NCERT Class 12 Chemistry Chapter 3 Notes Electrochemistry - Download PDF

NCERT Class 12 Chapter 2 Electrochemistry: Download PDF

Students can download the NCERT Class 12 Chapter 2 Electrochemistry notes pdf from the icon given below to make your learning simple and effective. These cover important topics like electrochemical cells, Nernst equation, and batteries, helping in quick revision and exam preparation.

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NCERT Notes for Class 12 Chapter 2 Electrochemistry

These notes cover 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. These Class 12 Electrochemistry notes are the best resource for quick revision, also they help build a clear understanding of fundamental principles and their real-life applications, such as batteries and corrosion. Detailed notes of the chapter on electrochemistry are given below.

Conductance of Electrolytic Solutions;

Conductance:

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

The inverse of resistance, R, is called conductance, G:

G=1/R = A/ρl = κA/l (R= ρl/A)

Where R is resistance, l is length, A is an area of cross-section, ρ (rho) is resistivity, and κ(kappa) is conductivity.

Conductivity:

The inverse of resistivity is called conductivity. The SI units of conductivity are S m^-1, but quite often, κ is expressed in S cm^–1.

κ = 1/ρ

The conductivity of a material in S m^–1 is its conductance when it is 1 m long and its area of cross-section is 1 m². It may be noted that 1 S cm^–1.

Molar conductivity:

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

Molar conductivity =${\mathrm{\Lambda }}_m=\frac{\kappa }{C}$

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 an electrolyte of one gram equivalent in a solution.

e= κ×1000/C_eq

Measurement of the Conductivity of Ionic Solutions:

We know that accurate measurement of an unknown resistance can be performed on a Wheatstone bridge.

Wheatstone bridge:

It consists of two resistors, R₁ and R₃, a variable resistor, R₂, and the unknown resistor, R₄. The Wheatstone bridge is fed by an oscillator $\epsilon$ (a source of a.c. power). G is a suitable detector (an electronic device, such as a galvanometer), and the bridge is balanced when no current passes through the detector. Under these conditions:

Unknwon resistance $R_4=\frac{R_2{R}_3}{R_1}$

Once the cell constant and the resistance of the solution in the cell are determined, the conductivity of the solution is given by the equation:

$\kappa$ = cell constant/R =G*/R

Variation of conductivity and molar conductivity with concentration:

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

For strong electrolytes: For strong electrolytes, ${\mathrm{\Lambda }}_m$ increases slowly with dilution and can be represented by the equation:

${\mathrm{\Lambda }}_m={\mathrm{\Lambda }}_m^{{}^{\circ }}-A{c}^{1/2}$

Variation of molar conductivity with concentration for weak electrolytes:

Molar conductivity of weak electrolytes cannot be found for weak electrolytes because the dissociation of weak electrolytes is much lower compared to strong electrolytes.

For weak electrolytes, molar conductivity at infinite dilution can be found using Kohlrausch law.

At any concentration c, if α is the degree of dissociation, then it can be approximated by the ratio of molar conductivity ${\mathrm{\Lambda }}_m$ at the concentration c to the limiting molar conductivity ${\mathrm{\Lambda }}_m^{{}^{\circ }}$. Thus we have:

α = ${\mathrm{\Lambda }}_m$/ ${\mathrm{\Lambda }}_m^{{}^{\circ }}$

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 = ${\mathrm{\Lambda }}_m$/ ${\mathrm{\Lambda }}_m^{{}^{\circ }}$

• Calculation of the dissociation constant of weak electrolytes.

$K_a=\frac{C{\alpha }^2}{1-\alpha }$

• Calculation of the solubility of sparingly soluble salts.

solubility= κ×1000

Electrochemical Cells:

Electrochemical cells are devices that convert chemical energy into electrical energy or vice versa through redox reactions. Students can also refer to NCERT Solutions for Class 12 Chapter 2 Electrochemistry to practise and solve questions from these topics effectively.

Galvanic cells:

Converts the chemical energy of a 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.

Measurement of Electrode Potential: EMF (Electromotive Force):

Emf Of Cell is the potential difference between the anode and the cathode when no current is drawn through the cell.

E_cell= E_right-E_left

Feasibility of a reaction-

E_cell= E_right-E_left

Reduction Half-$2{\mathrm{Ag}}^{+}(aq)+2e^{-}\to 2\mathrm{Ag}(s)$

Oxidation Half-$\mathrm{Cu}(s)\to {\mathrm{Cu}}^{2+}(aq)+2e^{-}$

For the above reaction, the reaction is feasible if

E_cell=E_Ag⁺_/Ag -E_Cu⁺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)│H_2(g)│H⁺_(aq)

Nernst equation:

For reaction-

Mn⁺_aq+ne^-→M_s

E(Mn⁺│M) =E_oM-RT/nF ln[{M(s)}/{Mn⁺(aq)}]

E(Mn⁺│M)=E_oM-2.303RT/nFlog[{M(s)}/{Mn⁺(aq)}]

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

E(Mn+│M) =E_oM-0.059nlog[{M(s)}/{Mn⁺(aq)}]

Here [M_(s)] is taken as zero

E(Mn⁺│M) =E_oM-0.059nlog[1/{Mn⁺_(aq)}]

E=E₀-0.059nlog[1/{Mn⁺_(aq)}] at 25⁰C

Applications of the Nernst equation-

1. Determining the cell potential using the Nernst equation-

Equilibrium Constant from Nernst Equation:

For a chemical reaction-

aA+bB→cC+dD

E_cell = E₀cell - (RT/nF) ln(Q)

$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0591}{n}\log \left(\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}\right)$
This is the Nernst equation at 298 K, where:
- $E_{\text{cell }}$ is the cell potential under non-standard conditions
- $E_{\text{cell }}^{\circ }$ is the standard cell potential
- $n$ is the number of moles of electrons transferred
- $[A],[B],[C],[D]$ are concentrations of the respective species
- $a,b,c,d$ are the stoichiometric coefficients in the balanced redox equation.

2. Determination of the concentration of a solution of a half-cell

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

3. To find the equilibrium constant using the Nernst equation-

At equilibrium, the Nernst equation takes the form of –

E⁰cell=2.303 (RT/nF)log(K)

Electrochemical Cell and Gibbs Energy of the Reaction:

∆_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:

There are two Faraday’s Laws Of Electrolysis, Faraday’s first law of electrolysis says that the quantity of substance deposited at the electrode is in direct proportion to the amount the electricity passed through the solution.

$w\propto ZQ$

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 an equal amount of electricity is passed through different solutions lined up in series, the mass of the substance deposited at the electrodes is in direct proportion to the equivalent weight. Students can also download these NCERT Class 12 Chapter 2 Electrochemistry notes pdf to study offline anytime and anywhere.

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

2. 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+SO₄^2-_(aq)→PbSO_4(s)+ 2e^-

Cathode-

PbO₂(s)+SO₄^2-(aq)+ 4H⁺ (aq)+ 2e^-→PbSO₄+2H₂O

Overall-

Pb+ PbO₂+2H₂SO₄(aq)→ 2PbSO₄+ 2H₂O

2. 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-

2H_2(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

Electrochemistry Previous Year Questions and Answers

Given below previous years NCERT Chapter 2 Electrochemistry questions and answers to help you understand important concepts and exam trends. Practising these will improve your accuracy.

Question 1. Give reason: In the experimental determination of electrolytic conductance, Direct Current (DC) is not used.

Answer:

Direct Current (DC) is not used in the experimental determination of electrolytic conductance because it causes electrolysis and polarization at the electrodes.

Explanation:
1. Electrolysis occurs with DC:
Because DC causes continuous movement of ions in one direction, this leads to chemical changes at the electrodes, i.e., electrolysis, which interferes with accurate conductance measurement.
2. Polarization of electrodes:

• Ions accumulate at the electrodes, forming concentration gradients or even gas bubbles (like ${\mathrm{H}}_2$ or ${\mathrm{O}}_2$).
• This creates an extra resistance called electrode polarization, which distorts the actual conductance reading.

3. AC prevents these issues:

• Alternating Current (AC) changes direction rapidly, so ions oscillate rather than accumulate.
• This prevents electrolysis and minimizes polarization, allowing accurate and stable measurement of conductance.

DC is avoided because it causes electrolysis and electrode polarization, which interfere with the correct determination of electrolytic conductance. AC is used instead for accurate results.

Question 2. Define the following: Fuel cell

Answer:

A fuel cell is an electrochemical device that converts the chemical energy of a fuel, typically hydrogen, and an oxidizing agent, usually oxygen, directly into electricity through electrochemical reactions.

Unlike batteries, fuel cells can continuously generate electricity as long as they have a supply of fuel and oxidant. They are known for their efficiency and clean operation, producing only electricity, water, and heat when hydrogen is used as the fuel. Fuel cells are used in various applications, including transportation and stationary power generation.

Question 3. The electrical resistance of a column of 0.05 M NaOH solution of area $0.8{\mathrm{cm}}^2$ and length $\mathbf{40}\mathrm{cm}$ is $5\times {10}^3\mathrm{ohm}$. Calculate its resistivity, conductivity, and molar conductivity.

Answer:

We are given:
- Concentration of NaOH solution: $C=0.05\mathrm{mol}/\mathrm{L}$
- Area of cross-section: $A=0.8{\mathrm{cm}}^2$
- Length of solution column: $l=40\mathrm{cm}$
- Resistance: $R=5\times {10}^3\mathrm{\Omega }$

We are to calculate:
1. Resistivity ( $\rho$ )
2. Conductivity ( $\kappa$ )
3. Molar conductivity ( ${\mathrm{\Lambda }}_m$ )

1. Resistivity ( $\rho$ )

Resistivity is given by:

$\rho =R\cdot \frac{A}{l}$
Substitute values:

$\rho =(5\times {10}^3)\cdot \frac{0.8}{40}=(5\times {10}^3)\cdot 2\times {10}^{-2}=100\mathrm{\Omega }\cdot \mathrm{cm}$

2. Conductivity ( $\kappa$ )

$\kappa =\frac{1}{\rho }=\frac{1}{100}=0.01\mathrm{S}{\mathrm{cm}}^{-1}$

3. Molar conductivity ( ${\mathrm{\Lambda }}_m$ )

${\mathrm{\Lambda }}_m=\frac{\kappa \times 1000}{C}$

${\mathrm{\Lambda }}_m=\frac{0.01\times 1000}{0.05}=200\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$
Hence, the answer is resistivity is 100 ohm cm, conductivity is 0.01 Scm^-1, and molar conductivity is (200 Scm² mol^-1).

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