Class 12 Chemistry Chapter 2: Electrochemistry | Complete NCERT Notes

Introduction

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical reactions. It explains how chemical energy can be converted into electrical energy and how electrical energy can be used to bring about chemical changes.

Many processes of daily life such as operation of batteries, corrosion of metals, electroplating, and functioning of fuel cells are governed by electrochemical principles. In this chapter, the concepts of electrochemical cells, electrode potential, Nernst equation, conductance of electrolytic solutions, electrolysis, corrosion, and electrochemical power sources are discussed in a systematic manner as prescribed by the NCERT syllabus.

The study of electrochemistry requires a clear understanding of redox reactions, concentration terms of solutions, and thermodynamic concepts. For solution-related concepts used in this chapter, students are advised to refer to Class 12 Chemistry Chapter 1: Solutions .

Basic Concepts of Electrochemistry

Electrochemistry is based on the concept of redox (reduction–oxidation) reactions, in which transfer of electrons takes place between reacting species. Whenever electrons are transferred, an electrical effect is produced.

Oxidation and Reduction

In electrochemical processes, oxidation and reduction always occur simultaneously. These processes are defined as follows:

• Oxidation: Loss of electrons by a species.
• Reduction: Gain of electrons by a species.

For example, consider the reaction:

Zn → Zn²⁺ + 2e⁻

In this reaction, zinc loses electrons and is therefore oxidised. The electrons released in oxidation are accepted by another species, which undergoes reduction.

Redox Reactions

A reaction in which oxidation and reduction occur simultaneously is called a redox reaction. Such reactions are the foundation of electrochemical cells.

In electrochemical cells, redox reactions are separated into two half-reactions:

• Oxidation half-reaction (occurs at anode)
• Reduction half-reaction (occurs at cathode)

The separation of these half-reactions allows the flow of electrons through an external circuit, producing electrical energy.

Oxidation Number

The oxidation number of an element is a number assigned to it to indicate the degree of oxidation or reduction during a chemical reaction.

The increase in oxidation number indicates oxidation, while the decrease in oxidation number indicates reduction.

For example, in the reaction:

Cu²⁺ + 2e⁻ → Cu

The oxidation number of copper decreases from +2 to 0, showing that copper ions undergo reduction.

Electrochemical Reactions

Electrochemical reactions are those redox reactions in which:

• Chemical energy is converted into electrical energy, or
• Electrical energy is used to bring about chemical change.

Based on this principle, electrochemical processes are classified into:

• Electrochemical cells – where chemical energy is converted into electrical energy.
• Electrolytic cells – where electrical energy is used to drive non-spontaneous chemical reactions.

These concepts form the theoretical foundation for understanding electrochemical cells, electrode potential, and electrolysis, which are discussed in the following sections.

Electrochemical Cells

An electrochemical cell is a device that converts chemical energy into electrical energy by means of a redox reaction. The redox reaction occurring in an electrochemical cell is spontaneous in nature.

Galvanic (Voltaic) Cell

A galvanic cell, also known as a voltaic cell, is an electrochemical cell in which a spontaneous redox reaction produces electrical energy. The oxidation and reduction reactions occur in separate compartments, allowing electrons to flow through an external circuit.

The flow of electrons from one electrode to another results in the generation of electric current.

Daniell Cell

The Daniell cell is a commonly studied galvanic cell. It consists of two half-cells:

• A zinc electrode dipped in zinc sulphate solution.
• A copper electrode dipped in copper sulphate solution.

The two solutions are connected by a salt bridge, which maintains electrical neutrality by allowing the movement of ions.

Electrode Reactions in Daniell Cell

The half-reactions occurring at the electrodes are:

At Anode (Oxidation):

Zn(s) → Zn²⁺(aq) + 2e⁻

At Cathode (Reduction):

Cu²⁺(aq) + 2e⁻ → Cu(s)

Zinc undergoes oxidation and acts as the anode, while copper ions undergo reduction at the cathode.

Anode and Cathode

In an electrochemical (galvanic) cell:

• Anode: The electrode at which oxidation occurs.
• Cathode: The electrode at which reduction occurs.

Electrons always flow from anode to cathode through the external circuit.

Salt Bridge

A salt bridge is a U-shaped tube containing an inert electrolyte such as potassium chloride or potassium nitrate in agar-agar gel.

The main functions of a salt bridge are:

• To complete the electrical circuit.
• To maintain electrical neutrality in the two half-cells.
• To prevent direct mixing of the electrolytes.

Cell Notation

An electrochemical cell is represented symbolically using cell notation. For a Daniell cell, the cell notation is written as:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

In cell notation:

• The anode is written on the left-hand side.
• The cathode is written on the right-hand side.
• A single vertical line (|) represents a phase boundary.
• A double vertical line (||) represents the salt bridge.

The concept of electrochemical cells forms the basis for understanding electrode potential and electromotive force, which are discussed in the next section.

Electrode Potential and Electromotive Force (EMF)

When a metal is dipped into a solution of its own ions, a potential difference develops between the metal and the solution. This potential difference is known as electrode potential.

Single Electrode Potential

The electrode potential of an electrode is defined as the potential difference between the electrode and the electrolyte with which it is in contact.

Since it is not possible to measure the absolute potential of a single electrode, electrode potentials are always measured relative to a reference electrode.

Standard Electrode Potential

The standard electrode potential is the electrode potential measured under standard conditions:

• Concentration of ions = 1 mol L⁻¹
• Pressure of gases = 1 bar
• Temperature = 298 K

Standard electrode potentials are denoted by the symbol E°.

Standard Hydrogen Electrode (SHE)

The standard hydrogen electrode (SHE) is used as the reference electrode to measure standard electrode potentials.

It consists of a platinum electrode coated with finely divided platinum black, immersed in an acidic solution having hydrogen ion concentration of 1 mol L⁻¹. Hydrogen gas is bubbled at a pressure of 1 bar.

The standard electrode potential of the standard hydrogen electrode is arbitrarily assigned a value of:

E° = 0.00 V

Electrode Potential Series

When different electrodes are arranged in order of increasing standard reduction potentials, the arrangement is known as the electrode potential series.

The electrode potential series helps in:

• Predicting the feasibility of redox reactions.
• Comparing the oxidising and reducing strengths of substances.
• Calculating the EMF of electrochemical cells.

Electromotive Force (EMF) of a Cell

The electromotive force (EMF) of an electrochemical cell is defined as the maximum potential difference between the two electrodes of the cell when no current is flowing.

The EMF of a cell is given by the difference between the electrode potentials of the cathode and the anode:

E_cell = E_cathode − E_anode

Under standard conditions, the EMF of the cell is expressed as:

E°_cell = E°_cathode − E°_anode

If the value of E°_cell is positive, the cell reaction is spontaneous. If it is negative, the reaction is non-spontaneous.

The concept of electrode potential and EMF provides the foundation for understanding the effect of concentration on cell potential, which is explained by the Nernst equation.

Nernst Equation

The electrode potential of an electrode and the EMF of an electrochemical cell depend on the concentration of ions in solution. This dependence is explained by the Nernst equation.

Need for Nernst Equation

Standard electrode potentials are measured under standard conditions only. In practical situations, concentrations of ions are usually different from standard values. Therefore, the Nernst equation is used to calculate electrode potential and cell EMF under non-standard conditions.

General Form of Nernst Equation

For a general redox reaction:

aA + bB ⇌ cC + dD

The Nernst equation is given by:

E = E° − (RT / nF) ln Q

where:

• E = electrode potential under non-standard conditions
• E° = standard electrode potential
• R = gas constant (8.314 J mol⁻¹ K⁻¹)
• T = temperature in Kelvin
• n = number of electrons transferred
• F = Faraday constant (96485 C mol⁻¹)
• Q = reaction quotient

Nernst Equation at 298 K

At 298 K, the Nernst equation is simplified as:

E = E° − (0.0591 / n) log Q

Nernst Equation for an Electrochemical Cell

For a complete electrochemical cell, the equation becomes:

E_cell = E°_cell − (0.0591 / n) log Q

This equation helps in calculating the EMF of a cell under non-standard concentration conditions.

Worked Numerical Example 1

Problem:

Calculate the electrode potential of Zn²⁺/Zn electrode at 298 K when the concentration of Zn²⁺ ions is 0.01 M. (Given: E° = −0.76 V)

Solution:

Given:

• E° = −0.76 V
• n = 2
• [Zn²⁺] = 0.01 M

Formula used:

E = E° − (0.0591 / n) log [Zn²⁺]

Substitution:

E = −0.76 − (0.0591 / 2) log(0.01)

log(0.01) = −2

E = −0.76 − (0.02955 × −2)

E = −0.76 + 0.0591

Final Answer:

E = −0.7009 V

Worked Numerical Example 2

Problem:

Calculate the EMF of a Daniell cell at 298 K. Given:

• E°_cell = 1.10 V
• [Zn²⁺] = 0.1 M
• [Cu²⁺] = 1.0 M
• n = 2

Solution:

Reaction Quotient:

Q = [Zn²⁺] / [Cu²⁺]

Q = 0.1 / 1.0 = 0.1

Nernst Equation:

E_cell = E°_cell − (0.0591 / 2) log Q

Substitution:

E_cell = 1.10 − (0.02955 × log 0.1)

log 0.1 = −1

E_cell = 1.10 − (−0.02955)

Final Answer:

E_cell = 1.1296 V

Conceptual Link

The effect of concentration on electrode potential is closely related to solution properties discussed in Solutions – Class 12 Chemistry , particularly molarity and concentration terms.

The Nernst equation forms the basis for understanding electrochemical equilibrium, corrosion, and electrolysis, which are discussed in the subsequent sections.

Electrical Conductance

The flow of electricity through an electrolytic solution occurs due to the movement of ions. The ease with which electric current flows through a solution is measured in terms of its electrical conductance.

Resistance and Conductance

The opposition offered by a conductor to the flow of electric current is called resistance and is denoted by R.

R = V / I

where V is the potential difference and I is the current.

The reciprocal of resistance is called conductance and is denoted by G.

G = 1 / R

The unit of conductance is siemens (S).

Specific Conductance (Conductivity)

Specific conductance or conductivity (κ) is defined as the conductance of a solution placed between two electrodes 1 cm apart with a cross-sectional area of 1 cm².

κ = G × (l / A)

where:

• κ = specific conductance (S cm⁻¹)
• G = conductance (S)
• l = distance between electrodes (cm)
• A = area of cross-section (cm²)

The unit of specific conductance is siemens per centimetre (S cm⁻¹).

Molar Conductivity

Molar conductivity (Λ_m) is defined as the conductance of all the ions produced by dissolving one mole of an electrolyte in solution.

Λ_m = κ × (1000 / C)

where C is the concentration of the electrolyte in mol L⁻¹.

The unit of molar conductivity is S cm² mol⁻¹.

Variation of Molar Conductivity with Concentration

The molar conductivity of an electrolyte increases with dilution. This increase occurs because the number of ions per unit volume increases and inter-ionic attractions decrease.

Strong Electrolytes

For strong electrolytes, molar conductivity increases slowly with dilution. This is because strong electrolytes are almost completely ionised in solution.

Weak Electrolytes

For weak electrolytes, molar conductivity increases sharply with dilution. This is due to an increase in the degree of ionisation.

Worked Numerical Example 1

Problem:

Calculate the molar conductivity of a solution having specific conductance 0.02 S cm⁻¹ and concentration 0.5 mol L⁻¹.

Solution:

Given:

• κ = 0.02 S cm⁻¹
• C = 0.5 mol L⁻¹

Formula used:

Λ_m = κ × (1000 / C)

Substitution:

Λ_m = 0.02 × (1000 / 0.5)

Λ_m = 0.02 × 2000

Final Answer:

Λ_m = 40 S cm² mol⁻¹

Worked Numerical Example 2

Problem:

The resistance of a solution in a conductivity cell is 50 Ω. The cell constant is 1 cm⁻¹. Calculate the specific conductance.

Solution:

Given:

• R = 50 Ω
• Cell constant (l/A) = 1 cm⁻¹

Formula used:

κ = (1 / R) × (l / A)

Substitution:

κ = (1 / 50) × 1

Final Answer:

κ = 0.02 S cm⁻¹

Conceptual Link

The dependence of molar conductivity on concentration is directly related to molarity and dilution concepts explained in Solutions – Class 12 Chemistry .

Faraday’s Laws of Electrolysis

Electrolysis is the process in which electrical energy is used to bring about a chemical change. The quantitative relationship between the amount of electricity passed and the chemical change produced is governed by Faraday’s laws of electrolysis.

Faraday’s First Law of Electrolysis

The first law states that the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

m ∝ Q

Since electric charge Q = I × t,

m ∝ I × t

where:

• m = mass of substance deposited (g)
• I = current (A)
• t = time (s)

Faraday’s Second Law of Electrolysis

According to the second law, when the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their equivalent weights.

m ∝ Equivalent weight

Mathematical Expression

Combining both laws, the mass deposited during electrolysis is given by:

m = (Q × E) / F

where:

• Q = charge passed (C)
• E = equivalent weight of the substance
• F = Faraday constant (96500 C mol⁻¹)

The equivalent weight (E) is calculated as:

E = M / n

where M is molar mass and n is the number of electrons involved.

Worked Numerical Example 1

Problem:

Calculate the mass of silver deposited when a current of 2 A is passed through a solution of silver nitrate for 30 minutes. (Atomic mass of Ag = 108, n = 1)

Solution:

Given:

• I = 2 A
• t = 30 min = 1800 s
• M = 108 g mol⁻¹
• n = 1
• F = 96500 C mol⁻¹

Charge passed:

Q = I × t = 2 × 1800 = 3600 C

Equivalent weight:

E = M / n = 108 / 1 = 108

Formula used:

m = (Q × E) / F

Substitution:

m = (3600 × 108) / 96500

Final Answer:

m ≈ 4.03 g

Worked Numerical Example 2

Problem:

How much charge is required to deposit 1.08 g of silver from AgNO₃ solution?

Solution:

Given:

• m = 1.08 g
• E = 108
• F = 96500 C mol⁻¹

Formula used:

Q = (m × F) / E

Substitution:

Q = (1.08 × 96500) / 108

Final Answer:

Q = 965 C

Faraday’s laws are widely used in calculating quantities involved in electrolysis, electroplating, and refining of metals.

Applications of Electrochemistry

Electrochemical principles have wide applications in industry, technology, and daily life. Important applications include corrosion of metals, electrochemical power sources such as batteries, and fuel cells.

Corrosion

Corrosion is the gradual destruction of metals due to chemical or electrochemical reactions with the environment. Rusting of iron is the most common example of corrosion.

Electrochemical Theory of Corrosion

According to the electrochemical theory, corrosion occurs due to the formation of small electrochemical cells on the surface of the metal.

At anodic regions, oxidation occurs:

Fe → Fe²⁺ + 2e⁻

At cathodic regions, reduction of oxygen takes place:

O₂ + 2H₂O + 4e⁻ → 4OH⁻

The ferrous ions further react to form hydrated ferric oxide, commonly known as rust.

Prevention of Corrosion

• Coating the metal surface with paint or grease.
• Galvanisation (coating iron with zinc).
• Electroplating with corrosion-resistant metals.
• Cathodic protection using sacrificial anodes.

Batteries

A battery is a device consisting of one or more electrochemical cells connected together to supply electrical energy.

Primary Batteries

Primary batteries are those in which the cell reactions are irreversible. Once discharged, they cannot be recharged.

Example: Dry cell

• Anode: Zinc container
• Cathode: Carbon rod
• Electrolyte: Paste of NH₄Cl and ZnCl₂

Secondary Batteries

Secondary batteries are rechargeable because the cell reactions are reversible.

Example: Lead storage battery

• Anode: Lead (Pb)
• Cathode: Lead dioxide (PbO₂)
• Electrolyte: Dilute sulphuric acid

Lead storage batteries are widely used in automobiles.

Fuel Cells

Fuel cells are electrochemical cells that convert the chemical energy of a fuel directly into electrical energy.

The hydrogen–oxygen fuel cell is an important example.

In this cell:

• Hydrogen is oxidised at the anode.
• Oxygen is reduced at the cathode.
• Water is the only product formed.

Fuel cells are highly efficient and environmentally friendly and are used in spacecrafts and future energy technologies.

These applications highlight the importance of electrochemistry in practical and industrial processes.

Formula Summary Table

The following table summarises the important formulae used in Electrochemistry. These formulae are strictly based on the NCERT syllabus and should be used along with proper theoretical understanding.

Students are advised to memorise these formulae only after understanding their derivation and application in numerical problems.

Solved Numerical Examples

The following numerical problems are strictly based on the NCERT syllabus. Each problem is solved step-by-step to ensure conceptual clarity and proper exam-oriented presentation.

Numerical Example 1: EMF of an Electrochemical Cell

Problem:

Calculate the standard EMF of a cell having the following standard electrode potentials:

• E° (Zn²⁺/Zn) = −0.76 V
• E° (Cu²⁺/Cu) = +0.34 V

Solution:

Given:

• E°_cathode = +0.34 V
• E°_anode = −0.76 V

Required: E°_cell

Formula used:

E°_cell = E°_cathode − E°_anode

Substitution:

E°_cell = 0.34 − (−0.76)

Final Answer:

E°_cell = 1.10 V

Numerical Example 2: EMF Using Nernst Equation

Problem:

Calculate the EMF of a Daniell cell at 298 K if the concentration of Zn²⁺ ions is 0.01 M and Cu²⁺ ions is 1.0 M. (E°_cell = 1.10 V)

Solution:

Given:

• E°_cell = 1.10 V
• [Zn²⁺] = 0.01 M
• [Cu²⁺] = 1.0 M
• n = 2

Reaction quotient:

Q = [Zn²⁺] / [Cu²⁺] = 0.01

Formula used:

E_cell = E°_cell − (0.0591 / n) log Q

Substitution:

E_cell = 1.10 − (0.0591 / 2) log(0.01)

log(0.01) = −2

E_cell = 1.10 − (0.02955 × −2)

Final Answer:

E_cell = 1.1591 V

Numerical Example 3: Molar Conductivity

Problem:

Calculate the molar conductivity of a solution having conductivity 0.01 S cm⁻¹ and concentration 0.2 mol L⁻¹.

Solution:

Given:

• κ = 0.01 S cm⁻¹
• C = 0.2 mol L⁻¹

Formula used:

Λ_m = κ × (1000 / C)

Substitution:

Λ_m = 0.01 × (1000 / 0.2)

Final Answer:

Λ_m = 50 S cm² mol⁻¹

Numerical Example 4: Faraday’s Law of Electrolysis

Problem:

Calculate the mass of copper deposited when a current of 3 A is passed through CuSO₄ solution for 20 minutes. (Atomic mass of Cu = 63.5, n = 2)

Solution:

Given:

• I = 3 A
• t = 20 min = 1200 s
• M = 63.5 g mol⁻¹
• n = 2
• F = 96500 C mol⁻¹

Charge passed:

Q = I × t = 3 × 1200 = 3600 C

Equivalent weight:

E = M / n = 63.5 / 2 = 31.75

Formula used:

m = (Q × E) / F

Substitution:

m = (3600 × 31.75) / 96500

Final Answer:

m ≈ 1.18 g

These solved examples demonstrate the application of electrochemical principles in quantitative problems commonly asked in board examinations.

Practice Problems

(A) Very Short Answer Questions

1. Define electrochemistry.
Answer
Electrochemistry is the branch of chemistry that deals with the interconversion of chemical energy and electrical energy.
2. What is an electrochemical cell?
Answer
An electrochemical cell is a device that converts chemical energy into electrical energy by a spontaneous redox reaction.
3. State the unit of molar conductivity.
Answer
The unit of molar conductivity is S cm² mol⁻¹.
4. What is the value of standard electrode potential of hydrogen?
Answer
The standard electrode potential of hydrogen is taken as 0.00 V.

(B) Short Answer Questions

1. Explain why oxidation and reduction always occur simultaneously.
   Answer

   Oxidation involves loss of electrons while reduction involves gain of electrons. Since electrons lost by one species must be gained by another, both processes always occur together.
2. What is the role of salt bridge in a Daniell cell?
   Answer

   A salt bridge maintains electrical neutrality in the two half-cells and completes the electrical circuit.
3. Why does molar conductivity increase on dilution?
   Answer

   On dilution, inter-ionic attractions decrease and degree of ionisation increases, resulting in higher molar conductivity.

(C) Numerical Problems

1. Calculate the EMF of a cell if E°_cathode = +0.80 V and E°_anode = −0.44 V.
   Answer

   E°_cell = 0.80 − (−0.44) = 1.24 V
2. Calculate the molar conductivity of a solution having conductivity 0.015 S cm⁻¹ and concentration 0.25 mol L⁻¹.
   Answer

   Λ_m = 0.015 × (1000 / 0.25) = 60 S cm² mol⁻¹
3. How much charge is required to deposit 5.4 g of silver from AgNO₃ solution? (Atomic mass of Ag = 108)
   Answer

   Q = (m × F) / E = (5.4 × 96500) / 108 = 4825 C

These practice problems help students assess their understanding of theoretical concepts and numerical applications of electrochemistry.

Multiple Choice Questions (MCQs)

1. The standard electrode potential of hydrogen electrode is:
   (a) +1.00 V   (b) −1.00 V   (c) 0.00 V   (d) +0.76 V
   Answer

   (c) 0.00 V
2. Which equation relates EMF of a cell with concentration?
   (a) Faraday equation   (b) Arrhenius equation
   (c) Nernst equation   (d) Gibbs equation
   Answer

   (c) Nernst equation
3. The unit of molar conductivity is:
   (a) S cm⁻¹   (b) S cm² mol⁻¹
   (c) Ω cm   (d) V cm⁻¹
   Answer

   (b) S cm² mol⁻¹
4. Rusting of iron is an example of:
   (a) Physical change   (b) Chemical equilibrium
   (c) Electrochemical corrosion   (d) Thermal decomposition
   Answer

   (c) Electrochemical corrosion
5. Which of the following is a secondary battery?
   (a) Dry cell   (b) Mercury cell
   (c) Lead storage battery   (d) Fuel cell
   Answer

   (c) Lead storage battery

Frequently Asked Questions (FAQ)

Conclusion

The chapter on Electrochemistry provides a detailed understanding of the relationship between chemical reactions and electrical energy. Concepts such as electrochemical cells, electrode potential, Nernst equation, conductance of electrolytic solutions, electrolysis, corrosion, batteries, and fuel cells form the foundation of many industrial and technological applications.

A thorough understanding of this chapter is essential not only for board examinations but also for advanced studies in chemistry and related fields.

For solution concentration terms repeatedly used in electrochemical calculations, students are encouraged to revise Class 12 Chemistry Chapter 1: Solutions .