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NCERT Class 12 Chemistry Chapter 3 Notes Electrochemistry - Download PDF

NCERT Class 12 Chemistry Chapter 3 Notes Electrochemistry - Download PDF

Edited By Shivani Poonia | Updated on Mar 26, 2025 02:43 PM IST

Electrochemistry is one of the important branches of chemistry that deals with the relationship between electrical energy and chemical reactions. Electrochemistry forms the basis that is going to help students understand complex topics. In our daily lives, we often use Batteries in Smartphones and electric vehicles for their charging. This phenomenon is based on Electrochemistry. Industries depending on Electrochemistry for the refining of metals, wastewater treatment, and impurity removal. Electrochemistry notes class 12 covers a brief outline of topics such as electrochemical cells, the Nernst equation, the Gibbs energy of cell reaction, conductivity, Kohlrausch law and its applications, electrolysis, etc.

This Story also Contains
  1. Electrochemistry Class 12 Notes:
  2. Conductance-
  3. Kohlrausch’s Law-
  4. Galvanic cells-
  5. EMF (electromotive force)-
  6. Nernst equation-
  7. Faraday’s Law of electrolysis-
  8. Batteries-
  9. Corrosion-
  10. Significance of NCERT Class 12 Chemistry Chapter 3 Notes

Electrochemistry Class 12 Chemistry notes will be helpful for a quick revision of topics. Chapter 2 notes are designed by our subject experts which ensures the credibility of the content provided. Class 12 Chemistry Chapter 2 notes contain all the important formulas of electrochemistry. The galvanic cell converts the chemical energy of the spontaneous reaction into electrical energy. Electrochemical principles are widely used in energy storage systems like lead-acid batteries in vehicles, fuel cells, lithium-ion batteries in smartphones, etc.

Background wave

Electrochemistry is widely used in Manufacturing, Healthcare, and the protection of the environment. It becomes difficult and time-consuming for students to read NCERT book texts, point to point. So, to solve this problem, we are providing NCERT notes that cover all the topics and concepts provided in the NCERT textbook in a very clear and comprehensive way.


Also, students can refer,


Electrochemistry Class 12 Notes:

Conductance-

Conductance is the measure of the ease with which current flows through a conductor.

Conductivity-

It is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section.

=1

Molar conductivity-

Molar conductivity is described as the conducting power produced by the ions by dissolving one mole of electrolyte in a solvent.

m = / C

Conductivity and molar conductivity-

m= κ×1000/M

Equivalent conductivity- Equivalent conductivity is defined as the conductivity power of combining ions formed by the dissolution of electrolyte of one gram equivalent in a solution.

e= κ×1000/Ceq

Variation of conductivity and molar conductivity with concentration-

Electrolytic conductance decreases with an increase in concentration or increases with the increase in dilution.

Molar conductivity increases with dilution.

The equation of Debye Huckel Onsager shows the variation of molar conductivity along with concentration for strong electrolytes.

m=m⁰-AC1/2

  • Variation of molar conductivity with concentration for weak electrolytes- Molar conductivity of weak electrolytes cannot be found for weak electrolytes because dissociation of weak electrolytes is much lesser compared to strong electrolytes.
  • For weak electrolytes, molar conductivity at infinite dilution can be found using Kohlrausch law.

Kohlrausch’s Law-

At infinite dilution when ions are completely dissociated, every ion makes its unique contribution to the molar conductivity of the electrolyte irrespective of the nature of the other ion with which it is associated.

Applications of Kohlrausch’s law-

  • Calculation of molar conductance at infinite dilution for weak electrolytes
  • The degree of dissociation of weak electrolytes is calculated as

α= c⁰

  • Calculation of dissociation constant of weak electrolytes.

K=C(0)2(1-0)

  • Calculation of solubility of sparingly soluble salts.

solubility= κ×1000

Galvanic cells-

Converts chemical energy of spontaneous reaction into electrical energy.

Two half cells -

Cu+2 +2e-→Cu(s) (reduction half cell)

Zn(s)→Zn2++2e- (oxidation half cell)

Overall cell reaction-

Zn(s)+ Cu+2(aq)→ Zn2+(aq)+Cu(s)

Cell potential-

The Potential difference between two electrodes of a galvanic cell is called cell potential.

EMF (electromotive force)-

The potential difference between anode and cathode when no current is drawn through the cell.

Feasibility of a reaction-

Ecell= Eright-Eleft

Reduction Half-2Ag+(aq)+2e2Ag(s)

Oxidation Half-Cu(s)Cu2+(aq)+2e

For the above reaction, the reaction is feasible if

Ecell=EAg+/Ag -ECu+2/Cu is positive.

  • SHE (Standard hydrogen electrode) is assigned a zero potential to determine the potential of individual half cells.

-It is denoted by Pt(s)│H2(g)│H+(aq)

Nernst equation-

For reaction-

Mn+aq+ne-→Ms

E(Mn+│M) =EoM-RTnF ln[M(s)][Mn+(aq)]

E(Mn+│M) =EoM-2.303RTnF log[M(s)][Mn+(aq)]

E(Mn+│M) =EoM-2.303×8.314×298n×96500 log[M(s)][Mn+aq]

E(Mn+│M) =EoM-0.059nlog[M(s)][Mn+(aq)]

Here [M(s)] is taken as zero

E(Mn+│M) =EoM-0.059nlog1[Mn+(aq)]

E=E0-0.059nlog1[Mn+(aq)] at 25⁰C

Applications of Nernst equation-

  1. Determining the cell potential using Nernst equation-

For a chemical reaction-

aA+bB→cC+dD

Ecell=E0cell-RTnFlnQ

At 298K, Ecell=E⁰cell-0.059nlog[C]c[D]d/[A]a[B]b

  1. Determination of concentration of a solution of half-cell-

Using the Nernst equation, the concentration of the unknown species can be found out.

  1. To find equilibrium constant using Nernst equation-

At equilibrium, Nernst equation takes the form of –

E⁰cell=2.303 RTnF logK

Electrochemical cell and Gibbs Energy-

rG=-nFEcell

This equation can help to predict the feasibility of the reaction.

Electrolysis- The process in which chemical changes take place due to the passage of current.

Faraday’s Law of electrolysis-

Faraday’s first law of electrolysis – It says that the quantity of substance settled at the electrode is in direct proportion with the amount the electricity passed through the solution.

w α ZQ

where w is the gram of substance deposited on passing Q coulombs of electricity if a current of 1 ampere is passed for t seconds.

Faraday’s second law of electrolysis- It says that when the equal amount of electricity is passed through different solutions lined up in series, the mass of the substance settled at the electrodes is in direct proportion with the equivalent weight.

Weight of Cu deposited Weight of Ag deposited=Eq. wt. of Cu Eq. wt. of Ag

Batteries-

  • Primary batteries

  1. Dry cells- Found in torches, flashlights, calculators, tape recorders, and many other devices.

Reactions occurring at the electrode are-

Anode

Zn→Zn+2+2e-

Cathode

2NH4+(aq) + 2MnO2+2e-→Zn+2 + 2MnOOH+2NH3

Overall-

Zn+2NH4+ (aq) + 2MnO2 + Zn+2 + 2MnOOH+2NH3

  1. Mercury cell-

-Found in electrical circuits.

-Reactions occurring at the electrodes are-

Anode-

ZnHg+2OH-→ZnOs+H2O+2e-

Cathode-

HgO(s)+ H2O+2e-→Hgl+2OH-

Overall-

ZnHg+ HgO(s)→ ZnO(s)+2OH-

  • Secondary batteries-

  1. Lead storage batteries-

-Battery used in automobiles.

-Reactions taking place at electrodes-

Anode-

Pbs+SO42-(aq)→PbSO4(s)+ 2e-

Cathode-

PbO2(s)+SO42-(aq)+ 4H+ (aq)+ 2e-→PbSO4+2H2O

Overall-

Pb+ PbO2+2H2SO4(aq)→ 2PbSO4+ 2H2O

  1. Nickel-cadmium storage cell-

-Has a longer life than the lead storage battery.

-Reactions occurring at electrodes-

Anode-

Cd+2OH-→CdO + H2O+2e-

Cathode−

2Ni(OH)3 + 2e-→2Ni(OH)2+2OH-(aq)

Overall-

Cd+2Ni(OH)3→CdO+2Ni(OH)2(s) + H2O(l)

Fuel Cells-

  • Features
  • Reactants are supplied continuously.
  • The energy of Combustion of fuels such as H2, CO, CH4, etc. is converted to electrical energy.
  • Reactions taking place at electrodes-

Anode-

2[H2+2OH-(aq)→2H2O+2e-]

Cathode-

O2+2H2O + 4e-→4OH-(aq)

Overall-

2H2(g)+ O2→2H2O

Corrosion-

-Deterioration of metal over time due to its reaction with air and water.

-Except gold, platinum, and palladium all other metals undergo corrosion.

Rusting of iron-

At anode-

[Fe→Fe2+(aq) + 2e-]×2

At cathode-

4H+ + O2 + 4e-→2H2O

Overall reaction-

2Fe + 4H+ + O2→2Fe+2(aq)+ 2H2O

Prevention of corrosion-

  • Barrier protection
  • Sacrificial protection
  • electrical protection

Significance of NCERT Class 12 Chemistry Chapter 3 Notes

Electrochemistry Class 12 notes will help the student to revise and score good marks in the coming 12 board examination. Electrochemistry class 12 notes cover all the important topics of electrochemistry. CBSE class 12 Electrochemistry notes
are also helpful for competitive exams like VITEEE, BITSAT, JEE Main, NEET, etc. .


NCERT Class 12 Notes Chapter-Wise


Subject Wise NCERT Exemplar Solutions


Subject Wise NCERT Solutions


NCERT Books and Syllabus



Frequently Asked Questions (FAQs)

1. What are the topics covered in NCERT Class 12 Chapter 3 note?

Here are the topics covered in NCERT Class 12 chapter 3 note

  • Kohlrausch’s Law-

  • Batteries

  • Fuel Cells-

  • Corrosion

2. What is the weightage of NCERT Class 12 Chemistry chapter 3 in NEET?

In NEET, NCERT Class 12 Chemistry Chapter 3, "Electrochemistry," generally carries a moderate weightage, with approximately 4-6% of the total chemistry questions coming from this chapter

3. What is the weightage of NCERT Class 12 Chemistry chapter 3 in CBSE board exam ?

In the CBSE Class 12 Chemistry board exam, NCERT Chapter 3, "Chemical Kinetics," generally carries a weightage of 7 marks

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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