CLASS 12 CHEMISTRY, CHAPTER 2 – ELECTROCHEMISTRY

PART 1: INTRODUCTION & ELECTROCHEMICAL CELLS

ELECTROCHEMISTRY

Definition

Electrochemistry is the branch of chemistry that studies:

• Conversion of chemical energy → electrical energy
• Conversion of electrical energy → chemical energy

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Applications

• Manufacture of NaOH, Cl₂, F₂ etc.
• Batteries
• Fuel cells
• Electroplating
• Extraction of metals
• Corrosion studies

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Types of Electrochemical Cells

1. Galvanic (Voltaic) Cell

Definition

A galvanic cell converts chemical energy into electrical energy through a spontaneous redox reaction.

Important Points

• Reaction is spontaneous.
• Produces electricity.
• No external power source required.
• Gibbs free energy decreases.

Example

Daniell Cell

Cell notation:

Zn | Zn²⁺ || Cu²⁺ | Cu

Overall reaction

Zn + Cu²⁺ → Zn²⁺ + Cu

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2. Electrolytic Cell

Definition

An electrolytic cell converts electrical energy into chemical energy by carrying out a non-spontaneous reaction.

Important Points

• External battery required.
• Electrical energy is consumed.
• Used in electroplating and electrolysis.

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Difference Between Galvanic & Electrolytic Cell

Galvanic Cell	Electrolytic Cell
Chemical → Electrical energy	Electrical → Chemical energy
Spontaneous reaction	Non-spontaneous reaction
No external battery	Battery required
Produces electricity	Consumes electricity
ΔG < 0	ΔG > 0

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Daniell Cell

Construction

Consists of

• Zinc electrode dipped in ZnSO₄ solution
• Copper electrode dipped in CuSO₄ solution
• Salt bridge
• External wire with voltmeter

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Cell Representation

Zn | Zn²⁺ || Cu²⁺ | Cu

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Half Reactions

At Anode (Oxidation)

Zn → Zn²⁺ + 2e⁻

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At Cathode (Reduction)

Cu²⁺ + 2e⁻ → Cu

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Overall Reaction

Zn + Cu²⁺ → Zn²⁺ + Cu

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Salt Bridge

Definition

A U-shaped tube containing an inert electrolyte (KCl/KNO₃/NH₄NO₃ in agar gel) connecting the two half cells.

Functions

• Maintains electrical neutrality.
• Completes the circuit.
• Prevents mixing of solutions.
• Minimizes liquid junction potential.

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Electrode

A conductor through which electrons enter or leave the electrolyte.

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Anode

Definition

Electrode where oxidation occurs.

Remember

• Oxidation
• Electrons released
• Negative in galvanic cell

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Cathode

Definition

Electrode where reduction occurs.

Remember

• Reduction
• Electrons accepted
• Positive in galvanic cell

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Electron Flow

Electrons always move

Anode → Cathode

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Conventional Current

Current flows

Cathode → Anode

(Opposite to electron flow.)

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Electrode Potential

Definition

The potential difference developed between an electrode and the electrolyte when they are in contact.

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Standard Electrode Potential (E°)

Electrode potential measured under standard conditions:

• Temperature = 298 K
• Concentration = 1 M
• Gas pressure = 1 bar

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Standard Hydrogen Electrode (SHE)

Standard Conditions

• Platinum electrode
• Hydrogen gas at 1 bar
• H⁺ concentration = 1 M
• Temperature = 298 K

Standard Potential

E° = 0.00 V

Used as the reference electrode for measuring other electrode potentials.

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Important Formula

Cell EMF

E°cell = E°cathode − E°anode

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Cell Notation Rules

• Anode is written on the left
• Cathode is written on the right
• Single line (|) → Phase boundary
• Double line (||) → Salt bridge

Example:

Zn | Zn²⁺ || Cu²⁺ | Cu

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Quick Board Revision

✔ Oxidation → Anode

✔ Reduction → Cathode

✔ Electrons → Anode to Cathode

✔ Current → Cathode to Anode

✔ Salt bridge maintains neutrality

✔ SHE potential = 0.00 V

✔ E°cell = E°Cathode − E°Anode

PART 2: NERNST EQUATION, CELL EMF & GIBBS ENERGY

NERNST EQUATION

Definition

The Nernst equation is used to calculate the electrode potential or cell potential when the concentration of ions is not under standard conditions.

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General Electrode Reaction

$$M^{n+}+ne^- \rightarrow M(s)$$Mn++ne−→M(s)

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Nernst Equation (General)

$$E=E^\circ-\frac{RT}{nF}\ln Q$$E=E∘−nFRT​lnQ

Where,

Symbol	Meaning
E	Electrode potential under given conditions
E°	Standard electrode potential
R	Gas constant = 8.314 J K⁻¹ mol⁻¹
T	Temperature (K)
n	Number of electrons transferred
F	Faraday constant = 96487 C mol⁻¹
Q	Reaction quotient

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At 298 K

$$E=E^\circ-\frac{0.0591}{n}\log Q$$E=E∘−n0.0591​logQ

This is the formula used in CBSE Board exams.

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Nernst Equation for Cell EMF

For a general reaction$$aA+bB \rightarrow cC+dD$$aA+bB→cC+dD $$E_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log\frac{[C]^c[D]^d}{[A]^a[B]^b}$$Ecell​=Ecell∘​−n0.0591​log[A]a[B]b[C]c[D]d​

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Nernst Equation for Daniell Cell

Cell

Zn | Zn²⁺ || Cu²⁺ | Cu

Reaction

Zn + Cu²⁺ → Zn²⁺ + Cu

Formula$$E_{cell}=E^\circ_{cell}-\frac{0.0591}{2}\log\frac{[Zn^{2+}]}{[Cu^{2+}]}$$Ecell​=Ecell∘​−20.0591​log[Cu2+][Zn2+]​

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Important Observations

Cell potential increases when

• Cu²⁺ concentration increases
• Zn²⁺ concentration decreases

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Cell potential decreases when

• Cu²⁺ concentration decreases
• Zn²⁺ concentration increases

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Standard Cell Potential

Formula

$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$Ecell∘​=Ecathode∘​−Eanode∘​

Important Rule

Always subtract

Cathode − Anode

Never reverse the order.

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Relationship Between EMF and Equilibrium Constant

At equilibrium,

• Cell reaction stops.
• No current flows.
• Cell potential becomes zero.

Hence,$$E^\circ_{cell}=\frac{0.0591}{n}\log K$$Ecell∘​=n0.0591​logK

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Conclusions

If E°cell is positive

• K is very large.
• Reaction is spontaneous.
• Products are favoured.

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If E°cell is zero

• System is at equilibrium.

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If E°cell is negative

• K is very small.
• Reaction is non-spontaneous.
• Reactants are favoured.

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Gibbs Free Energy (ΔG)

Formula

$$\Delta G=-nFE_{cell}$$ΔG=−nFEcell​

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Standard Gibbs Free Energy

$$\Delta G^\circ=-nFE^\circ_{cell}$$ΔG∘=−nFEcell∘​

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Relationship Between ΔG° and Equilibrium Constant

$$\Delta G^\circ=-RT\ln K$$ΔG∘=−RTlnK

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Combined Relation

$$E^\circ_{cell}=\frac{2.303RT}{nF}\log K$$Ecell∘​=nF2.303RT​logK

At 298 K$$E^\circ_{cell}=\frac{0.0591}{n}\log K$$Ecell∘​=n0.0591​logK

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Important Formula Sheet (Learn As It Is)

Cell EMF

$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$Ecell∘​=Ecathode∘​−Eanode∘​

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Nernst Equation

$$E=E^\circ-\frac{0.0591}{n}\log Q$$E=E∘−n0.0591​logQ

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Gibbs Energy

$$\Delta G=-nFE$$ΔG=−nFE

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Standard Gibbs Energy

$$\Delta G^\circ=-nFE^\circ$$ΔG∘=−nFE∘

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Equilibrium Constant

$$E^\circ=\frac{0.0591}{n}\log K$$E∘=n0.0591​logK

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Gibbs–Equilibrium Relation

$$\Delta G^\circ=-RT\ln K$$ΔG∘=−RTlnK

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Standard Hydrogen Electrode (SHE) – Quick Facts

• Standard electrode potential = 0.00 V
• Platinum electrode is used.
• H₂ gas pressure = 1 bar
• H⁺ concentration = 1 M
• Temperature = 298 K

Used as the reference electrode to measure standard electrode potentials.

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Board Exam Tips

✔ Learn these formulas exactly

• $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$Ecell∘​=Ecathode∘​−Eanode∘​
• $E=E^\circ-\frac{0.0591}{n}\log Q$E=E∘−n0.0591​logQ
• $\Delta G=-nFE$ΔG=−nFE
• $\Delta G^\circ=-nFE^\circ$ΔG∘=−nFE∘
• $E^\circ=\frac{0.0591}{n}\log K$E∘=n0.0591​logK

✔ Frequently Asked Numerical Topics

1. Calculate cell EMF using the Nernst equation.
2. Calculate equilibrium constant from $E^\circ_{cell}$Ecell∘​.
3. Calculate $\Delta G^\circ$ΔG∘ from $E^\circ_{cell}$Ecell∘​.
4. Calculate $E_{cell}$Ecell​ for the Daniell cell at non-standard concentrations.

PART 3: CONDUCTANCE OF ELECTROLYTIC SOLUTIONS

CONDUCTANCE OF ELECTROLYTIC SOLUTIONS

Definition

Conductance is the ability of a substance to allow the flow of electric current through it.

In electrolytic solutions, current is carried by ions.

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Resistance (R)

Definition

Resistance is the opposition offered to the flow of electric current.

Formula

$$R=\rho\frac{l}{A}$$R=ρAl​

Where

• R = Resistance (Ω)
• ρ = Resistivity
• l = Length of conductor
• A = Area of cross-section

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Unit

Ohm (Ω)

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Resistivity (ρ)

Definition

Resistance of a conductor having

• Length = 1 m
• Cross-sectional area = 1 m²

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Formula

$$\rho=\frac{RA}{l}$$ρ=lRA​

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SI Unit

Ω m

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Conductance (G)

Definition

Conductance is the reciprocal of resistance.

Formula

$$G=\frac{1}{R}$$G=R1​

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Unit

Siemens (S)

or

Ω⁻¹ (Mho)

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Conductivity (κ)

Definition

Conductivity is the conductance of a solution kept between two electrodes 1 m apart having 1 m² cross-sectional area.

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Formula

$$\kappa=\frac{1}{\rho}$$κ=ρ1​

or$$\kappa=G\frac{l}{A}$$κ=GAl​

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SI Unit

S m⁻¹

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Relation Among Quantities

Quantity	Formula
Resistance	$R=\rho\frac{l}{A}$R=ρAl​
Resistivity	$\rho=\frac{RA}{l}$ρ=lRA​
Conductance	$G=\frac1R$G=R1​
Conductivity	$\kappa=\frac1\rho$κ=ρ1​

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Conductivity Depends On

1. Nature of electrolyte
2. Size of ions
3. Solvent
4. Concentration
5. Temperature

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Metallic Conductance vs Electrolytic Conductance

Metallic Conductance	Electrolytic Conductance
Due to electrons	Due to ions
No chemical change	Chemical change may occur
In metals	In electrolyte solutions
Decreases with temperature	Increases with temperature

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Measurement of Conductivity

Conductivity is measured using a Conductivity Cell.

Construction

• Two platinum electrodes
• Connected to AC source
• Filled with electrolyte solution

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Why AC is Used Instead of DC?

If DC is used,

• Electrolysis occurs.
• Composition of solution changes.
• Accurate measurement becomes impossible.

Hence Alternating Current (AC) is used.

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Cell Constant

Formula

$$\text{Cell Constant}=\frac{l}{A}$$Cell Constant=Al​

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Unit

m⁻¹

or

cm⁻¹

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Relation Between Cell Constant & Conductivity

$$\kappa=\frac{\text{Cell Constant}}{R}$$κ=RCell Constant​

or$$\kappa=G\times\text{Cell Constant}$$κ=G×Cell Constant

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Molar Conductivity (Λₘ)

Definition

Conductance of the volume of solution containing 1 mole of electrolyte placed between electrodes 1 m apart.

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Formula

$$\Lambda_m=\frac{\kappa}{c}$$Λm​=cκ​

Where

• κ = Conductivity
• c = Concentration (mol m⁻³)

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Practical Formula

If κ is in S cm⁻¹ and concentration is in mol L⁻¹$$\Lambda_m=\frac{\kappa\times1000}{M}$$Λm​=Mκ×1000​

Where

M = Molarity

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Units

SI Unit

S m² mol⁻¹

Common Unit

S cm² mol⁻¹

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Variation of Conductivity with Concentration

On Dilution

Conductivity decreases

Reason

Number of ions per unit volume decreases.

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Variation of Molar Conductivity with Concentration

On Dilution

Molar conductivity increases

Reason

Although ions become fewer per unit volume, each ion moves more freely because interionic attraction decreases.

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Strong Electrolytes

Examples

• HCl
• NaCl
• KCl
• NaOH

Characteristics

• Almost completely ionised.
• High conductivity.
• Λₘ increases slowly with dilution.

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Formula

$$\Lambda_m=\Lambda_m^\circ-A\sqrt{c}$$Λm​=Λm∘​−Ac​

Where

• Λ°ₘ = Limiting molar conductivity
• A = Constant
• c = Concentration

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Weak Electrolytes

Examples

• CH₃COOH
• NH₄OH

Characteristics

• Partially ionised.
• Λₘ increases rapidly on dilution because the degree of ionisation increases.

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Limiting Molar Conductivity (Λ°ₘ)

Definition

Molar conductivity at infinite dilution (c → 0).

At infinite dilution,

• Interionic attraction is negligible.
• Electrolyte is completely dissociated.

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Comparison

Property	Conductivity	Molar Conductivity
Symbol	κ	Λₘ
Depends on concentration	Yes	Yes
Effect of dilution	Decreases	Increases
Unit	S m⁻¹	S m² mol⁻¹

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Formula Sheet (Learn These)

Resistance

$$R=\rho\frac{l}{A}$$R=ρAl​

Resistivity

$$\rho=\frac{RA}{l}$$ρ=lRA​

Conductance

$$G=\frac1R$$G=R1​

Conductivity

$$\kappa=\frac1\rho$$κ=ρ1​

Cell Constant

$$\frac{l}{A}$$Al​

Conductivity

$$\kappa=\frac{\text{Cell Constant}}{R}$$κ=RCell Constant​

Molar Conductivity

$$\Lambda_m=\frac{\kappa}{c}$$Λm​=cκ​

or$$\Lambda_m=\frac{\kappa\times1000}{M}$$Λm​=Mκ×1000​

Strong Electrolyte

$$\Lambda_m=\Lambda_m^\circ-A\sqrt{c}$$Λm​=Λm∘​−Ac​

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Board Exam Points

✔ Conductivity decreases on dilution.

✔ Molar conductivity increases on dilution.

✔ Strong electrolytes are almost completely dissociated.

✔ Weak electrolytes show a sharp increase in molar conductivity on dilution due to increased ionisation.

✔ AC is used in conductivity measurements to prevent electrolysis and changes in solution composition.

PART 4: KOHLRAUSCH’S LAW & ITS APPLICATIONS

KOHLRAUSCH’S LAW OF INDEPENDENT MIGRATION OF IONS

Statement

Kohlrausch’s law states that at infinite dilution, each ion contributes independently to the total molar conductivity of an electrolyte.

In other words:

• At very high dilution, ions move independently.
• The contribution of each ion remains constant.
• Total molar conductivity is the sum of contributions of individual ions.

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Mathematical Expression

For an electrolyte:$$A_xB_y \rightarrow xA^{+}+yB^{-}$$Ax​By​→xA++yB−

The limiting molar conductivity is:$$\Lambda_m^\circ=x\lambda^\circ_A+y\lambda^\circ_B$$Λm∘​=xλA∘​+yλB∘​

Where:

• $\Lambda_m^\circ$Λm∘​ = Limiting molar conductivity
• $\lambda^\circ_A$λA∘​ = Ionic conductivity of cation
• $\lambda^\circ_B$λB∘​ = Ionic conductivity of anion

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Example

For KCl:$$KCl \rightarrow K^+ + Cl^-$$KCl→K++Cl−

Therefore,$$\Lambda_m^\circ(KCl)=\lambda^\circ_{K^+}+\lambda^\circ_{Cl^-}$$Λm∘​(KCl)=λK+∘​+λCl−∘​

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For MgCl₂

$$MgCl_2 \rightarrow Mg^{2+}+2Cl^-$$MgCl2​→Mg2++2Cl−

Therefore,$$\Lambda_m^\circ(MgCl_2)=\lambda^\circ_{Mg^{2+}}+2\lambda^\circ_{Cl^-}$$Λm∘​(MgCl2​)=λMg2+∘​+2λCl−∘​

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Applications of Kohlrausch’s Law

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1. Calculation of Limiting Molar Conductivity of Weak Electrolytes

Weak electrolytes cannot be directly measured at infinite dilution because they are not completely ionised at ordinary concentrations.

Example:

Acetic acid (CH₃COOH)

Its limiting molar conductivity can be calculated using strong electrolytes.

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Example

For CH₃COOH:$$\Lambda_m^\circ(CH_3COOH)$$Λm∘​(CH3​COOH)

can be calculated as:$$\Lambda_m^\circ(CH_3COOH) = \Lambda_m^\circ(CH_3COONa) + \Lambda_m^\circ(HCl) – \Lambda_m^\circ(NaCl)$$Λm∘​(CH3​COOH)=Λm∘​(CH3​COONa)+Λm∘​(HCl)−Λm∘​(NaCl)

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Reason

Because:

CH₃COONa provides:$$CH_3COO^- + Na^+$$CH3​COO−+Na+

HCl provides:$$H^+ + Cl^-$$H++Cl−

NaCl provides:$$Na^+ + Cl^-$$Na++Cl−

Na⁺ and Cl⁻ cancel, leaving:$$H^+ + CH_3COO^-$$H++CH3​COO−

which gives CH₃COOH.

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2. Calculation of Degree of Dissociation

For weak electrolytes:$$\alpha=\frac{\Lambda_m}{\Lambda_m^\circ}$$α=Λm∘​Λm​​

Where:

• α = Degree of dissociation
• Λₘ = Molar conductivity at given concentration
• Λ°ₘ = Limiting molar conductivity

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3. Calculation of Dissociation Constant

For weak electrolyte:$$K_a=\frac{c\alpha^2}{1-\alpha}$$Ka​=1−αcα2​

Where:

• Ka = Dissociation constant
• c = Concentration
• α = Degree of dissociation

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4. Calculation of Solubility of Sparingly Soluble Salts

Examples:

• AgCl
• BaSO₄
• PbSO₄

These salts have very low solubility and their conductivity can be used to calculate solubility.

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Relation Between Solubility and Conductivity

For a sparingly soluble salt:$$S=\frac{1000\kappa}{\Lambda_m^\circ}$$S=Λm∘​1000κ​

Where:

• S = Solubility (mol L⁻¹)
• κ = Conductivity
• Λ°ₘ = Limiting molar conductivity

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Ionic Conductivity

Definition

The contribution of an individual ion towards the total conductivity of an electrolyte solution.

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Factors Affecting Ionic Conductivity

1. Size of Ion

• Smaller ions generally move faster.
• Hydrated ions may behave differently.

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2. Charge on Ion

Higher charge increases attraction with solvent molecules and affects mobility.

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3. Temperature

Increase in temperature:

• Decreases viscosity.
• Increases ionic movement.
• Increases conductivity.

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Important Ionic Conductivity Order

Hydrogen ion and hydroxide ion have exceptionally high conductivity due to their special movement mechanism.

Approximate order:$$H^+ > K^+ > Na^+$$H+>K+>Na+

and$$OH^- > Cl^-$$OH−>Cl−

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Strong vs Weak Electrolytes (Revision)

Strong Electrolytes	Weak Electrolytes
Completely ionised	Partially ionised
High conductivity	Low conductivity
Small increase in Λₘ on dilution	Large increase in Λₘ on dilution
Example: NaCl, HCl	Example: CH₃COOH

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Important Formula Sheet

Kohlrausch’s Law

$$\Lambda_m^\circ=\lambda^\circ_+ + \lambda^\circ_-$$Λm∘​=λ+∘​+λ−∘​

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Degree of Dissociation

$$\alpha=\frac{\Lambda_m}{\Lambda_m^\circ}$$α=Λm∘​Λm​​

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Dissociation Constant

$$K_a=\frac{c\alpha^2}{1-\alpha}$$Ka​=1−αcα2​

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Solubility of Sparingly Soluble Salt

$$S=\frac{1000\kappa}{\Lambda_m^\circ}$$S=Λm∘​1000κ​

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Board Exam Important Points

✔ Kohlrausch’s law is valid at infinite dilution.

✔ Every ion contributes independently to limiting molar conductivity.

✔ Weak electrolytes require Kohlrausch’s law to calculate Λ°ₘ.

✔ Degree of dissociation increases with dilution.

✔ Ionic mobility increases with temperature.

✔ Hydrogen ion has the highest ionic conductivity among common ions.

PART 5: ELECTROLYSIS & FARADAY’S LAWS OF ELECTROLYSIS

ELECTROLYSIS

Definition

Electrolysis is the process in which electrical energy is used to bring about a non-spontaneous chemical reaction.

Example:

Electrolysis of molten NaCl:$$NaCl(l) \rightarrow Na^+ + Cl^-$$NaCl(l)→Na++Cl−

At cathode:$$Na^+ + e^- \rightarrow Na$$Na++e−→Na

At anode:$$2Cl^- \rightarrow Cl_2 + 2e^-$$2Cl−→Cl2​+2e−

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Electrolytic Cell

Components

An electrolytic cell contains:

1. Electrolyte
2. Two electrodes
3. External power source

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Electrodes in Electrolytic Cell

Cathode

• Connected to negative terminal of battery.
• Reduction occurs.
• Positive ions move towards cathode.

Example:$$Cu^{2+}+2e^- \rightarrow Cu$$Cu2++2e−→Cu

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Anode

• Connected to positive terminal of battery.
• Oxidation occurs.
• Negative ions move towards anode.

Example:$$2Cl^- \rightarrow Cl_2+2e^-$$2Cl−→Cl2​+2e−

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Important Difference

Cathode	Anode
Negative electrode	Positive electrode
Reduction occurs	Oxidation occurs
Cations move towards it	Anions move towards it

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Electrolysis of Aqueous Solutions

When an aqueous electrolyte is electrolysed, water may also participate in the reaction.

The products depend on:

• Nature of ions
• Electrode material
• Concentration of ions

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Electrolysis of Aqueous NaCl

Solution contains:$$Na^+, Cl^-, H^+, OH^-$$Na+,Cl−,H+,OH−

At Cathode:

Water is reduced:$$2H_2O+2e^- \rightarrow H_2+2OH^-$$2H2​O+2e−→H2​+2OH−

Hydrogen gas is produced.

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At Anode:

Chloride ions are oxidised:$$2Cl^- \rightarrow Cl_2+2e^-$$2Cl−→Cl2​+2e−

Chlorine gas is produced.

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Overall Reaction:

$$2NaCl+2H_2O\rightarrow2NaOH+H_2+Cl_2$$2NaCl+2H2​O→2NaOH+H2​+Cl2​

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Electrolysis of Copper Sulphate Solution

With Copper Electrodes

Cathode:

$$Cu^{2+}+2e^- \rightarrow Cu$$Cu2++2e−→Cu

Copper deposits on cathode.

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Anode:

$$Cu\rightarrow Cu^{2+}+2e^-$$Cu→Cu2++2e−

Copper dissolves from anode.

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Result:

• Concentration of Cu²⁺ remains constant.
• Copper is transferred from anode to cathode.

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With Platinum Electrodes

Cathode:

$$Cu^{2+}+2e^- \rightarrow Cu$$Cu2++2e−→Cu

Copper deposits.

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Anode:

Water is oxidised:$$2H_2O\rightarrow O_2+4H^++4e^-$$2H2​O→O2​+4H++4e−

Oxygen gas is released.

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FARADAY’S LAWS OF ELECTROLYSIS

Faraday’s First Law

Statement

The mass of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.$$m \propto Q$$m∝Q

Since,$$Q=It$$Q=It

Therefore,$$m\propto It$$m∝It

or$$m=ZIt$$m=ZIt

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Where:

• m = Mass deposited
• I = Current
• t = Time
• Z = Electrochemical equivalent

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Electrochemical Equivalent (Z)

Definition

The mass of substance deposited by passing 1 coulomb of electricity.$$Z=\frac{m}{Q}$$Z=Qm​

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Faraday’s Second Law

Statement

When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their equivalent masses.$$m\propto Equivalent\ mass$$m∝Equivalent mass

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Equivalent Mass

$$Equivalent\ mass=\frac{Molar\ mass}{n}$$Equivalent mass=nMolar mass​

Where:

• Molar mass = Atomic/Molecular mass
• n = Number of electrons involved

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Combined Faraday Equation

$$m=\frac{MIt}{nF}$$m=nFMIt​

Where:

Symbol	Meaning
m	Mass deposited
M	Molar mass
I	Current
t	Time
n	Number of electrons
F	Faraday constant

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Faraday Constant

$$F=96500 \ C\ mol^{-1}$$F=96500 C mol−1

Meaning:

One mole of electrons carries 96500 coulombs of charge.

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Important Numerical Formula

Charge Passed

$$Q=It$$Q=It

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Mass Deposited

$$m=\frac{MIt}{nF}$$m=nFMIt​

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Number of Moles of Electrons

$$\text{Moles of electrons}=\frac{Q}{F}$$Moles of electrons=FQ​

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Electroplating

Definition

The process of depositing a thin layer of one metal over another metal using electrolysis.

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Uses of Electroplating

• Protection from corrosion
• Improving appearance
• Increasing durability

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Examples

Chromium plating

Used on:

• Automobile parts
• Bathroom fittings

Silver plating

Used on:

• Cutlery
• Decorative items

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Important Board Points

✔ Electrolysis converts electrical energy into chemical energy.

✔ In electrolytic cells:

• Cathode = negative
• Anode = positive

✔ Reduction always occurs at cathode.

✔ Oxidation always occurs at anode.

✔ Faraday’s first law: mass ∝ charge passed.

✔ Faraday’s second law: deposited mass depends on equivalent mass.

✔ 1 Faraday = 96500 C mol⁻¹.

✔ Formula for electrochemical deposition:$$m=\frac{MIt}{nF}$$m=nFMIt​

PART 6: BATTERIES, FUEL CELLS & CORROSION

BATTERIES

Definition

A battery is an electrochemical device that converts chemical energy into electrical energy.

A battery consists of one or more electrochemical cells connected together.

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Types of Batteries

Batteries are mainly of two types:

1. Primary Batteries
2. Secondary Batteries

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1. Primary Batteries

Definition

Primary batteries are batteries in which the chemical reactions are irreversible.

They cannot be recharged after use.

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Examples

• Dry cell
• Mercury cell

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Dry Cell (Leclanché Cell)

Construction

A dry cell contains:

• Zinc container → acts as anode
• Carbon rod → acts as cathode
• Paste of NH₄Cl and ZnCl₂ → electrolyte
• MnO₂ → depolariser

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Working

At Anode:

Zinc undergoes oxidation:$$Zn \rightarrow Zn^{2+}+2e^-$$Zn→Zn2++2e−

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At Cathode:

Reduction of MnO₂ occurs.

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Features

• Portable
• Low cost
• Used in torches, radios and toys

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Mercury Cell

Construction

Contains:

• Zinc–mercury amalgam as anode
• Mercury oxide as cathode
• KOH or NaOH as electrolyte

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Electrode Reactions

Anode:

$$Zn(Hg)+2OH^- \rightarrow ZnO+H_2O+2e^-$$Zn(Hg)+2OH−→ZnO+H2​O+2e−

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Cathode:

$$HgO+H_2O+2e^- \rightarrow Hg+2OH^-$$HgO+H2​O+2e−→Hg+2OH−

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Overall Reaction:

$$Zn(Hg)+HgO \rightarrow ZnO+Hg$$Zn(Hg)+HgO→ZnO+Hg

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Advantages

• Constant voltage output
• Small size
• Used in watches and hearing aids

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2. Secondary Batteries

Definition

Secondary batteries are rechargeable batteries.

The reactions are reversible, so they can be used repeatedly.

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Examples

• Lead storage battery
• Lithium-ion battery

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Lead Storage Battery

Uses

• Automobiles
• Inverters
• Backup power systems

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Construction

Contains:

• Lead plates → anode
• Lead dioxide plates → cathode
• Sulphuric acid solution → electrolyte

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During Discharge

At Anode:

Lead is oxidised:$$Pb+SO_4^{2-}\rightarrow PbSO_4+2e^-$$Pb+SO42−​→PbSO4​+2e−

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At Cathode:

Lead dioxide is reduced:$$PbO_2+4H^++SO_4^{2-}+2e^- \rightarrow PbSO_4+2H_2O$$PbO2​+4H++SO42−​+2e−→PbSO4​+2H2​O

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Overall Reaction:

$$Pb+PbO_2+2H_2SO_4 \rightarrow2PbSO_4+2H_2O$$Pb+PbO2​+2H2​SO4​→2PbSO4​+2H2​O

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Charging of Lead Battery

During charging:

• The above reaction is reversed.
• PbSO₄ converts back into Pb and PbO₂.

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Lithium-Ion Battery

Features

• Rechargeable battery
• High energy density
• Lightweight
• Long life

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Uses

• Mobile phones
• Laptops
• Electric vehicles

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FUEL CELLS

Definition

A fuel cell is an electrochemical cell that converts the energy of a fuel directly into electrical energy.

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Hydrogen-Oxygen Fuel Cell

The most common fuel cell uses:

• Hydrogen as fuel
• Oxygen as oxidising agent

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Construction

Contains:

• Porous carbon electrodes
• Electrolyte solution
• Continuous supply of H₂ and O₂ gases

---

Electrode Reactions

At Anode:

Hydrogen is oxidised:$$2H_2+4OH^- \rightarrow4H_2O+4e^-$$2H2​+4OH−→4H2​O+4e−

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At Cathode:

Oxygen is reduced:$$O_2+2H_2O+4e^- \rightarrow4OH^-$$O2​+2H2​O+4e−→4OH−

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Overall Reaction:

$$2H_2+O_2\rightarrow2H_2O$$2H2​+O2​→2H2​O

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Advantages of Fuel Cells

✔ High efficiency

✔ Environment friendly

✔ Water is the only product in hydrogen fuel cells

✔ Continuous production of electricity possible

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CORROSION

Definition

Corrosion is the slow deterioration of a metal due to chemical or electrochemical reactions with the environment.

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Example

Rusting of iron.

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Electrochemical Theory of Rusting

Rusting occurs due to formation of tiny electrochemical cells on the iron surface.

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Steps in Rust Formation

1. Oxidation of Iron (Anode)

$$Fe\rightarrow Fe^{2+}+2e^-$$Fe→Fe2++2e−

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2. Reduction of Oxygen (Cathode)

$$O_2+2H_2O+4e^- \rightarrow4OH^-$$O2​+2H2​O+4e−→4OH−

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3. Formation of Rust

Iron ions react with hydroxide ions:$$Fe^{2+}+2OH^-\rightarrow Fe(OH)_2$$Fe2++2OH−→Fe(OH)2​

Further oxidation produces hydrated ferric oxide:$$Fe_2O_3.xH_2O$$Fe2​O3​.xH2​O

This is called rust.

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Factors Affecting Corrosion

1. Moisture

More moisture increases corrosion.

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2. Presence of Electrolytes

Salt water increases corrosion rate.

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3. Temperature

Higher temperature generally increases corrosion.

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4. Impurities

Impurities create electrochemical cells and increase corrosion.

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Prevention of Corrosion

1. Painting

Forms a protective layer over metal.

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2. Oiling and Greasing

Prevents contact with air and moisture.

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3. Galvanisation

Coating iron with zinc.

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4. Electroplating

Depositing a protective metal layer.

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5. Alloy Formation

Making corrosion-resistant alloys.

Example:

Stainless steel

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Important Revision Table

Device	Type	Energy Conversion
Galvanic Cell	Electrochemical cell	Chemical → Electrical
Electrolytic Cell	Electrochemical cell	Electrical → Chemical
Battery	Electrochemical device	Chemical → Electrical
Fuel Cell	Electrochemical device	Fuel energy → Electrical

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Chapter Formula Revision

Nernst Equation

$$E=E^\circ-\frac{0.0591}{n}\log Q$$E=E∘−n0.0591​logQ

Gibbs Energy

$$\Delta G=-nFE$$ΔG=−nFE

Equilibrium Relation

$$E^\circ=\frac{0.0591}{n}\log K$$E∘=n0.0591​logK

Faraday Law

$$m=\frac{MIt}{nF}$$m=nFMIt​

Conductivity

$$\kappa=\frac{1}{\rho}$$κ=ρ1​

Molar Conductivity

$$\Lambda_m=\frac{\kappa\times1000}{M}$$Λm​=Mκ×1000​

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FINAL BOARD REVISION POINTS

✔ Primary batteries cannot be recharged.

✔ Secondary batteries are rechargeable.

✔ Fuel cells produce electricity continuously as long as fuel is supplied.

✔ Hydrogen-oxygen fuel cell produces water as the only product.

✔ Rust is hydrated ferric oxide.

✔ Corrosion is an electrochemical process.

✔ Galvanisation protects iron by coating it with zinc.

PART 7: IMPORTANT NUMERICAL FORMULAS + BOARD EXAM QUESTIONS

ELECTROCHEMISTRY FORMULA SHEET

1. Resistance

$$R=\rho\frac{l}{A}$$R=ρAl​

Where:

• R = Resistance
• ρ = Resistivity
• l = Length of conductor
• A = Area of cross-section

Unit:

Ω

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2. Conductance

$$G=\frac{1}{R}$$G=R1​

Unit:

Siemens (S)

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3. Conductivity

$$\kappa=\frac{1}{\rho}$$κ=ρ1​

or$$\kappa=G\frac{l}{A}$$κ=GAl​

Unit:

S m⁻¹

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4. Cell Constant

$$\text{Cell Constant}=\frac{l}{A}$$Cell Constant=Al​

Unit:

m⁻¹

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5. Conductivity Using Cell Constant

$$\kappa=\frac{\text{Cell Constant}}{R}$$κ=RCell Constant​

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6. Molar Conductivity

$$\Lambda_m=\frac{\kappa\times1000}{M}$$Λm​=Mκ×1000​

Where:

• Λₘ = Molar conductivity
• κ = Conductivity
• M = Molarity

Unit:

S cm² mol⁻¹

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7. Nernst Equation

At 298 K:$$E=E^\circ-\frac{0.0591}{n}\log Q$$E=E∘−n0.0591​logQ

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8. Cell Potential

$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$Ecell∘​=Ecathode∘​−Eanode∘​

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9. Gibbs Energy

$$\Delta G=-nFE$$ΔG=−nFE

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10. Equilibrium Constant

$$E^\circ_{cell}=\frac{0.0591}{n}\log K$$Ecell∘​=n0.0591​logK

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11. Faraday Law

$$m=\frac{MIt}{nF}$$m=nFMIt​

Where:

• m = Mass deposited
• M = Molar mass
• I = Current
• t = Time
• n = Number of electrons
• F = 96500 C mol⁻¹

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IMPORTANT BOARD QUESTIONS

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Q1. Define electrochemistry.

Answer:

Electrochemistry is the branch of chemistry that deals with the conversion of chemical energy into electrical energy and electrical energy into chemical energy.

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Q2. What is the function of a salt bridge?

Answer:

Functions of salt bridge:

1. Maintains electrical neutrality of both solutions.
2. Completes the electrical circuit.
3. Prevents direct mixing of electrolytes.
4. Minimises liquid junction potential.

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Q3. Why is a salt bridge filled with KCl generally used?

Answer:

KCl is used because:

• K⁺ and Cl⁻ ions have almost similar ionic mobility.
• They do not usually react with most electrolytes.

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Q4. Write the electrode reactions of Daniell cell.

Anode:

$$Zn\rightarrow Zn^{2+}+2e^-$$Zn→Zn2++2e−

Cathode:

$$Cu^{2+}+2e^-\rightarrow Cu$$Cu2++2e−→Cu

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Q5. State the Nernst equation.

Answer:

The Nernst equation relates electrode potential with concentration of ions.

At 298 K:$$E=E^\circ-\frac{0.0591}{n}\log Q$$E=E∘−n0.0591​logQ

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Q6. What happens to conductivity on dilution?

Answer:

Conductivity decreases on dilution because the number of ions per unit volume decreases.

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Q7. What happens to molar conductivity on dilution?

Answer:

Molar conductivity increases on dilution because:

• Ion mobility increases.
• Interionic attraction decreases.
• Degree of ionisation increases (especially for weak electrolytes).

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Q8. State Kohlrausch’s law.

Answer:

At infinite dilution, each ion contributes independently to the total molar conductivity of an electrolyte.

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Q9. Why does hydrogen ion have high ionic conductivity?

Answer:

Hydrogen ion has high conductivity because it moves through the solution by a special proton transfer mechanism.

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Q10. State Faraday’s first law of electrolysis.

Answer:

The mass of a substance deposited during electrolysis is directly proportional to the quantity of electricity passed.$$m\propto Q$$m∝Q

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IMPORTANT NUMERICAL TYPES

Type 1: Calculate Cell EMF

Given:

• E°cathode
• E°anode

Formula:$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$Ecell∘​=Ecathode∘​−Eanode∘​

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Type 2: Calculate EMF Using Nernst Equation

Steps:

1. Write balanced reaction.
2. Find number of electrons (n).
3. Calculate reaction quotient (Q).
4. Apply:

$$E=E^\circ-\frac{0.0591}{n}\log Q$$E=E∘−n0.0591​logQ

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Type 3: Find ΔG°

Formula:$$\Delta G^\circ=-nFE^\circ_{cell}$$ΔG∘=−nFEcell∘​

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Type 4: Find Equilibrium Constant

Formula:$$E^\circ_{cell}=\frac{0.0591}{n}\log K$$Ecell∘​=n0.0591​logK

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Type 5: Electrolysis Calculation

Formula:$$m=\frac{MIt}{nF}$$m=nFMIt​

Steps:

1. Find charge:

$$Q=It$$Q=It

2. Find electrons:

$$\text{Moles of electrons}=\frac{Q}{F}$$Moles of electrons=FQ​

3. Calculate deposited mass.

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MOST IMPORTANT ONE-LINERS FOR EXAM

⭐ Oxidation always occurs at anode.

⭐ Reduction always occurs at cathode.

⭐ Electrons flow from anode to cathode.

⭐ Salt bridge maintains electrical neutrality.

⭐ E° of SHE = 0.00 V.

⭐ Positive E°cell indicates spontaneous reaction.

⭐ Primary batteries are not rechargeable.

⭐ Secondary batteries are rechargeable.

⭐ Fuel cells work continuously when fuel is supplied.

⭐ Rust is hydrated ferric oxide (Fe₂O₃·xH₂O).

PART 8: IMPORTANT NUMERICALS WITH SOLUTIONS

NUMERICAL 1: Calculate Standard Cell Potential

Question:

Given:$$E^\circ_{Cu^{2+}/Cu}=+0.34V$$ECu2+/Cu∘​=+0.34V $$E^\circ_{Zn^{2+}/Zn}=-0.76V$$EZn2+/Zn∘​=−0.76V

Calculate $E^\circ_{cell}$Ecell∘​.

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Solution:

Formula:$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$Ecell∘​=Ecathode∘​−Eanode∘​

Cathode = Cu

Anode = Zn$$E^\circ_{cell}=0.34-(-0.76)$$Ecell∘​=0.34−(−0.76) $$E^\circ_{cell}=1.10V$$Ecell∘​=1.10V

Answer:

$$\boxed{E^\circ_{cell}=1.10V}$$Ecell∘​=1.10V​

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NUMERICAL 2: Calculate Gibbs Energy

Question:

For a cell:$$E^\circ_{cell}=1.10V$$Ecell∘​=1.10V

Number of electrons:$$n=2$$n=2

Find ΔG°.

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Solution:

Formula:$$\Delta G^\circ=-nFE^\circ_{cell}$$ΔG∘=−nFEcell∘​

Given:

F = 96500 C mol⁻¹$$\Delta G^\circ=-2\times96500\times1.10$$ΔG∘=−2×96500×1.10 $$\Delta G^\circ=-212300J$$ΔG∘=−212300J

Convert into kJ:$$=-212.3kJ$$=−212.3kJ

Answer:

$$\boxed{\Delta G^\circ=-212.3kJ}$$ΔG∘=−212.3kJ​

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NUMERICAL 3: Nernst Equation

Question:

Calculate cell potential at 298 K for:$$Zn|Zn^{2+}(0.1M)||Cu^{2+}(1M)|Cu$$Zn∣Zn2+(0.1M)∣∣Cu2+(1M)∣Cu

Given:$$E^\circ_{cell}=1.10V$$Ecell∘​=1.10V

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Solution:

Reaction:$$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$$Zn+Cu2+→Zn2++Cu

Number of electrons:$$n=2$$n=2

Reaction quotient:$$Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}$$Q=[Cu2+][Zn2+]​ $$Q=\frac{0.1}{1}=0.1$$Q=10.1​=0.1

Nernst equation:$$E=E^\circ-\frac{0.0591}{n}\log Q$$E=E∘−n0.0591​logQ $$E=1.10-\frac{0.0591}{2}\log(0.1)$$E=1.10−20.0591​log(0.1)

Since:$$\log(0.1)=-1$$log(0.1)=−1 $$E=1.10+0.02955$$E=1.10+0.02955 $$E=1.1295V$$E=1.1295V

Answer:

$$\boxed{E=1.13V}$$E=1.13V​

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NUMERICAL 4: Calculate Equilibrium Constant

Question:

For a cell:$$E^\circ_{cell}=1.10V$$Ecell∘​=1.10V

Number of electrons = 2

Find K.

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Solution:

Formula:$$E^\circ_{cell}=\frac{0.0591}{n}\log K$$Ecell∘​=n0.0591​logK

Substitute:$$1.10=\frac{0.0591}{2}\log K$$1.10=20.0591​logK $$\log K=\frac{1.10\times2}{0.0591}$$logK=0.05911.10×2​ $$\log K=37.22$$logK=37.22

Therefore:$$K=10^{37.22}$$K=1037.22

Answer:

$$\boxed{K=1.66\times10^{37}}$$K=1.66×1037​

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NUMERICAL 5: Faraday Law

Question:

Calculate mass of copper deposited when a current of 2 A is passed for 30 minutes.

Given:

Atomic mass of Cu = 63.5$$n=2$$n=2

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Solution:

Formula:$$m=\frac{MIt}{nF}$$m=nFMIt​

Time:$$30\times60=1800s$$30×60=1800s

Substitute:$$m=\frac{63.5\times2\times1800}{2\times96500}$$m=2×9650063.5×2×1800​ $$m=1.18g$$m=1.18g

Answer:

$$\boxed{1.18g\text{ copper deposited}}$$1.18g copper deposited​

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HIGH FREQUENCY BOARD QUESTIONS

Q1. Why does a galvanic cell stop working after some time?

Answer:

A galvanic cell stops working because:

• Reactants are consumed.
• Concentration changes.
• Cell potential becomes zero when equilibrium is reached.

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Q2. Why is conductivity of an electrolyte solution affected by temperature?

Answer:

With increase in temperature:

• Viscosity decreases.
• Ion mobility increases.
• Conductivity increases.

---

Q3. Why are weak electrolytes highly affected by dilution?

Answer:

Because dilution increases their degree of ionisation, producing more ions and increasing molar conductivity.

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Q4. Why is lithium used in lithium-ion batteries?

Answer:

Lithium is used because:

• It is lightweight.
• It has high electrode potential.
• It provides high energy density.

---

Q5. Explain corrosion of iron.

Answer:

Corrosion occurs through electrochemical reactions:

Anode:$$Fe\rightarrow Fe^{2+}+2e^-$$Fe→Fe2++2e−

Cathode:$$O_2+2H_2O+4e^-\rightarrow4OH^-$$O2​+2H2​O+4e−→4OH−

The final product formed is hydrated ferric oxide, called rust.