CBSE Chemistry Class 12 Electrochemistry Notes (Chapter-3)

www.free-education.in is a platform where you can get pdf notes from 6th to 12th class notes, General Knowledge post, Engineering post, Career Guidelines , English Speaking Trick , How to crack interview and lots more. ( Class 12 Electrochemistry Notes )

free education

Class 12 Electrochemistry , In this post We will Study, Electrochemistry Notes & Important Topic for Student to excel in exam.

NCERT/CBSE class 12th Chemistry notes provided by free-education.in (Wisdom Education Academy).

Here We are providing all subject wise pdf notes to student for their help to get good marks in exam.

In this post you will get Download CBSE 2020-21 Chemistry PDF notes given below by free-education.in to excel in the exam.

www.free-education.in is a platform where you can get pdf notes from 6th to 12th class notes, General Knowledge post, Engineering post, Career Guidelines , English Speaking Trick , How to crack interview and lots more.

Class 12 Electrochemistry

Introduction

  • Electrochemistry refers to the conversion of chemical energy to electrical energy and vice versa. It is basically the study of Production of electricity from energy released during spontaneous reaction and use of electrical energy to bring about non-spontaneous chemical transformation.
  • A spontaneous chemical reaction is a reaction which happens on its own and releases free energy. This reaction produces electric energy from chemical reaction. For example, burning of coal, rusting of iron, melting of ice, etc.
  • On the other hand non-spontaneous reaction occurs by providing an external source like electricity. For example, Hydrolysis of water.
  • Electrochemistry is used for the following purposes.
  • Production of metals like sodium hydroxide, chlorine, fluorine and many other chemicals.
  • It is also used for purification of metals.
  • The process is used in batteries as well as fuel cells which converts the chemical energy into electrical energy and is used in several instruments and devices.
  • This process is used in electroplating.
  • The reactions carried out using the process of electrochemistry are energy effective and less polluting.

Download Previous Year Question Papers..

Daniel cell

  • The cell that converts the chemical energy liberatedas a result of redox reaction to electrical energy is called a Daniel cell.
  • It has anelectrical potential of 1.1 V.
Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes
  • The setup for Daniel cell is as follows:
    • In a beaker a plate of zinc is dipped in a solution of zinc sulfate (ZnSO4).
    • In another beaker a plate of copper is dipped in a solution of copper (II) sulfate in another container. These plates of metal are called the electrodes of the cell.
    • These electrodes behave as terminal to hold the electrons.
    • The two electrodes are connected via wire.
    • A salt bridge is placed between the two beakers. This provides a path for the movement of ions from one beaker to the other in order to maintain electrical neutrality.
    • Zinc electrode gets oxidized and hence releases electrons that flow through the wire towards the copper electrode.
    • The copper (II) sulfate solution releases copper ions Cu2+.                                                                                                                                                                                                                                          
    • At the anode:

      Oxidation —————- loss of electrons.
      Zn –> Zn2+ + 2e

      At cathode,

      Reduction ————-gain of electrons.

      Cu2+ + 2e –> Cu
    • Zinc atoms being more reactive have a greater tendency to lose electrons than that of copper.
      The electrons in this cell moves from zinc anode to copper cathode through the wire connecting the two electrodes in the external circuit
    • A bulb placed within this circuit will glow and a voltmeter connected within this circuit will show deflection.
    • The net reaction of this cell is the sum of two half-cell reactions.
      Zn(s) + Cu2+ (aq) –> Zn2+ (aq) + Cu(s)


In a Daniel cell a salt bridge is placed between the two beakers containing a solution of zinc sulfate (ZnSO4) and a solution of copper (II) sulfate respectively. This provides a path for the movement of ions from one beaker to the other in order to maintain electrical neutrality.

Salt Bridge

Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes
  • Salt bridge is an inverted U-tube like structure.
  • The tube is filled with concentrated solution of an inert electrolyte.
  • The electrolyte does not undergo any sort of chemical reaction nor does it reacts with concentrated solution in the two half cells.
  • Most of the time salts like KCl, KNO3, NH4NO3 are used as electrolyte.
  • Salt bridge is constructed by filing the U-tube with a mixture prepared by mixing agar-agar or gelatin with a hot concentrated solution of the electrolyte.
  • The apparatus is then cooled and consequently the solution fixes within the U-tube in the form of gel.
  • This prevents the intermixing of fluids.
  • The either ends of the tube are then sealed using cotton wool in order to reduce diffusion.
  • The main advantage of suing this bridge is that it prevents the arousal of potential difference (also called liquid junction potential) when two solutions are in contact with each other.
  • It completes the electrical circuit as it connects the electrolytes in the two half cells.
  • It avoids the chances of diffusion of solutions between the half cells.
  • It maintains electrical neutrality by providing a path for the movement of ions.

Measurement of electrode potential in a cell

  • The electrode is connected to a standard hydrogen electrode (SHE) to constitute a cell.
  • The electrode forming the negative terminal of the cell is allotted negative value of electrode potential whereas the electrode forming the positive terminal of the cell is allotted a positive value of electrode potential.
  • The potential difference developed between the two terminals is measured using a potentiometer.
  • The direction of the flow of electric current in the external circuit is identified using a galvanometer.
  • This enables us to identify the positive and the negative terminal of the set up as the current flows from positive terminal to negative terminal.
  • This in turn will help us mark the anode and the cathode electrodes due to the fact that the electrons will flow from anode to cathode.
Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes

Ecell = Ecathode – Eanode

  • In case of Daniel cell

At the anode:

Oxidation —————- loss of electrons.

Zn –> Zn2+ + 2e

At cathode,

Reduction ————-gain of electrons.

Cu2+ + 2e  –> Cu

The net reaction of this cell is the sum of two half-cell reactions.
Zn(s) + Cu2+ (aq) –> Zn2+ (aq) + Cu(s)

Emf of the cell = Ecell = Ecathode – Eanode

                                      = 0.34 V – (- 0.76) V = 1.10 V

(Measured Emf of Cu is 0.34 V and that of Zn is 0.76 V).  

Standard hydrogen electrode

  • The electrode is connected to a standard hydrogen electrode (SHE) to constitute a cell
  • It consists of a platinum electrodecoated with a layer of platinum black.
Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes
  • The electrode is immersed in an acidicsolution and the pure hydrogen gas is bubbled through it.
  • Theconcentration of the reduced form and the oxidized form of hydrogen issustained at unity with following conditions:
  • Pressure of hydrogen gas = 1 bar
  • Concentration of hydrogen ion in the solution = 1 molar

Ecell = Ecathode – Eanode

Ecell = Ecathode – 0 = Ecathode

  • The measured Emf of the cell:

Pt| H2 (1 bar)| H+ (1M) || Cu2+ (1M)| Cu is 0.34 V.

The positive value of the standard electrode potential signifies the easy reduction of Cu2+ ions than H+ ions.

  • The measured Emf of the cell

Pt| H2 (1 bar)| H+ (1M) || Zn2+ (1M)| Zn is -0.76 V.

The negative value of the standard electrode potential signifies that the hydrogen ions oxidizes the zinc (or it can be said that zinc can reduce hydrogen ions).

  • An electrode with standard electrode potential greater than zero is stable in its reduced form compared tohydrogen gas.
  • Whereas an electrode with negative standard electrode potential is less stable in its reduced form compared to hydrogen gas.
  • This decreases the standard electrodepotential which in turn decreases the oxidizing power ofthe specific electrode on the left and increases the reducing power of the electrodeto the right of the reaction.

Nernst Equation

  • This equation was named after a German physicist Walther Nernst.
Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes
  • The Nernst Equation empowers the assurance of cell potential under non-standard conditions and relates the measured cell potential to the reaction quotient and permits the exact measurement of equilibrium constants.
  • Let us consider an electrochemical reaction of the following type:

aA +bB  –> cC + dD

  • Nernst equation for this can be written as follows:
  • If the circuit in Daniel cell is closed:

Zn(s) + Cu2+ (aq) →Zn2+ (aq) + Cu(s)

Class 12 Electrochemistry Notes

Problem:

Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag+ (0.002 M) → Ni2+ (0.160 M) + 2Ag(s). Given that Eøcell = 1.05 V.

Electrochemical Cell and Gibbs Energy of the Reaction

Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes
  • Electrical work done (1 second) = Electrical potentialX Total charge passed
  • Passing the charges reversibly through the galvanic cell results in maximum work.
  • Reversible work done by galvanic cell = Decrease in Gibbs energy
  • Let E = Emf of the cell

           nF = Amount of charge passed

        Δr= Gibbs energy of the reaction

        ΔrG = -nFEcell

  For the reaction,

           Zn(s) + Cu2+ (aq) –> Zn2+ (aq) + Cu(s)

            [ΔrG = -2FEcell  ]

But when the equation becomes

2Zn(s) + 2Cu2+ (aq) –> 2Zn2+ (aq) + 2Cu(s)

  [ΔrG = -4FEcell  ]

Problem:

The cell in which the following reactions occurs:

Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes

Electrical resistance

  • It is denoted by the symbol R’.
  • It is measured in a unit called ohm (Ω) with the help of a Wheatstone bridge
  • The electrical resistance is:
  1. Directly proportional to its length, l
  2. Inversely proportional to its area of cross section, A.
Class 12 Electrochemistry Notes

Electrical conductance

Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes
  • The free ions Na+ and Cl present in the solution are responsible for the conductance in a solution.
  • The inverse of resistivity is termed as conductivity.
  • Ω = 1/k
  • It is represented by the symbol (Greek, kappa).
  • It is measured in a unit called Sm–1.
  • When length = 1m.

Cross sectional area = 1m2

            Then conductivity becomes the conductance.

The conductivity of an electrolytic solution depends on:

1.Nature of the electrolyte added
2.Size of the ions produced and their solvation
3.Nature of the solvent and its viscosity
4.Concentration of the electrolyte
5.It increases with the increase of temperature.
6.Pressure

  • Matters can be classified into conductors, insulators andsemiconductors depending on the magnitude of their conductivity.

Conductors:

  • Solids with conductivities ranging between 104 to 107 ohm–1m–1 are conductors.
  • Metals have conductivities in the order of 107 ohm–1m–1 is good conductors.
  • For example, Iron, Copper, Aluminum.
Class 12 Electrochemistry Notes

Fig. Wood and plastics are also solids but are insulators

Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes

A man touching the electric pole with a metal rod will get an electric shock because metal rod is a conductor whereas a man touching the same with a wooden plank will be safe because wood is an insulator.

Semiconductors:

  • Solids with conductivities in the intermediate range from 10–6 to 104 ohm–1 m–1.
  • For example, Gallium, Germanium, Silicon

Measurement of resistance of a solution of an electrolyte

Class 12 Electrochemistry Notes

PROBLEM:

THE RESISTANCE OF A CONDUCTIVITY CELL CONTAINING 0.001M KCL SOLUTION AT 298 K IS 1500 Ω. WHAT IS THE CELL CONSTANT IF CONDUCTIVITY OF 0.001M KCL SOLUTION AT 298 K IS 0.146 X 10-3 S CM-1.

SOLUTION:

Conductivity, k = 0.146 × 10 – 3 S cm – 1

Resistance, R = 1500 Ω

Cell constant = k × R = 0.146 × 10 – 3 × 1500 = 0.219 cm – 1

Arrangement for measurement of Resistance

Class 12 Electrochemistry Notes
  • The arrangement consists of two resistances R3 and R4.
  • There is a variable resistance R1 and a conductivity cell with unknown resistance R2.
  • The Wheatstone bridge is provided with an oscillator O that acts as source of a.c. power.
  • The arrangement has a suitable detector P.
  • The Wheatstone bridge is balanced when there is no flow of current through the detector.

Unknown resistance = R2 = R1R4 /R3

  • After calculating the resistance the conductiviry can be easily calculated using the formula:
  • κ = G’ /R

Molar conductivity

  • It is denoted by the symbolIt is related to the conductivity of the solution by the following equation:
Class 12 Electrochemistry Notes
  • The units of is S mmol-1.

Problem:

THE CONDUCTIVITY OF 0.20 M SOLUTION OF KCL AT 298 K IS 0.0248 SCM-1. CALCULATE ITS MOLAR CONDUCTIVITY.

Solution:

k = 0.0248 S cm – 1

c = 0.20 M

Molar conductivity,Λm =  (k x 1000) / c

= 0.0248 x1000 / 0.20

= 124 Scmmol – 1

Variation of Conductivity and Molar Conductivity with Concentration

  • They depend on the concentration of the electrolyte. The Conductivity and Molar Conductivity of both weak and strong electrolytes decreases withdecrease in concentration as the number of ions per unitvolume carrying the current in a solution decreases on dilution.
  • Conductivity of a solution at a specific concentration = Conductance of solution placed in between the two platinum electrodes where
  • Volume of solution = 1 unit

 Cross sectional area of electrodes = 1unit

Class 12 Electrochemistry Notes

Strong electrolytes

  • A solute or substances that completely ionize or dissociates in a solution are known as strong electrolyte. These ions are good conductors of electricity in the solution.
  • For example, HCl, HBr, HI, HNO3, NaOH, KOH, etc.
  • For strong electrolytes, Λm  increases slowly with dilution and can berepresented by the equation:
Class 12 Electrochemistry Notes

Kohlrausch law of independent migration of ions

  • The Kohlrausch law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
Class 12 Electrochemistry Notes

Weak electrolytes

  • A substance which forms ions in an aqueous solution do not dissociate completely at moderate concentrations is known as weak electrolyte.
  • For example, HC2H3O2 (acetic acid), H2CO3 (carbonic acid), etc.
  • The conductance of the solution increases with dilution of solution.
  • The interionic forces of attraction are not strong at a low concentration.
  • The slope for Λm vs c1/2 is not linear even at a lower concentrations.
Class 12 Electrochemistry Notes
  • These electrolytes have lower degree of dissociation at higher concentrations.
  • The value of Λchanges with dilution due to increase in the degree of dissociation.
  • Ëm increases sharply on dilution exclusively at lower concentrations.
  • At infinite dilutionwhen concentration approaches to zero, the electrolyte dissociates completely. But at lower concentration the conductivity of a solution is low to an extent that cannot be even measured.

Electrolysis

Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes

Fig. Electrolysis

  • Two copper strips are dipped in an aqueous solution ofcopper sulphate (CuSO4).
  • A DC voltage is applied to both the electrodes.
  • Cu 2+ ions discharge at the cathode

Cu2+(aq) + 2e –> Cu (s)

  • Copper metal is deposited on the cathode and copper is converted into Cu2+ ions at the anode by the following reaction:

Cu –> Cu2+(aq) + 2e

  • The impure copper is the anode that dissolves on passing current whereas the pure copper gets deposited at the cathode. Metals like Na, Mg, Al, etc. are produced using electrochemical reduction of their respective cations.

Faraday’s Laws of Electrolysis

  • Michael Faraday described the quantitative aspects of electrolysis and came forward with two laws of electrolysis:
  • 1st Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).
  • 2nd Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights
  • Mathematically, Atomic Mass of Metal ÷ Number of electrons required to reduce the cation.

It
where Q 
is in coloumb

is in ampere and

is in second.

  • This quantity of electricity is known as Faraday and is represented by the symbol F.
Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes
  • In the above example application of more current results in the deposition of more amount of copper from the anode to the cathode.

PROBLEM:

IF A CURRENT OF 0.5 AMPERE FLOWS THROUGH A METALLIC WIRE FOR 2 HOURS, THEN HOW MANY ELECTRONS WOULD FLOW THROUGH THE WIRE?

Solution:

I = 0.5 A

t = 2 hours = 7200 s

By using the formula, Q = It

= 0.5 A × 7200 s

= 3600 C

We know that 96487C = 6.023 X 1023 number of electrons.

Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes

Products of Electrolysis

  • The product of electrolysis produced depends on the nature of material being that is being electrolyzed as well as the type of electrodes that is being used.
  • An inert electrode e.g., platinum or gold does not participate in chemicalreaction and acts as a source or sink for electrons.
  • Whereas a reactive electrode participates in the electrode reaction.
  • It also depends on the different oxidizingas well as reducing species that are present in the electrolytic cell and their standardelectrode potentials.
  • Electrolysis of molten NaCl results in the production of sodium metal and Cl2
Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes

Problem:

Given the standard electrode potentials,

K+/ K = – 2.93V, 

Ag+/ Ag = 0.80V,

Hg2+ / Hg = 0.79V

Mg2+ / Mg = – 2.37 V,

Cr3+ / Cr = – 0.74V 

Arrange these metals in their increasing order of reducing power.

Solution:

Lower the reduction potential leads to higher reducing power. The given standard electrode potentials increases in the following order:

K+/ K < Mg2+ / Mg < Cr3+ / Cr < Hg2+ / Hg < Ag/ Ag.

Hence, reducing power of the given metals increases in the following order:

Ag < Hg < Cr < Mg < K

Battery:

Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes

Primary battery:

  • The reaction in a primary battery occurs only once.
  • The battery becomes dead after used onceand cannot be reused
  • Example of primary cell is dry cell and mercury cell.
  • The dry cell consists of a zincanode and the carbon (graphite) cathode surrounded by powdered manganesedioxide and carbon.

Secondary Battery

  • The reaction in a secondary battery occurs many times.
  • Once it exhausts it can be recharged and used again.
  • They are recharged by passing electric current through it in opposite direction.
  • Example of secondary cell is lead and nickel cadmium cell.
Class 12 Electrochemistry Notes

Fig. Nickel-Cadmium cell

  • It is used extensively in automobiles and invertors.
  • The lead cell contains a lead anode and a lead dioxide (PbO2) cathode and sulphuric acid is used as an electrolyte.
    Anode: Pb(s) + SO42–(aq) –> PbSO4(s) + 2e
    Cathode: PbO2(s) + SO42–(aq) + 4H+(aq) + 2e –> PbSO4 (s) + 2H2O (l)
  • The net cell reaction is:
    Pb(s)+PbO2(s)+2H2SO4(aq) –> 2PbSO4(s) + 2H2O(l)
  • While charging, the reaction within the battery is reversed and PbSO4(s) on anode and cathode is converted into Pb and PbO2, respectively.
  • It has a longer life thanlead storage cell and it requires more expenses to
  • The net reaction is:
    Cd (s)+2Ni(OH)3 (s) –> CdO (s) +2Ni(OH)2 (s) +H2O(l)

Fuel cells

Class 12 Electrochemistry Notes
Class 12 Electrochemistry Notes

Cathode: O2 (g) + 2H2O (l) + 4e –> 4OH(aq)
Anode: 2H2 (g) + 4OH(aq) ⎯ –> 4H2O(l) + 4e

  • The net reaction is:
    2H2(g) + O2(g) →2 H2O(l)

Corrosion

We all must have observed newly bought iron, silver or coper articles appears very shiny but with passage of time they get dull. This is due to the layer of metal oxide that develops on their surface. Rusting of iron, silver jewellery getting tarnished or copper articles getting covered by green layer. Metals react with atmospheric oxygen and produces metal oxides that are basic in nature because they react with water to form bases.

  • In case of rusting of iron, the iron reacts with the oxygen present in air and moisture and develops rust (hydrated iron (III) oxide).

Fig. Copper developing green coloured rust on exposure to moist air

  • In case of tarnishing of silver articles, the metallic silver reacts with hydrogen sulphide or sulphur present in air and gets tarnished.

Download Previous Year Question Papers..

Download pdf notes

Leave a Reply

Your email address will not be published. Required fields are marked *