NCERT Class 12 Chemistry Chapter 3 Notes Electrochemistry - Download PDF

Electrochemistry is one of the important branches of chemistry that deals with the relationship between electrical energy and chemical reactions. It 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. Many industries depend on Electrochemistry for the refining of metals, wastewater treatment, and impurity removal. These notes 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.

NCERT notes will be helpful for a quick revision of topics. These notes are designed by our subject experts, which ensures the credibility of the content provided. It contains 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. It becomes difficult and time-consuming for students to read the NCERT book texts point-to-point. So, to solve this problem, we are providing these notes that cover all the topics and concepts provided in the NCERT textbook in a very clear and comprehensive way.

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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 an electrolyte of one gram equivalent in a solution.

e= κ×1000/C_eq

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 - AC^1/2

• 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 the anode and the cathode when no current is drawn through the cell.

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-RTnF ln[M(s)][Mn⁺(aq)]

E(Mn⁺│M) =E_oM-2.303RTnF log[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-0.059nlog1[Mn⁺_(aq)]

E=E₀-0.059nlog1[Mn⁺_(aq)] at 25⁰C

Applications of the Nernst equation-

1. Determining the cell potential using the Nernst equation-

For a chemical reaction-

aA+bB→cC+dD

E_cell = E₀cell - RTnF lnQ

$$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0591}{n}\log \left(\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}\right)$$

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 out.

3. To find equilibrium constant using Nernst equation-

At equilibrium, the 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 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.

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

3. 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

Most Important Questions of Class 12 Chemistry Chapter 2

Question: The molar conductance of an infinitely dilute solution of ammonium chloride was found to be $185\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and $70\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^1$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is $85.5\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$, its degree of dissociation is given by $\mathrm{x}\times {10}^{-1}$. The value of $x$ is _______ (Nearest integer)

Answer:

$\begin{array}{rl}& {\lambda }_{\mathrm{m}}^{\prime \prime }\text{ of }{\mathrm{NH}}_4\mathrm{Cl}=185\\ & {\left({\lambda }_{\mathrm{m}}^{\circ }\right)}_{{\mathrm{NH}}_4}+{\left({\lambda }_{\mathrm{m}}^{\circ }\right)}_{{\mathrm{Cl}}^{-}}=185\\ & {\left({\lambda }_{\mathrm{m}}^{\circ }\right)}_{{\mathrm{NH}}_4}=185-70=115{\mathrm{Scm}}^2{\mathrm{mol}}^{-1}\\ & {\left({\lambda }_{\mathrm{m}}^{\circ }\right)}_{{\mathrm{NH}}_4\mathrm{OH}}={\left({\lambda }_{\mathrm{m}}^{\circ }\right)}_{{\mathrm{NH}}_4^{\circ }}+{\left({\lambda }_{\mathrm{m}}^{\circ }\right)}_{{\mathrm{OH}}^{-}}\\ & =115+170\\ & {\left({\lambda }_{\mathrm{m}}^0\right)}_{{\mathrm{NH}}_4\mathrm{OH}}=285{\mathrm{Scm}}^2{\mathrm{mol}}^{-1}\\ & \text{ degree of dissociation }=\frac{{\left({\lambda }_{\mathrm{m}}\right)}_{{\mathrm{NH}}_4\mathrm{OH}}}{{\left({\lambda }_{\mathrm{m}}^{\circ }\right)}_{{\mathrm{NH}}_4\mathrm{OH}}}\\ & =\frac{85.5}{285}\\ & =0.3\\ & =3\times {10}^{-1}\end{array}$

Hence, the answer is 3.

Question: The correct order of limiting molar conductivity for cations in water at 298 K is

1) ${\mathrm{H}}^{+}>{\mathrm{Na}}^{+}>{\mathrm{K}}^{+}>{\mathrm{Ca}}^{2+}>{\mathrm{Mg}}^{2+}$

2) ${\mathrm{H}}^{+}>{\mathrm{Ca}}^{2+}>{\mathrm{Mg}}^{2+}>{\mathrm{K}}^{+}>{\mathrm{Na}}^{+}$

3) ${\mathrm{Mg}}^{2+}>{\mathrm{H}}^{+}>{\mathrm{Ca}}^{2+}>{\mathrm{K}}^{+}>{\mathrm{Na}}^{+}$

4) ${\mathrm{H}}^{+}>{\mathrm{Na}}^{+}>{\mathrm{Ca}}^{2+}>{\mathrm{Mg}}^{2+}>{\mathrm{K}}^{+}$

Answer:

The limiting molar conductivities of ions are important in determining how well they conduct electricity in solution. Here are the values for some common cations at 298 K :
${\mathrm{H}}^{+}:349.8\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$
${\mathrm{Na}}^{+}:50.11\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$
${\mathbf{K}}^{+}:73.52\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$
${\mathbf{C}\mathbf{a}}^{\mathbf{2}\mathbf{+}}:119\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-\mathbf{1}}$
${\mathbf{M}\mathbf{g}}^{\mathbf{2}\mathbf{+}}:\mathbf{1}\mathbf{0}\mathbf{6}\mathbf{.}\mathbf{1}\mathbf{2}\mathrm{S}{\mathrm{cm}}^{\mathbf{2}}{\mathrm{mol}}^{-\mathbf{1}}$

Therefore correct order of limiting molar conductivity of cations will be -$\{{\mathrm{H}}^{+}>{\mathrm{Ca}}^{+2}>{\mathrm{Mg}}^{+2}>{\mathrm{K}}^{+}>{\mathrm{Na}}^{+}$}

Hence, the correct answer is option (2).

Approach to Solve Questions of Class 12 Chemistry Chapter 2 Electrochemistry

The following are the points that can help you build a good approach to solve the questions effectively

1. Know the key terms and concepts
This chapter has a lot of key terms and concepts that needs to be revised regularly.
Electrolytic vs. Galvanic cells – identify type of cell, electrodes and direction of electron flow. Generally, oxidation occurs at anode and reduction at cathode.
Salt Bridge – completes the circuit and maintains charge neutrality.

2. Standard Electrode Potentials
Use electrochemical series to predict feasibility and direction of redox reactions.
Standard EMF of Cell-
$$E_{\text{cell }}^{\circ }=E_{\text{cathode }}^{\circ }-E_{\text{anode }}^{\circ }$$

3. Apply the Nernst Equation
Use it when conditions are not standard (i.e., concentrations $\ne 1\mathrm{M}$ ):
$$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0591}{n}\log \left(\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}\right)$$

4. Electrolysis and Faraday's Laws
First Law: $m=\frac{ZIt}{1000}$
Second Law: masses of different substances deposited are proportional to their equivalent weights.
Use molar mass and n-factor to calculate Z (electrochemical equivalent).
$n$ : number of electrons
Concentrations must be in mol/L

6. Conductance and Kohlrausch’s Law
Try to understand the key terms and the effect of dilution-
Conductivity (κ): decreases with dilution
Molar Conductivity (Λm): increases with dilution

Use Kohlrausch’s law for
Determining molar conductivity at infinite dilution (Λ⁰m)
Calculating the degree of dissociation

7. Solve Numericals Step-by-Step
Keep all units consistent.
Always write what's given, what's asked and apply the formula.
Include proper values for constants like -
F=96500 C/mol

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