only for class xii students

1. UNIT:3(Electro Chemistry)
(L.D.SINGH )VKV-ROING
Study of production of electricity from energy release during
spontaneous chemical reaction and use of electrical energy to
bring a non spontaneous chemical transformations is called
Electro chemistry.
Electrochemical cell: This cell convert the chemical energy
liberated during a chemical reaction to electrical energy.
E.g, Galvanic cell or voltaic cell.
Electrolytic cell: This cell convert the electrical energy to
carry out a chemical reaction to give chemical energy.

2. Electrochemical cell

3. Daniell cell: It convert chemical energy to electrical potential of 1.1 V and it is also called as
Voltaic cell or Galvanic cell.
It is made by dipping Zn solid in ZnSO4 and Cu in CuSO4 to give a redox reaction
Zn(s) + CU2+
(aq) → Zn2+
(Aq) +CU(s)
If external opposite potential is applied and increase slowly the reaction continue till
the opposing voltage reaches to 1.1V When the reaction stop ,no current flow through
the circuit. Further increase in potential start reaction oppositely to give an electrolytic
cell. When EExt=1.1 V,then (i) No flow of current (ii) No chemical reaction
When EExt<1.1 V ,then (i)flow of electron from Zn to Cu thus current flow from Cu to
Zn (ii) Zn dissolved at anode and Cu deposited at cathode.
When EExt>1.1 V ,then (i)flow of electron from Cu to Zn thus current flow from Cu to Zn
(ii) Cu dissolved at Cu electrode and Zn deposited at Zn electrode.
• Galvanic cell : The Gibbs energy of a spontaneous redox reaction is converted into
electrical work. In the redox reaction: Zn(s) + CU2+
(aq) → Zn2+
(Aq) +CU(s) which can
be divide into two half
• (i) Cu2+ + 2e- →Cu (reduction half cell reaction)
• (ii)Zn(S) →Zn2+ +2e- (oxidation half cell reaction
• The electrolyte of the 2 half cell are connected by a salt bridge(if electrolyte are
different if not no need of salt bridge)

4. At the electrolyte - electrolyte interface there is a tendency of metal ion from the
solution to deposit on metal electrode trying to make its +ve charge. At the same
time,metal atom of the electrode have a tendency to go into the solution as ions and
leave behind the electrons at the electrode trying to make a –ve charged . At
equilibrium, There is a separation of charge depending on tendencies of two opposing
reaction which bring a potential difference between the electrode and electrolyte
which is called electrode potential . When the concentration of of all species is unity
then electrode potential is called as standard electrode potential. Half cell which
undergo oxidation is anode and it has –ve potential w.r.t solution where as the one
which undergo reduction is called cathode. and it has +ve potential w.r.t solution .The
electron flow from –ve to +ve electrode while current flow from +ve to –ve.
It is measured in volt. The cell potential is the difference between electrode potential
of cathode and anode and called as emf when no current is passed in the circuit. By
convention anode is on left while cathode is on right. A Galvanic cell is represented by
putting vertical line between metal and electrolyte solution and a parallel line to
indicate salt bridge. For the above cell:
Zn(S) Zn2+
(Aq) Cu2+
(Aq) Cu(S)
E cell= E right– E left

5. Measurement of electrode potential
•Standard Hydrogen electrode:
Pt(s) H2(g) H+(aq) is called as standard hydrogen electrode and
is assigned as zero potential at all temperature corresponding
to the reaction, H+ (Aq) +e- → H2(g)
The standard hydrogen electrode consist of pt –electrode
coated with platinum black and dipped at acidic solution and pure
hydrogen gas bubbled through it . The concentration of both
reduce and oxidized hydrogen is maintain as 1.The pressure of
hydrogen gas is 1 bar and concentration of H+ in the
solution is 1 molar. At 298K the e m f of the cell, SHE Second half cell by taking SHE
as anode and the other as cathode. If concentration of oxidized and reduced form of
species in RHS are unity, then the cell potential is equal to standard electrode
potential E0
R of the given half cell.
E0= E0
R - E0
L = E0
R-0= E0
R
The + ve value of E0
R indicates that metal ion get reduced easily than H+ ions. In case of
cu, E0
R is +ve thus H+ cannot oxidize cu i.e +0.34V.Hydrogen gas can reduce cu ion
under standard condition. Thus Cu does not dissolve in HCl. But in nitric acid it is
oxidized to nitrate ion and not Hydrogen ion. The - ve value of E0
R indicates that
Hydrogen ion can oxidized Zn or Zn can reduced Hydrogen ion as Standard potential
with SHE and Zn half cell is – ve ie,-0.76V
Pt or gold are also used as inert electrode as they do not participate in the reaction
but provide their surface for oxidation or reduction and for conduction of electrons.

6. If Standard electrode potential of an electrode is
greater than Zero then its reduced form is more stable
compared to Hydrogen gas. If Standard electrode
potential of an electrode is less than Zero then
Hydrogen gas is more stable than reduced form of the
species.F2 gas has maximum tendency to get reduced
to F- thus fluorine gas is the strongest oxidizing agent
and fluoride ion is the weakest reducing agent.
Lithium is the lowest electrode potential indicating
that lithium ion is the weakest oxidizing agent while
lithium metal is the strongest reducing agent in
aqueous solution.

7. • Nernst equation:
For a reaction : Mn+(aq) + ne- → M(s)
The electrode potential at any concentration measured wrt SHE can be represented as
E(M
n+/
M) = E0
(Mn+/M) - RT ln 1
nF [Mn+]
Here R =8.314JK-1mol-1, F=96487Cmol-1 ,T= temperature in Kelvin and [Mn+] is the
concentration of the species Mn+
For a general equation:
aA +bB ne- cC + dD
Nernst equation can be written as
E(cell) = E0
(cell) - RT lnQ
nF
E(cell) = E0
(cell) - RT ln [C] c[D]d
nF [A]a [B]b
E(cell) = E0
(cell) - 2.303 RT log [C] c[D]d
nF [A]a [B]b
E(cell) = E0
(cell) - 0.059v log [C] c[D]d
nF [A]a [B]b
For a pure substance molar concentration is 1.
•

8. Equilibrium constant from Nernst equation.
In a Dainell cell ,the cell reaction; Zn(S) + CU2+
(aq) → Zn2+
(Aq) +CU(s)
take place and as time passes ,Zn2+ concentration keep on increasing while CU2+
concentration keeps on decrease while voltage read at voltmeter keeps decreasing
After some time , there is no change in concentration for both and the voltage reaches
zero which indicate the equilibrium state. In this state Nernst equation can be written
as: E(cell) =0= E0
(cell) - 2.303 RT log [Zn 2+] [Cu]
2F [CU2+] [Zn]
or E0
(cell) =2.303 RT log [Zn 2+]
2F [CU2+]
But at equilibrium, [Zn 2+] = K c
[CU2+]
At 298K it can be written as or E0
(cell) = 0.059log K c =1.1V
2
Thus ,log Kc =(1.1VX2) =37.228; Hence K c=2x1037 at 298K
0.059V
In General , E0
(cell) = 2.303RT log K c
nF

9. Electrochemical cell and Gibbs energy of the reaction
Electrical work done in 1sec is equal to electrical
potential multiplied by total charge passed . To
obtained maximum work from a galvanic cell ,charge
has passed reversibly. Thus reversible work done is
equal to decrease in Gibbs free energy.
∆rG= - n F E(cell)
If concentration of all the species is 1
then E(cell)= E0
(cell)
Thus, ∆rG0= - n F E0
(cell)
By using the equation we can calculate equilibrium
constant as ∆rG0= - RT lnK.

10. Conductance and Electrolytic solution
Factors affecting electrolytic conduction.
1.Nature of electrolyte:-Strong electrolyte ionize completely thus conduct electricity
completely while weak electrolyte ionize partly and hence conduct small extent of
electricity .
2. Size of ion and their solvation (mixing with solvent):Greater the size of the ion or
solvation less is the conductance.
3.Nature of solvent & viscosity:-Electrolyte ionizes more in polar solvent. Greater the
polarity of the solvent , greater is the ionization and hence greater is the conductance
Similarly. greater the viscosity of solvent less is the conductance
4.Concentration of the solution : Higher the concentration less is the conductance.
5.Temperature: on increases temperature , the dissociation increases and hence
conduction increases
Factors affecting metallic or electronic conductance.
1 . Nature and structure of the metal
2. Number of valence electrons per atom
3.Temperature :it decrease with increase of temperature.

11. • Difference between Metallic conductors and electrolytic conductors:-
Metallic conductors Electrolytic conductors
1. Flow of electricity takes place
without the decomposition of the
substance.
1. Flow of electricity takes place
accompanied by decomposition of
the substance.
2. Flow of electricity is due to flow
of electrons
2. Flow of electricity is due to flow
of ions.
3.The conductance decreases with
increase of temperature.
3.The conductance increases with
increase of temperature.
4.The resistance offered by metal is
due to vibrating kernels.
4.The resistance offered by metal is
due to inter ionic attraction,
viscosity of solvent etc.

12. Electrical Resistance and Conductance:
Ohm’s law state that If to the ends of a conductor is applied a voltage
‘E’ and a current ‘I’ flow through it, then the resistance ‘R’ of the
conductor is E/I.Generally current is measured in amperes, where as
voltage is measured in volts.
1ampere current flows through a conductor when voltage of one volt
is applied to it,the resistance of the conductor is taken as 1 ohm
Thus R=E/I , In term of SI unit ,Ω= (kgm2)/(S3A2)
Ω=V/A =Work per unit charge/A
= work x 1 = F x l x 1 = m x a x l = kg x ms-2x m = Kg m2
Charge x A A x s A A2S A2S S3A2
The reciprocal of the electrical resistance is called conductance and
represented by G .Thus G=1/R its unit is Ω-1 or mho or siemens (S)

13. Specific , equivalent Molar conductivities
Specific conductivity or conductivity:
It is observed that resistance R of a conductor is
(i) directly proportional to its length
(ii) inversely proportional to its area of cross section(a)
R α L or R = ρ l ………………………....(1)
A A
ρ (RESISTIVITY) is defined as the resistance of a conductor whose length is 1cm and area of
cross section is 1cm2,i.e,it is the resistance of 1cm3 of the conductor or in terms of SI units , it
is the resistance of 1m3 of the conductor.
The reciprocal of resistivity is known as specific conductance or simply conductivity . It is
denoted by k(kappa).Thus ,if k is the conductivity and G is the conductance of the solution then
R=1/G and ρ =1/k
Thus, 1 = 1 X L . Thus, K= G x L
G K A A
Hence, Conductivity of a solution is defined as the conductance of a solution of 1 cm length and
having 1 sq.cm as the area of cross section.
Alternatively it may be defined as conductance of 1cm3of the solution of the electrolyte.

14. Equivalent conductivity: It is defined as the conductance of all the ions produced
from 1gm equivalent of the electrolyte dissolved in vcm3 of the solution when the
distance between the electrode is one cm and the area of the electrode is so large that
whole of the solution is contained between them. It is represented by Л e q (lambda).
• Relation between Equivalent conductivity and specific conductivity :
Equivalent conductivity is usually calculated from specific conductivity.
In general if the volume of the solution containing 1 gm equivalent of the
electrolyte is v cm3,we have Equivalent conductivity=Specific conductivity X V
ᴧ e q=Kv XV
If the solution has a concentration of c gram equivalent per litre i.e., c gm are
present in 1000cm3of solution, then the volume of the solution containing 1gm
eq will be 1000/C .Hence ᴧ e q= Ke x 1000 =K e x 1000
C eq Normality
ᴧe q= k vX V= ohm-1cm-1X cm3 = Sm2eq-1 =104Scm2eq-1
Gram Eq.

15. Molar conductivity: Molar conductivity of a solution at a dilution V is the
conductance of all the ions produced from 1 mole of the electrolyte dissolved in vcm3 of
the solution when the electrode are 1 cm apart and the area of the electrode is so large
that the whole of the solution is contained between them .
Molar conductivity = Л m.= K/C
Relation between molar conductivity and specific conductivity :
ᴧ m.= kv X V,
ᴧ m= Kc x 1000 =K c x 1000
C Molarity
Unit= Sm2mol-1 =104Scm2mol-1
= Scm2mol-1 =10-4S m2mol-1
ᴧ m(Scm2mol-1)= K(s cm-1) X1000(cm3L-1)
Molarity (mol L-1)

16. Measurement of the conductivity of ionic solution.
For measuring the resistance of an ionic solution suffer 2 problem
1. passing DC changes the composition 2. A solution cannot connect to a conductor or wire
To solve 1st AC is applied .To solve 2nd a conductivity cell is used.
Pt have cross sectional area “A” and are separated by distance “l” and the solution confined between the
electrodes is a column of length l and area A. The resistance of such a column of solution is then given by
the equation; R=ρ X l/A = K x l /A………(1) where l/A is called cell constant and denoted by G*.
Cell constant is determined by measuring the resistance of the cell containing a solution whose conductivity
is already known. Once the cell constant is known , then we used it for measuring
the resistance of any solution. The instrument consist of 2 resistance R3 and R4,a variable resistance R1 and
the conductivity cell having unknown resistance R2.The wheatstone bridge is fed by an oscillator Q.(AC
power of 550 to 5000Cycle per Second ), P is a detector (Headphone) and the bridge is balance when no
current passes through the detector. Then Unknown Resistance R2=R1R4/R3……(2)
Once cell constant and resistance is known then conductivity is given by formula K=Cell constant/R=G*/R di
Conductivity differ due to charge and size of ions in which they disssociate.

17. Variation of conductivity and molar conductivity with concentration
Conductivity decreases with decrease in concentration due to decrease in ions .The
conductivity of a solution at any given concentration is the conductance of 1 unit
volume of solution kept between two pt-electrode with unit cross section area and
at a distance of unit length. Thus G=K A/l =K.
Molar conductivity at a given concentration is conductance of v volume of solution
containing 1 mole of electrolyte kept between 2 electrodes with area of cross section
A and distance of unit length. Thus ᴧ m.= k X A/l =K if l=1 A=1 then ᴧ m.= kv X V
Molar conductivity increases with decrease in concentration because the total volume
v of solution containing 1 mole of electrolyte also increases.
When concentration approach to zero, the molar conductivity is known as limiting
molar conductivity and represented as ᴧm
0
Strong electrolyte: Л increases slowly with dilution and represented by the equation:
ᴧ m =ᴧ m
0 - AC1/2 . if we plot Л m vs C1/2 a straight line with intercept ᴧ m
0 and slope
equal to –A.The value of A for a given solvent and temperature depends on type of
electrolyte. Kohlrausch law of independent migration of ion state that limitting
molar conductivity of an electrolyte can be represented as the sum of the individual
contribution of anion and cation of the electrolyte.
ᴧ m
0 =V+ ᴧ +
0 + V- ᴧ -
0

18. Weak electrolyte
Weak electrolyte have lower degree of dissociation at higher
concentrations. The change in ᴧm with dilution is due to increase in the
degree of dissociation and the number of ions in total volume of
solution that contain 1 mole of electrolyte. ᴧm increases steeply on
dilution near lower concentrations. Лm cannot be obtained by
Extrapolation of ᴧm to zero concentration. At infinite dilution electrolyte
dissociates completely(ἀ=1) but at low concentration the conductivity of
the solution is so low Thus ,ᴧ m
0 for weak electrolytes is obtained by
using Kohlrausch law of independent migration of ions. At any
concentration c, if α is the degree of dissociation then it can be
approximate to the ratio of molar conductivity ᴧm at the concentration c
to limiting molar conductivity, ᴧ m
0 . Thus , α = ᴧ m / ᴧ m
0
• Ka= c α2 = c α2 = c ᴧ m
2
(1- α) ᴧ m
0 2(1 - ᴧ m ) ᴧ m
0 (ᴧ m
0 - ᴧ m)
• ᴧ m
0

19. FARADAY,S LAWS OF ELECTROLYSIS
• 1ST LAW: The amount of chemical reaction which occurs at any electrode during
electrolysis by a current is proportional to quantity of electricity passed through the
electrolyte.Q=It
• 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 ( at.mass of metal+ No of e- required to reduce the cation).
Charge on 1 electron=1.6021 x 10-19C
Charge on 1 mole of electron= NA X1.6021 x 10-19C=6.02x1.6021 x 10-19
C=96487Cmol-1.
This quantity of electricity is called Faraday.
• ELECTROLYSIS OF NaCl:-At cathode:Na+(aq) +e- → Na(s) E0
cell=-2.71v
H+(aq) +e- → ½ H2(g) E0
cell=0.00v
H2O(l) +e- → ½ H2(g) +OH-
NaCl H2O Na+ (aq) +Cl-(aq)
Catode: H2O(l) +e- → ½ H2(g) +OH-
Anode:Cl-(aq) → ½ Cl2(g) +e-

20. • Electrolysis of H2SO4: - 2H2O(l)→O2(g) +4H+(aq) +4e-, E0
cell=+1.23v
2SO4
2-(aq)→S2O8
2-(aq) +2e-, E0
cell=+1.96v
Primary cell: Cannot recharge.
• Dry cell:
• Anode:Zn(S) → Zn2+ +2e-
• Cathode:MnO2 + NH4
+ +e- → MnO(OH) +NH3
Dry cell Mercury cell
• Mercury cell: Zn(Hg) +2OH- → ZnO(s)+H2O +2e-
HgO2 + H2O+ 2e- → Hg(l) +2OH-

21. • Secondary cell: can recharge.eg,battery
• Anode:-Pb(S) + SO4
2-(Aq) → PbSO4(S) +2e-
• Cathode: PbO2 (s) +SO4
2-(Aq) +4H+ +2e- →PbSO4(S) +2H2O
• On charging the battery the reaction is reversed and PbSO4 on
anode and cathode is converted into pb and PbO2
• Nickel cadmium cell:-Discharge reaction is
• Cd(S) + 2Ni(OH)3 (S) →CdO(S) +2Ni(OH)3(S) +H2O(l)

22. Hydrogen oxygen fuel cell
Cathode :-O2(g) +H2O +4e-→ 4OH-
Anode:-2H2(g) +4OH- (aq) → 4H2O(l) +4e-
Overall: 2H2 +O2 → 2H2O(l)
Corrosion of Iron:
Cathode :-O2(g) +4H+
(aq) +4e-→ 2H2O-,
E0
(H+/O2/H2O)=-1.23v
Anode:-2Fe(s) → 2Fe2
++4e-, E0
(fe2+/fe)=-0.44v
Overall: 2H2 +O2 → 2H2O(l)

23. ‘[Thank you for seeing this]’