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Electrochemistry JEE Advanced 2023 Revision Notes

Last updated date: 28th Nov 2023
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MVSAT Dec 2023

Electrochemistry JEE Advanced 2023 Revision Notes - Free PDF Download

Electrochemistry is a crucial topic covered in the JEE Advanced syllabus. This chapter teaches students how chemical reactions occur during electrolysis. The different factors controlling an electrochemical reaction are explained in these notes. Different types of chemical reactions take place in an electrochemical system and can be expressed in balanced equations. Download and refer to the Electrochemistry JEE Advanced revision notes to understand the various types of electrochemical reactions.


JEE Advanced Revision Notes


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The subject experts at Vedantu have prepared these notes to aid your JEE Advanced preparation. The subject matter experts have explained all the concepts covered in this chapter in a simpler manner. So students can learn the concepts and Electrochemistry formulas for JEE Advanced and develop the best skills to answer fundamental questions from this chapter.

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Access JEE Advanced 2023 Revision Notes Chemistry Electrochemistry

Electrochemistry Deals With 

  • Oxidation-reduction responses that either produce or use electrical energy and electrochemical responses take place in cells. 

  • Each cell has two electrodes, operators through which electrons enter or leave the cell.

Three effects are:

  • Either oxidation or reduction occurs at each electrode.

  • Electrons flow through an external captain.

  • Ions flow in the electrolyte result.

There are Two Types of Electrochemical Cells: Voltaic Cells and Electrolytic Cells. 

  • A voltaic cell derives its energy from spontaneous redox reactions, while electrolytic cells involve non-spontaneous reactions and thus require an external electron source like a DC battery or an AC power source. 

  • An electrolytic cell is an electrochemical cell in which the energy from an external power source is used to drive a normally non-spontaneous reaction, i.e. apply a reverse voltage to a voltaic cell. 

Electric current is considered a flow of electrons. The substance which allows the electric current to pass through it is called a conductor.

Two Types of Conductors:


Electronic Conductors

Electrolytic Conductor


Conducts electricity by the flow of electrons.

Conduct electricity by the movement of ions.


Electrons flow from the negative end to the positive end.

Charge species (ions) move toward the oppositely charged electrodes.


No transfer of matter takes place.

Transfer of matter takes place.


In this there is no chemical change.

Chemical decomposition takes place at the electrodes.


It decreases with rising temperatures.

It increases with the increase in temperature.


Metals, alloys, graphite, solid such as Silver, Copper etc.

Liquids of molten salt, an aqueous solution of salts, and acids.

Electrochemical Cells:

An electrochemical cell is a device that can induce electrical energy from the chemical responses being in it, or use the electrical energy supplied to it to grease chemical responses in it. 

  • This device is able of converting chemical energy into electrical energy, or vice versa. 

  • A common illustration of an electrochemical cell is a standard 1.5- volt cell which is used to power numerous electrical appliances similar to television remotes and timepieces.

Daniel Cell:

  • Example of a galvanic cell is the Daniel cell.

  • It consists of zinc and copper half-cells.

  • This cell converts the chemical energy liberated during the redox reaction.

$Z{{n}_{(s)}}+C{{u}^{2+}}_{(aq)}\to Z{{n}^{2+}}_{(aq)}+C{{u}_{(s)}}$ 

  • Electron flow from Cu to Zn

  • Zinc is deposited at the zinc electrode and Copper dissolves at copper electrode.

Galvanic Cell:

  • A Galvanic cell and a voltaic cell are the same thing. There are other cells, however, similar to an electrolytic cell.

  • A Galvanic/ voltaic cell converts chemical energy into electrical energy. An electrolytic cell uses electrical energy to drive a non-spontaneous response i.e., converts electrical energy into chemical energy. 

  • Example: Zn rod is dipped in zinc sulphate solution

  • Zinc undergoes oxidation, gives electrons, it is electron-rich, so represented by a negative sign. $Zn\to Z{{n}^{2+}}+2{{e}^{-}}$

  • $C{{u}^{+2}}$present in the solution undergoes reduction on Cu rod surface. Cu rod is electron-poor, so it is represented by a positive sign. $C{{u}^{2+}}+2{{e}^{-}}\to Cu$

  • Electron flow in the external circuit from Zn to Cu.

  • A salt bridge is used to prevent the accumulation of charges, at the electrodes.

  • The reaction taking place in the electrochemical cell is$Z{{n}_{(s)}}+CuS{{O}_{4}}_{(aq)}\to ZnS{{O}_{4}}_{(aq)}+C{{u}_{(s)}}$

Electromotive Force of the Cell:

The difference between two electrode potential present in the electrochemical cell is the electromotive force of the cell.

Ecell = Standard electrode potential of right hand electrode-Standard electrode potential of left hand electrode

  • If the EMF of the cell is positive, the cell reaction is spontaneous or irreversible. We can draw current from it.

  • If EMF of the cell is negative, cell reaction is nonspontaneous or reversible. We cannot draw current from it.

  • If the EMF of the cell is zero, the cell reaction is at equilibrium.

EMF of a cell

EMF of a Cell

The reduction potential of an electrode is exactly equal in magnitude but opposite in sign to its oxidation potential.

Electrochemical Series:

  • In the IUPAC system of referring to electrode potentials, standard reducing potentials are simply called standard potential.

  • In the electrochemical series, the metals are arranged in the increasing order to reduce potential.

  • Metals placed above can displace the below metal ions from their solution.

  • Oxidation power ∝ reduction potential ∝ $\dfrac{{\text{1}}}{{{\text{Oxidation potential}}}}$

Reducing power ∝ oxidation potential ∝ $\dfrac{{\text{1}}}{{{\text{reduction potential}}}}$

Note: In two half-cell reaction having electrode potential $E_1^\circ $and $E_2^\circ $ are combined to give a third half-cell reaction having an electrode potential $E_3^\circ$ then,

$\Delta G_3^\circ  = \Delta G_1^\circ  + \Delta G_2^\circ$

$- {n_3}FE_3^\circ  =  - {n_1}FE_1^\circ  - {n_2}FE_2^\circ$

$E_3^\circ  = \dfrac{{{n_1}E_1^\circ  + {n_2}E_2^\circ }}{{{n_3}}}$ 

If the number of electrons involved are equal $E_3^\circ  = E_1^\circ  + E_2^\circ $

EMF of Galvanic Cell - Nernst Equation:

The variation of electrode potential and cell potential with a concentration of ions in solutions is given by Nernst equation which is derived from Vant’s Hoff isotherm equation.

$\Delta G = \Delta G^\circ  + 2.303RT\log Q$

R is the univeral gas constant $\left( 8.314J/K/mole \right)$.

T is the temperature in kelvin.

n= number of electrons involved in the reaction

F= Faraday =96500 coulombs

Q= reaction quotient

${{\text{E}}_{\text{cell}}}\text{=E}_{\text{cell}}^{\text{ }\!\!{}^\circ\!\!\text{ }}\text{-}\dfrac{\text{2}\text{.303RT}}{\text{nF}}\text{log}\dfrac{\left[ \text{Product} \right]}{\left[ \text{Reactants} \right]}......\left( 1 \right)$ 

By substituting the value of F, R and T(298 K) in the above equation(1) we get,

${{\text{E}}_{\text{cell}}}\text{=E}_{\text{cell}}^{\text{ }\!\!{}^\circ\!\!\text{ }}\text{-}\dfrac{\text{0}\text{.059}}{\text{n}}\text{log}\dfrac{\left[ \text{Product} \right]}{\left[ \text{Reactants} \right]}$

For Cation Electrodes:

The Nernst equation for a single electrode potential $\left( {{E_{{M^{n + }}/M}}} \right)$ at 298K is $M^{n+}(aq)+ne^-\longrightarrow M(s)$

The equilibrium state EMF is given by:

$M^{n+}(aq)+ne^-\longrightarrow M(s)$


$E=E{}^\circ -\dfrac{2.303RT}{nF}\log \dfrac{1}{c}$

$\therefore c=\left[ {{M}^{n+}} \right]$

$\left[ {{M}^{n+}} \right]$is the molar concentration of the metal ion .

R is the univeral gas constant $\left( 8.314J/K/mole \right)$.

T is the temperature in kelvin.

n is the numnver of electrons involved in the reaction.

F is the faraday’s constant (96500 C/mole).

The concentration of pure liquid and solid are taken as one so, the expression will be:

(or) $E=E{}^\circ +\dfrac{0.059}{n}\log \left[ {{M}^{n+}} \right]$

Electrolytic Solution Resistance and Conductance:

Ohm’s Law: It states the current (I) flowing through a conductor at a given temperature is proportional to the potential difference (V) and proportional to resistance. 

Ohm’s law formula is written as:

V ∝ I

Thus, V = RI where R is a constant called resistance. R depends on the dimensions of the conductor and also on the material of the conductor. Its SI unit is Ohm( Ω)

Resistance(R): It is the opposition to the flow of current offered by the electrolytic solution. Its unit are Ohm( Ω.)

The two parallel electrodes of the cell and inversely proportional to the area of cross-section(A) of electrode i.e.,${{R\alpha }}\dfrac{{\text{l}}}{{\text{A}}}{\text{OR R =}} \rho {\text{.}}\dfrac{{\text{l}}}{{\text{A}}}$


It is the reciprocal of the electrical resistance(R).

$G = \dfrac{1}{R}$

It measures the ease with which the current flows through a conductor.

$G = \dfrac{1}{R}$

$G = \dfrac{1}{\rho } \times \dfrac{1}{{\dfrac{l}{A}}} = k \times \dfrac{1}{{\dfrac{l}{A}}}\because k = G \times \dfrac{l}{A}$ 

Where, G= Conductance.

$\rho$= Specific conductance (Conductivity).

$\dfrac{l}{A}$= Cell Constant 

S.I unit of conductance = Siemens(S).

The conductance or the current conducting capacity of an electrolytic solution can be expressed in terms of 

  • Specific conductance (k) Kappa

  • Molar conductance $\left( {{\Lambda _m}} \right)$

  • Equivalent conductance $\left( {{\Lambda }} \right)$

Specific conductance(K): It is reciprocal of specific resistance $\left( \rho  \right)$

$k = \dfrac{1}{\rho }$

$R = \rho .\dfrac{l}{A};\dfrac{1}{G} = \dfrac{1}{k}.\dfrac{l}{A}$

$k = G \times \dfrac{l}{A}$

G= conductance, $\dfrac{l}{A}$=cell constant

Molar conductance $\left( {{\Lambda _m}} \right)$

It is the conducting power of all the ions produced by dissolving 


$\Lambda_{m}(Scm^2 mol^{−1})=\dfrac{k(Scm^{−1})×1000(cm^3/L)}{molarity (mole/L)}$

Equivalent Conductance:

It is the conducting power of all the ions produced by dissolving 1 gram equivalent of an electrolyte (or) The conductance of a solution containing 1 gram equivalent of the electrolyte placed between two parallel electrodes separated by unit length and area large enough to contain that 1 mole is called Equivalent conductance.

$\left( {{ \wedge _{eq}}} \right)\left( {oh{m^{ - 1}}c{m^2}e{q^{ - 1}}} \right) = \dfrac{{k\left( {oh{m^{ - 1}}c{m^{ - 1}}} \right) \times 1000}}{{N(eq.wts/L)}}$

$\left( {{ \wedge _{eq}}} \right)\left( {oh{m^{ - 1}}{m^2}e{q^{ - 1}}} \right) = \dfrac{{k\left( {s{m^{ - 1}}} \right) \times {{10}^{ - 3}}}}{{N(eq.wts/L)}}$

Factors Affecting Molar and Equivalent Conductance

  • Nature of electrolyte

  • Nature of the solvent

  • Viscosity of solvent

  • Temperature

  • Concentration of electrolyte

  • Size of the ions produced and their solvation

Effect of Dilution: With both $\left( {{\Lambda _{eq}}} \right)$ as well as $\left( {{\Lambda _m}} \right)$ of both weak strong electrolytes increases.

Specific conductivity (k) decreases with dilution because of the decrease in the number of ions per $c{m^3}$of electrolyte.

Effect of Temperature: The conductivity $\left( {{\Lambda _{eq}}} \right)$ and $\left( {{\Lambda _m}} \right)$ of all electrolytes increases with increase in temperature.

Debye-Huckel-Onsager Equation:

$\wedge_{c}=\wedge_{o}-\left[\dfrac{82.4}{(\mathrm{DT})_{\eta}^{\dfrac{1}{2}}}+\dfrac{8.2 \times 10^{5}}{(\mathrm{DT})^{\dfrac{3}{2}}} \wedge_{\mathrm{o}}\right] \sqrt{C}$


D= Dielectric constant of water

T = Temperature in kelvin scale

${ \wedge _c}$ = Equivalent conductance at conc. ‘c’

${ \wedge _o}$ = equivalent conductance at almost zero concentration or infinite dilution.

$\eta$ = Viscosity coefficient of solvent

${\wedge _c} = { \wedge _o} - b\sqrt C$, where b is constant and depends on the nature of the solvent and temperature.


Non Electrolyte: The substance which does not conduct electricity either in its molten state or in aqueous solution is called non-electrolyte.

Electrolyte : A substance, that is in the molten state or in the dissolved state containing ions and is functioning as an electrically conducting medium is called an electrolyte.

Faraday’s Law of Electrolysis:

First Law:

The amount of the substance deposited or electrolysed is directly proportional to the quantity of electricity passed through the electrolyte.

  • The weight of the substance deposited on the electrode for 1 coulomb of electricity is called electrochemical equivalent(Z).

  • The unit for electrochemical equivalent is gram/coulomb (g/C).

  • Electrochemical equivalent depends only on the nature of the electrolyte.

Second Law: 

When the same quantity of electricity is passed through different electrolytes connected in series the weights the substances deposited will be in the ratio of their equivalent weight.

$\dfrac{{{\text{mass of the hydrogen}}}}{{{\text{mass of copper}}}}{\text{ = }}\dfrac{{{\text{Equivalent mass of hydrogen}}}}{{{\text{Equivalent mass of copper}}}}$

  • Chemical equivalent weight of an element.

$\left[ \dfrac{{{W}_{1}}}{{{W}_{2}}}=\dfrac{{{E}_{1}}}{{{E}_{2}}} \right]$Where,

 ${{W}_{1}}$and ${{W}_{2}}$ are weight deposited of two elements.

${{E}_{1}}$ and ${{E}_{2}}$are the equivalent weight of two element 

  • 1 Faraday of electric current or 96500 coulombs of electricity or one mole of electron, when passed through an electrolyte 1G.E.W of a substance will be deposited.

  • 1 coulomb deposited

${\text{ = }}\left[ {\dfrac{{{\text{gram equivalent weight}}}}{{{\text{96500}}}}} \right]$

${\text{ = }}\dfrac{{\text{E}}}{{\text{F}}}{\text{ = e (electrochemical equivalent)}}$ 

  • 1Faraday=charge present on $6.023 \times {10^{23}}$ electrons

  • 1 coulomb $ = 6.25 \times {10^{18}}$ electrons

Current Efficiency: % of current efficiency $= \dfrac{{{\text{mass actually produced}}}}{{{\text{mass of substance expected}}}}\times 100$

Electrochemical Corrosion:

The process of corrosion may be chemical (or) electrochemical in nature

  • The anodic dissolution of a metal under the conditions of corrosion is known as electrochemical corrosion

$M \to {M^{n + }} + n{e^ - }$


The phenomenon of a metal reaching a stage of non-reactive state in its reaction with concentrated acids may called passivity.

  • Iron group of metal and some other transition metals are rendered passive with concentrated acids.

  • Passivity of metal can be classified into 

  1. Chemical passivity

  2. Mechanical passivity

  3. Electro-chemical passivity

Chemical Passivity: 

  • Non-reactivity of metals after initial reaction with concentrated $HN{O_3}$ 

  • Passive iron doesn’t dissolve in dill $HN{O_3}$ and iron don’t displace Ag from $AgN{O_3}$

Mechanical Passivity:

  • In some cases, dissolution of metal stops due to visible oxide film formation.

Electrochemical Passivity:

  • Metal with more -ve potential function as anode in cell.

  • Generally, Fe, Ni, Co functions are anode.

  • They dissolve as $M \to {M^{n + }} + n{e^ - }$

At a particular stage anode stops dissolving due to formation of an invisible metal oxide film. This phenomenon is called electro-chemical passivity.

Importance of Electrochemistry - JEE Advanced

In this chapter, students will learn the advanced concepts of Chemistry related to electrolysis. In this section, you will find out how an electrochemical reaction is conducted in a system using electrodes and electrolytes.

Every reaction will be explained in this chapter using chemical reactions in different phases of electrolysis. It will help you understand the mechanism of an electrochemical reaction, so you will be able to figure out how an expected result is achieved using these methods.

This chapter is very important as quite a few questions related to electrochemical reactions and electrochemical series are expected in JEE Advanced. These revision notes explain how the ions behave in an electrochemical system and the roles of different parts of electrochemical reactions.

This chapter will also explain the associated terms of electrochemistry and their uses. Refer to the Electrochemistry formulas JEE Advanced revision notes prepared by our experts to get a detailed understanding of these topics.

Benefits of Vedantu’s Electrochemistry IIT JEE Notes PDF

  • You can download the PDF version and use it at your convenience. Make your study sessions more productive.

  • Use these notes for preparing this chapter faster by comprehending the concepts, fundamental principles and scientific formulas of electrochemistry. 

  • Use the simple description given in the notes to make your study sessions and revision time efficient. 

  • Refer to these simple explanations and learn how to accurately answer the fundamental questions during an exam.

  • Prefer using these Electrochemistry JEE notes to revise efficiently and recall the fundamental concepts during an exam. The easier format of these notes will help you to memorise and remember the formulas accurately.

Download Electrochemistry JEE Notes Free PDF

It is now easier to complete studying this chapter by downloading the free PDF version of the revision notes. You can use the simpler explanation to comprehend the concepts such as different types of cells, their differences, and learn how to calculate the electrode potentials. Find out the easiest way to remember the electrochemical series using these notes and answer fundamental questions related to electrochemistry correctly in the JEE Advanced exam.

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FAQs on Electrochemistry JEE Advanced 2023 Revision Notes

1. How many types of electrochemical cells are there?

There are two types of cells we can find in this chapter, galvanic and electrolytic cells. A galvanic cell converts chemical energy into electrical energy. An electrolytic cell converts electrical energy to chemical energy.

2. What is an electrochemical series?

It is an activity series that determines the position of radicals and elements based on their electrode potential values. The values are measured by using the standard hydrogen electrode as a reference frame.

3. What is the application of the electrochemical series?

The prime application of the electrochemical series is to determine the reactivity of elements; reducing and oxidising strengths of elements and ions. This series enables us to identify the reducing and oxidising agents in an electrochemical system.

4. What is the importance of an electrochemical cell diagram?

A cell diagram explains the different parts of an electrochemical cell. It is used to explain how an electrochemical reaction will take place in a system.