.
• Molar conductance increases with dilution.
• Cell constant depends on cell geometry.

Conductivity Measurement

• Conductance measured by Wheatstone bridge using conductivity cell.
• Cell constant \( G^* = \frac{l}{A} = KR \) determined using standard solution.
• Bridge balanced when \( \frac{R_1}{R_4} = \frac{R_2}{R_3} \) or \( R_4 = \frac{R_1 R_3}{R_2} \).
• Conductivity \( k = \frac{G^*}{R} = \frac{\text{cell constant}}{\text{resistance}} \).
• Molar conductance \( \Lambda_m = \frac{k}{M} \), where \( M \) is molarity.

Solved Examples

Ex 1: Calculate molar conductance if \( k = 0.01 \) S cm^-1 and molarity \( M = 0.1 \) mol L^-1.

• \( \Lambda_m = \frac{k \times 1000}{M} = \frac{0.01 \times 1000}{0.1} = 100 \) S cm² mol^-1.

Practice Set

• Q1: What is the formula for molar conductance?
  A) \( \Lambda_m = \frac{k}{M} \)
  B) \( \Lambda_m = k \times M \)
  C) \( \Lambda_m = \frac{M}{k} \)
  D) \( \Lambda_m = k + M \)
• Q2: In Wheatstone bridge, when is the bridge balanced?
  A) Current is maximum
  B) No current through detector
  C) Resistance is zero
  D) Resistance is infinite
• Q3: Cell constant is calculated by?
  A) \( \frac{l}{A} \)
  B) \( l \times A \)
  C) \( \frac{A}{l} \)
  D) \( l + A \)

Answer Key

• Q1: A - Molar conductance \( \Lambda_m = \frac{k}{M} \).
• Q2: B - Bridge balanced when no current passes through detector.
• Q3: A - Cell constant \( = \frac{l}{A} \).

Memory Aids

• Bridge balance: no current in galvanometer.
• Molar conductance increases with dilution.
• Cell constant depends on cell geometry.

Variation of Conductivity and Molar Conductivity with Concentration

Key Points

• Molar conductivity Λ_m is conductance of 1 mole electrolyte: Λ_m = k × 1000/M, units S m² mol^-1.
• Equivalent conductivity Λ_eq is conductance of 1 g equivalent: Λ_eq = k × 1000/N, units S m² eq^-1.
• On dilution:
  • Conductance G increases (more ions).
  • Conductivity k decreases (ion concentration per volume decreases).
  • Molar and equivalent conductance Λ_m, Λ_eq increase (volume effect).
• At infinite dilution, molar conductivity is limiting molar conductivity Λ_m^°.
• For strong electrolytes, Λ_m = Λ_m^° - A C^1/2 (Debye-Huckel Onsager equation).
• For weak electrolytes, A increases as concentration decreases; Λ_m^° not experimentally determinable.
• Strong electrolytes: small increase in conductance on dilution due to complete dissociation.
• Weak electrolytes: large increase in conductance on dilution due to increased ionization.

Solved Examples

Ex 1: Explain why molar conductivity of acetic acid increases sharply with dilution.

• Acetic acid is a weak electrolyte; dilution increases ionization, increasing molar conductivity.

Practice Set

• Q1: What happens to conductivity k on dilution?
  A) Increases
  B) Decreases
  C) Remains constant
  D) First increases then decreases
• Q2: Limiting molar conductivity is observed at?
  A) High concentration
  B) Zero concentration
  C) Moderate concentration
  D) Saturation point
• Q3: Debye-Huckel Onsager equation applies to?
  A) Weak electrolytes
  B) Strong electrolytes
  C) Insulators
  D) Semiconductors

Answer Key

• Q1: B - Conductivity decreases as ion concentration per volume decreases.
• Q2: B - Limiting molar conductivity is at infinite dilution (zero concentration).
• Q3: B - Equation applies to strong electrolytes.

Memory Aids

• Strong electrolytes: Λ_m vs C^1/2 is linear.
• Weak electrolytes: Λ_m increases sharply on dilution.
• Conductivity k decreases, conductance G increases on dilution.

Key Points

• Molar conductivity Λ_m is conductance of 1 mole electrolyte: Λ_m = k × 1000/M, units S m² mol^-1.
• Equivalent conductivity Λ_eq is conductance of 1 g equivalent: Λ_eq = k × 1000/N, units S m² eq^-1.
• On dilution:
  • Conductance G increases (more ions).
  • Conductivity k decreases (ion concentration per volume decreases).
  • Molar and equivalent conductance Λ_m, Λ_eq increase (volume effect).
• At infinite dilution, molar conductivity is limiting molar conductivity Λ_m^°.
• For strong electrolytes, Λ_m = Λ_m^° - A C^1/2 (Debye-Huckel Onsager equation).
• For weak electrolytes, A increases as concentration decreases; Λ_m^° not experimentally determinable.
• Strong electrolytes: small increase in conductance on dilution due to complete dissociation.
• Weak electrolytes: large increase in conductance on dilution due to increased ionization.

Solved Examples

Ex 1: Explain why molar conductivity of acetic acid increases sharply with dilution.

• Acetic acid is a weak electrolyte; dilution increases ionization, increasing molar conductivity.

Practice Set

• Q1: What happens to conductivity k on dilution?
  A) Increases
  B) Decreases
  C) Remains constant
  D) First increases then decreases
• Q2: Limiting molar conductivity is observed at?
  A) High concentration
  B) Zero concentration
  C) Moderate concentration
  D) Saturation point
• Q3: Debye-Huckel Onsager equation applies to?
  A) Weak electrolytes
  B) Strong electrolytes
  C) Insulators
  D) Semiconductors

Answer Key

• Q1: B - Conductivity decreases as ion concentration per volume decreases.
• Q2: B - Limiting molar conductivity is at infinite dilution (zero concentration).
• Q3: B - Equation applies to strong electrolytes.

Memory Aids

• Strong electrolytes: Λ_m vs C^1/2 is linear.
• Weak electrolytes: Λ_m increases sharply on dilution.
• Conductivity k decreases, conductance G increases on dilution.

Key Points

• Molar conductivity \( \Lambda_m \) is conductance of 1 mole electrolyte: \( \Lambda_m = \frac{k \times 1000}{M} \), units S m² mol^-1.
• Equivalent conductivity \( \Lambda_{eq} \) is conductance of 1 g equivalent: \( \Lambda_{eq} = \frac{k \times 1000}{N} \), units S m² eq^-1.
• On dilution:
  • Conductance \( G \) increases (more ions).
  • Conductivity \( k \) decreases (ion concentration per volume decreases).
  • Molar and equivalent conductance \( \Lambda_m, \Lambda_{eq} \) increase (volume effect).
• At infinite dilution, molar conductivity is limiting molar conductivity \( \Lambda_m^\circ \).
• For strong electrolytes, \( \Lambda_m = \Lambda_m^\circ - A C^{1/2} \) (Debye-Huckel Onsager equation).
• For weak electrolytes, \( A \) increases as concentration decreases; \( \Lambda_m^\circ \) not experimentally determinable.
• Strong electrolytes: small increase in conductance on dilution due to complete dissociation.
• Weak electrolytes: large increase in conductance on dilution due to increased ionization.

Solved Examples

Ex 1: Explain why molar conductivity of acetic acid increases sharply with dilution.

• Acetic acid is a weak electrolyte; dilution increases ionization, increasing molar conductivity.

Practice Set

• Q1: What happens to conductivity \( k \) on dilution?
  A) Increases
  B) Decreases
  C) Remains constant
  D) First increases then decreases
• Q2: Limiting molar conductivity is observed at?
  A) High concentration
  B) Zero concentration
  C) Moderate concentration
  D) Saturation point
• Q3: Debye-Huckel Onsager equation applies to?
  A) Weak electrolytes
  B) Strong electrolytes
  C) Insulators
  D) Semiconductors

Answer Key

• Q1: B - Conductivity decreases as ion concentration per volume decreases.
• Q2: B - Limiting molar conductivity is at infinite dilution (zero concentration).
• Q3: B - Equation applies to strong electrolytes.

Memory Aids

• Strong electrolytes: \( \Lambda_m \) vs \( C^{1/2} \) is linear.
• Weak electrolytes: \( \Lambda_m \) increases sharply on dilution.
• Conductivity \( k \) decreases, conductance \( G \) increases on dilution.

Kohlrausch Law of Independent Migration of Ions

Law and Formula

• Limiting molar conductivity Λm° = ν+λ+° + ν-λ-°, where λ+°, λ-° are limiting molar conductivities of cation and anion, ν+, ν- are their stoichiometric numbers.
• Example: Λm° (NaCl) = λNa+° + λCl-°.
• For weak electrolytes, degree of dissociation α = Λm/Λm°.
• Dissociation constant Ka = Cα2/(1-α) = CΛm2/[Λm°(Λm° - Λm)] .

Solved Examples

Ex 1: Calculate limiting molar conductivity of NaCl if λNa+° = 50 and λCl-° = 76 S cm2 mol-1.

• Λ_m^° = 50 + 76 = 126 S cm² mol^-1.

Practice Set

• Q1: Kohlrausch law relates to?
  A) Conductance of metals
  B) Limiting molar conductivity
  C) Resistance of insulators
  D) Temperature dependence of conductance
• Q2: Degree of dissociation α is?
  A) Λ_m^°/Λ_m
  B) Λ_m/Λ_m^°
  C) Λ_m + Λ_m^°
  D) Λ_m - Λ_m^°
• Q3: Dissociation constant K_a depends on?
  A) Concentration and α
  B) Only concentration
  C) Only temperature
  D) Only molar conductivity

Answer Key

• Q1: B - Kohlrausch law calculates limiting molar conductivity from ion contributions.
• Q2: B - Degree of dissociation α = Λ_m/Λ_m^°.
• Q3: A - K_a depends on concentration and degree of dissociation.

Memory Aids

• Limiting molar conductivity is sum of individual ion conductivities.
• Degree of dissociation α = Λ_m/Λ_m^°.
• Use Kohlrausch law to find K_a of weak electrolytes.

Law and Formula

• Limiting molar conductivity \( \Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ \), where \( \lambda_+^\circ, \lambda_-^\circ \) are limiting molar conductivities of cation and anion, \( \nu_+, \nu_- \) are their stoichiometric numbers.
• Example: \( \Lambda_m^\circ (NaCl) = \lambda_{Na^+}^\circ + \lambda_{Cl^-}^\circ \).
• For weak electrolytes, degree of dissociation \( \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \).
• Dissociation constant \( K_a = C \frac{\alpha^2}{1-\alpha} = C \frac{\Lambda_m^2}{\Lambda_m^\circ (\Lambda_m^\circ - \Lambda_m)} \).

Solved Examples

Ex 1: Calculate limiting molar conductivity of NaCl if \( \lambda_{Na^+}^\circ = 50 \) and \( \lambda_{Cl^-}^\circ = 76 \) S cm² mol^-1.

• \( \Lambda_m^\circ = 50 + 76 = 126 \) S cm² mol^-1.

Practice Set

• Q1: Kohlrausch law relates to?
  A) Conductance of metals
  B) Limiting molar conductivity
  C) Resistance of insulators
  D) Temperature dependence of conductance
• Q2: Degree of dissociation \( \alpha \) is?
  A) \( \frac{\Lambda_m^\circ}{\Lambda_m} \)
  B) \( \frac{\Lambda_m}{\Lambda_m^\circ} \)
  C) \( \Lambda_m + \Lambda_m^\circ \)
  D) \( \Lambda_m - \Lambda_m^\circ \)
• Q3: Dissociation constant \( K_a \) depends on?
  A) Concentration and \( \alpha \)
  B) Only concentration
  C) Only temperature
  D) Only molar conductivity

Answer Key

• Q1: B - Kohlrausch law calculates limiting molar conductivity from ion contributions.
• Q2: B - Degree of dissociation \( \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \).
• Q3: A - \( K_a \) depends on concentration and degree of dissociation.

Memory Aids

• Limiting molar conductivity is sum of individual ion conductivities.
• Degree of dissociation \( \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \).
• Use Kohlrausch law to find \( K_a \) of weak electrolytes.

Electrolysis

Explanation

• Electrolysis is the decomposition of an electrolyte by passing electric current through its aqueous solution or molten state.
• At electrodes, ions gain or lose electrons to form products.
• Example: Molten NaCl produces Na metal at cathode and Cl₂ gas at anode.

Solved Examples

• Ex 1: Electrolysis of molten NaCl
  • Cathode: Na⁺ + e^- → Na (metal)
  • Anode: 2Cl^- → Cl₂ + 2e^-

Practice Set

• Q1: What is produced at the cathode during electrolysis of aqueous NaCl?
  • A. Na metal
  • B. H₂ gas
  • C. Cl₂ gas
  • D. O₂ gas
• Q2: During electrolysis of dilute H₂SO₄, what forms at the anode?
  • A. Cl₂
  • B. O₂
  • C. H₂
  • D. SO₂
• Q3: In concentrated H₂SO₄ electrolysis, which gas is released at the anode?
  • A. O₂
  • B. SO₂
  • C. Cl₂
  • D. H₂

Answer Key

• Q1: B - H₂ gas forms at cathode due to reduction of H⁺ ions in aqueous NaCl.
• Q2: B - O₂ gas forms at anode by oxidation of water in dilute H₂SO₄.
• Q3: B - SO₂ gas is released at anode in concentrated H₂SO₄ due to oxidation of H₂SO₄.

Memory Aids

• Overvoltage causes Cl^- oxidation over H₂O in chloride solutions.
• Discharge potential: Ion with lower discharge potential discharges first at electrode.
• Remember: Molten NaCl → Na (cathode), Cl₂ (anode).

Faraday's Laws of Electrolysis

Explanation

• First Law: Amount of substance deposited or liberated at an electrode is proportional to the quantity of electricity passed, W ∝ Q.
• Mathematically, W = Z × Q = Z × I × t, where Z = electrochemical equivalent = Equivalent weight / 96500.
• Second Law: Amounts of different substances liberated by the same quantity of electricity are proportional to their equivalent weights.
• Mathematically, W₁/W₂ = E₁/E₂, where E = Atomic mass / number of electrons transferred.

Solved Examples

• Ex 1: Calculate mass of Cu deposited by 2A current for 30 minutes. (Eq. wt. Cu = 31.75)
  • Q = I × t = 2 × 1800 = 3600 C
  • W = Z × Q = (31.75/96500) × 3600 = 1.18 g

Practice Set

• Q1: According to Faraday's first law, the mass deposited is proportional to?
  • A. Voltage applied
  • B. Quantity of electricity passed
  • C. Time only
  • D. Current only
• Q2: If 96500 C deposits 1 g equivalent of a substance, what is Z for that substance?
  • A. 1
  • B. Equivalent weight
  • C. 1/96500
  • D. Equivalent weight/96500
• Q3: Two substances A and B are deposited by same charge. If W_A/W_B = 2/3, what is the ratio of their equivalent weights?
  • A. 3/2
  • B. 2/3
  • C. 1
  • D. 5/6

Answer Key

• Q1: B - Mass deposited is proportional to quantity of electricity passed (Q = I × t).
• Q2: D - Z = Equivalent weight / 96500.
• Q3: A - W₁/W₂ = E₁/E₂, so ratio of equivalent weights = 3/2.

Memory Aids

• Remember Z = Equivalent weight / 96500.
• Mass ∝ Charge passed (Q = I × t).
• Equivalent weight = Atomic mass / electrons transferred.

Overvoltage and Discharge Potential

Explanation

• Overvoltage: Extra potential needed to oxidize H₂O due to slow oxidation reaction.
• Causes preferential oxidation of Cl^- over H₂O at anode in chloride solutions.
• Discharge potential: Minimum potential required to discharge ions at electrodes.
• Determines which ion discharges first during electrolysis.

Solved Examples

• Ex 1: Explain why Cl^- oxidizes before H₂O in aqueous NaCl electrolysis.
  • Due to overvoltage, oxidation of H₂O is slow and requires extra potential.
  • Cl^- oxidizes at lower potential, so it discharges first.

Practice Set

• Q1: What is overvoltage?
  • A. Extra potential needed to reduce ions
  • B. Extra potential needed to oxidize H₂O
  • C. Minimum potential to start electrolysis
  • D. Potential difference between electrodes
• Q2: Discharge potential is the minimum potential to?
  • A. Start current flow
  • B. Discharge ions at electrodes
  • C. Heat the electrolyte
  • D. Break chemical bonds
• Q3: Why does Cl^- oxidize before H₂O in aqueous NaCl?
  • A. Cl^- has higher discharge potential
  • B. H₂O has overvoltage
  • C. Cl^- is more abundant
  • D. H₂O is reduced

Answer Key

• Q1: B - Overvoltage is extra potential needed to oxidize H₂O.
• Q2: B - Discharge potential is minimum potential to discharge ions at electrodes.
• Q3: B - Overvoltage makes H₂O oxidation slow, so Cl^- oxidizes first.

Memory Aids

• Overvoltage = extra potential for H₂O oxidation.
• Discharge potential decides ion discharge order.
• Cl^- oxidizes before H₂O due to overvoltage.

Electrolysis Products Table

Explanation

• Electrolysis products depend on electrolyte and electrode reactions.
• Molten NaCl: Na at cathode, Cl₂ at anode.
• Aqueous NaCl: H₂ at cathode, Cl₂ at anode (due to overvoltage).
• Dilute H₂SO₄: H₂ at cathode, O₂ at anode.
• Concentrated H₂SO₄: H₂ at cathode, SO₂ at anode.

Solved Examples

• Ex 1: Identify products of electrolysis of aqueous NaCl.
  • Cathode: 2H⁺ + 2e^- → H₂
  • Anode: 2Cl^- → Cl₂ + 2e^-

Practice Set

• Q1: What gas is released at anode during electrolysis of molten NaCl?
  • A. H₂
  • B. Cl₂
  • C. O₂
  • D. SO₂
• Q2: In dilute H₂SO₄ electrolysis, what forms at cathode?
  • A. H₂
  • B. O₂
  • C. SO₂
  • D. Cl₂
• Q3: Which product forms at anode in concentrated H₂SO₄ electrolysis?
  • A. O₂
  • B. SO₂
  • C. Cl₂
  • D. H₂

Answer Key

• Q1: B - Cl₂ gas is released at anode in molten NaCl electrolysis.
• Q2: A - H₂ gas forms at cathode in dilute H₂SO₄ electrolysis.
• Q3: B - SO₂ gas forms at anode in concentrated H₂SO₄ electrolysis.

Memory Aids

• Molten NaCl: Na (cathode), Cl₂ (anode).
• Aqueous NaCl: H₂ (cathode), Cl₂ (anode).
• Dilute H₂SO₄: H₂ (cathode), O₂ (anode).
• Concentrated H₂SO₄: H₂ (cathode), SO₂ (anode).

Batteries

Definition and Structure

• Battery is a 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.
  Answer: Dry cell is a primary cell because it cannot be recharged once exhausted.
• Ex 2: Name a rechargeable battery used in vehicles.
  Answer: Lead storage battery is a secondary cell used in vehicles.

Practice Set

• Q1: Which of the following is a primary cell?
  A. Nickel-cadmium cell
  B. Dry cell
  C. Lead storage battery
  D. Lithium-ion cell
• Q2: What happens when a secondary cell is exhausted?
  A. It is discarded
  B. It can be recharged and used again
  C. It explodes
  D. It loses all chemical properties permanently
• Q3: Which of the following batteries is rechargeable?
  A. Mercury cell
  B. Dry cell
  C. Nickel-cadmium cell
  D. Zinc-carbon cell

Answer Key

• Q1: B - Dry cell is a primary cell and cannot be recharged.
• Q2: B - Secondary cells can be recharged and reused multiple times.
• Q3: C - Nickel-cadmium cell is a rechargeable secondary cell.

Memory Aids

• Primary cells = "Use and throw" (e.g., Dry cell, Mercury cell).
• Secondary cells = "Recharge and reuse" (e.g., Nickel-cadmium, Lead storage battery).
• Remember: Lead storage battery is common in vehicles.

Fuel Cells

Definition and Components

• Fuel cells convert chemical energy of fossil fuel combustion directly into electrical energy.
• Example: Hydrogen-oxygen (H₂-O₂) fuel cell.
• Combustion in fuel cells is a redox reaction.
• Anode: Porous carbon with catalyst.
• Cathode: Porous carbon with catalyst.
• Electrolyte: Concentrated aqueous KOH or NaOH.

Reactions in Hydrogen-Oxygen Fuel Cell

• At anode: 2H₂(g) + 4OH^-(aq) → 4H₂O + 4e^-
• At cathode: O₂(g) + 2H₂O + 4e^- → 4OH^-(aq)
• Overall: 2H₂(g) + O₂(g) → 2H₂O(l)

Working Principle

• Hydrogen enters at the anode (negative side).
• Oxygen enters at the cathode (positive side).
• Electrolyte (aqueous NaOH) conducts ions.
• Hydrogen splits into protons and electrons at anode.
• Electrons flow through external circuit producing electric current.
• Protons move through electrolyte to cathode.
• Oxygen combines with electrons and protons to form water.
• Water is the only byproduct, exiting the cell.

Solved Examples

• Ex 1: Write the anode reaction in a hydrogen-oxygen fuel cell.
  Answer: 2H₂(g) + 4OH^-(aq) → 4H₂O + 4e^-
• Ex 2: Write the overall reaction of the hydrogen-oxygen fuel cell.
  Answer: 2H₂(g) + O₂(g) → 2H₂O(l)

Practice Set

• Q1: What is the electrolyte used in a hydrogen-oxygen fuel cell?
  A. Dilute H₂SO₄
  B. Concentrated KOH
  C. Dilute NaCl
  D. Concentrated HCl
• Q2: At the cathode of a hydrogen-oxygen fuel cell, which reaction occurs?
  A. 2H₂ → 4H⁺ + 4e^-
  B. O₂ + 4e^- + 2H₂O → 4OH^-
  C. H₂ + O₂ → H₂O
  D. 4OH^- → O₂ + 2H₂O + 4e^-
• Q3: What is the overall product formed in a hydrogen-oxygen fuel cell?
  A. H₂O₂
  B. H₂O
  C. O₂
  D. H₂

Answer Key

• Q1: B - Concentrated KOH is the electrolyte that conducts OH^- ions.
• Q2: B - Oxygen is reduced at cathode forming OH^- ions.
• Q3: B - Water (H₂O) is the main product of the fuel cell reaction.

Memory Aids

• Fuel cell reaction: Hydrogen + Oxygen → Water + Electricity.
• Electrolyte: Concentrated KOH/NaOH (alkaline medium).
• Anode: Hydrogen oxidation; Cathode: Oxygen reduction.

Commercially Important Electrochemical Cells

Overview and Reactions

• Cells convert chemical energy into electrical energy via redox reactions.
• Anode: Oxidation occurs; Cathode: Reduction occurs.
• Examples include dry cell, mercury cell, lead storage cell, nickel-cadmium cell, and fuel cell.
• Each cell has specific anode and cathode reactions affecting performance and use.

Solved Examples

• Ex 1: Identify the anode reaction in a dry cell.
  Answer: Zn → Zn²⁺ + 2e^-
• Ex 2: Identify the cathode reaction in a lead storage cell.
  Answer: PbO₂ + SO₄^2- + 4H⁺ + 2e^- → PbSO₄ + 2H₂O

Practice Set

• Q1: In which cell does Zn get oxidized?
  A. Mercury cell
  B. Dry cell
  C. Nickel-cadmium cell
  D. Fuel cell
• Q2: What is the cathode reaction in a mercury cell?
  A. HgO + H⁺ + e^- → Hg + H₂O
  B. Zn + 2OH^- → ZnO + H₂O + 2e^-
  C. PbO₂ + SO₄^2- + 4H⁺ + 2e^- → PbSO₄ + 2H₂O
  D. NiO(OH) + H₂O + e^- → Ni(OH)₂ + OH^-
• Q3: Which cell uses Ni-Cd electrodes?
  A. Dry cell
  B. Mercury cell
  C. Nickel-cadmium cell
  D. Lead storage cell

Answer Key

• Q1: B - Zinc is oxidized in dry cells.
• Q2: A - Mercury cell cathode reaction reduces HgO to Hg.
• Q3: C - Nickel-cadmium cell uses Ni-Cd electrodes.

Memory Aids

• Dry cell: Zn oxidized, carbon cathode.
• Mercury cell: HgO reduced at cathode.
• Lead storage cell: Pb/PbO₂ electrodes with H₂SO₄.
• Ni-Cd cell: Nickel oxide and cadmium electrodes.

Corrosion

Definition and Rusting

• Corrosion: Slow deterioration of metals exposed to atmosphere.
• Rusting of iron is an electrochemical process needing moisture and oxygen.

Electrochemical Reactions in Rusting

• Anode reaction: 2Fe_(s) 1rarr1 2Fe²⁺_(aq) + 4e^- (Eb0 = -0.44 V)
• Cathode reaction: O_2(g) + 4H⁺_(aq) + 4e^- 1rarr1 2H₂O_(l) (Eb0 = 1.23 V)
• Overall reaction: 2Fe_(s) + O_2(g) + 4H⁺_(aq) 1rarr1 2Fe²⁺_(aq) + 2H₂O_(l) (Eb0 = 1.67 V)

Prevention Methods

• Barrier protection: Coatings like paint, oil, or plastic to block air and moisture.
• Sacrificial protection: Using a more reactive metal (e.g., Zn) to protect iron.
• Anti-rust solutions: Chemicals that prevent rust formation.

Solved Examples

• Ex 1: Write the cathode reaction in rusting of iron.
  Answer: O_2(g) + 4H⁺_(aq) + 4e^- 1rarr1 2H₂O_(l)

Practice Set

• Q1: What is the anode reaction in iron rusting?
  A. 2Fe 1rarr1 2Fe²⁺ + 4e^-
  B. O₂ + 4H⁺ + 4e^- 1rarr1 2H₂O
  C. Fe³⁺ + e^- 1rarr1 Fe²⁺
  D. 2Fe³⁺ 1rarr1 2Fe
• Q2: Which method is NOT used to prevent corrosion?
  A. Barrier protection
  B. Sacrificial protection
  C. Electroplating with iron
  D. Anti-rust solutions
• Q3: Calculate the cell potential for rusting if Eb0 of Fe²⁺/Fe is -0.44 V and O₂/H⁺/H₂O is 1.23 V.
  A. 0.79 V
  B. 1.67 V
  C. 1.23 V
  D. -0.44 V

Answer Key

• Q1: A - Anode reaction involves iron oxidation to Fe²⁺.
• Q2: C - Electroplating with iron does not prevent corrosion; it may worsen it.
• Q3: B - Cell potential = Cathode Eb0 - Anode Eb0 = 1.23 - (-0.44) = 1.67 V.

Memory Aids

• Rusting needs moisture + oxygen.
• Zn protects Fe sacrificially (Zn is more reactive).
• Barrier = paint/oil; Anti-rust = chemicals.

Hydrogen Economy

Hydrogen as Energy Source

• Hydrogen is renewable and non-polluting fuel.
• Water splits into H₂ and O₂ using solar energy (photolysis).

Water Splitting Reaction

• H₂O (liquid) + hv (light energy) 1rarr1 H₂ (gas) + 1/2 O₂ (gas)
• Photolysis produces hydrogen fuel from water.

Hydrogen Combustion

• Hydrogen reacts with oxygen to form water.

Combustion Reaction

• H₂ (gas) + 1/2 O₂ (gas) 1rarr1 H₂O (liquid)
• Balanced equation shows water formation from hydrogen and oxygen.

Energy Production

• Both water splitting and hydrogen combustion are electrochemical processes.
• Hydrogen energy is economical compared to fossil fuels and nuclear energy.

Solved Examples

• Ex 1: Write the equation for water splitting using light.
  Answer: H₂O + hv 1rarr1 H₂ + 1/2 O₂
• Ex 2: Write the combustion reaction of hydrogen.
  Answer: H₂ + 1/2 O₂ 1rarr1 H₂O

Practice Set

• Q1: What is the product of hydrogen combustion?
  A. H₂O
  B. H₂O₂
  C. O₂
  D. CO₂
• Q2: Which energy source splits water into hydrogen?
  A. Thermal energy
  B. Solar energy
  C. Nuclear energy
  D. Mechanical energy
• Q3: Write the balanced equation for water splitting.
  A. H₂O 1rarr1 H₂ + O₂
  B. H₂O + hv 1rarr1 H₂ + 1/2 O₂
  C. 2H₂O 1rarr1 2H₂ + O₂
  D. H₂O + heat 1rarr1 H₂ + O₂

Answer Key

• Q1: A - Hydrogen combustion produces water (H₂O).
• Q2: B - Solar energy splits water into hydrogen and oxygen.
• Q3: B - Correct balanced photolysis equation with light energy.

Memory Aids

• Photolysis = water + light 1rarr1 H₂ + 1/2 O₂.
• Hydrogen combustion always forms water.
• Hydrogen energy is clean and renewable.