Contents: Electrolytic Conduction (G), Conductivity (k), Cell constant (G*), molar conductivity, variation with concentration & temperature, Molar conductivity of weak electrolytes, Debye Huckel Onsager equation, Limiting molar conductivity, Kohlrausch law & application, Electrolytic cell & electrolysis, Some examples of electrolysis of electrolytes in molten or aqueous state, Faraday's laws of electrolysis, its application, Corrosion

1. ELECTROCHEMISTRY (PART 2)
CLASS XII
BY: ARUNESH GUPTA
PGT (CHEMISTRY)
KENDRIYA VIDYALAYA, BARRACKPORE (AFS)

2. ELECTROCHEMISTRY
Contents:
• Resistance (R) & conductance (G) of a solution of an electrolyte
• Conductivity (k) of solution, Cell constant (G*) & their units,
• Molar conductivity (Λm) & its variation with concentration & temperature,
• Debye Huckel Onsager equation & Limiting molar conductivity,
• Kohlrausch’s law & its application & numerical problems.
• Electrolytic cells & electrolysis.
• Some examples of electrolysis of electrolytes in molten / aq. state.
• Faraday’s laws of electrolysis: First & second law- numerical problems.
• Corrosion
• Electrochemical theory of rusting.
• Prevention of rusting.

3. Conductance (G) (or electrolytic conductance)
It is the ease of flow of electric current through the conductor.
It is reciprocal of resistance (R).
G =
1
𝑅
Unit of G is ohm-1, mho, S, Ω−1. (S denotes seimens)
(i) Ohm’s law:” the strength of current(I) is directly proportional to the potential
difference applied across the conductor & inversely proportional to the
resistance of the conductor. We can directly write: V = I R
(ii) Resistivity ( 𝝆) of a conductor is its resistance of 1 cm length & having area of
cross section to 1 cm2.
R = 𝛒
𝐥
𝐚
or. 𝛒 = 𝐑
𝐚
𝐥
Unit of resistivity = ohm cm or 𝛀 𝐜𝐦
SI unit of resistivity = 𝛀 𝐦

4. Conductivity (k) is defined as the conductance of a Solution if 1 cm length and having
1 sq. cm as the area of cross section. (k  kappa)
Conductivity is the conductance of 1cm3 of a solution of an electrolyte.
Conductivity of a solution is the reciprocal of resistivity of a solution of an electrolyte.
K =
𝟏
𝝆
or k =
𝟏
𝐑
𝟏
𝐚
𝐥
k = G.G* where cell constant G* =
𝐥
𝐚
.
Unit of cell constant is cm-1 or m-1 (SI unit)

5. Conductivity
The unit of conductivity (k) of a solution in S cm-1 or S m-1
Here 1 S m-1 = 10-2 S cm-1.
Different materials have different conductivity (k) values.
Conductance (G) of a solution increases on dilution (ie. by adding water) but its
conductivity (k) decreases, as number of ions per unit volume decreases on
dilution.
Molar Conductivity (Λm)
The conductivity of all the ions produced when 1 mole of an electrolyte is
dissolved in V mL of solution is known as molar conductivity.
It is the conductance of 1 mole of an electrolyte placed between two electrodes 1
cm apart having V cm2 area of cross section.
It is related to conductance as Λm =
𝟏𝟎𝟎𝟎 𝐤
𝐌
where M is the molarity of solution
of electrolyte in mol L-1.
It units are Ω-1 cm2 mol-1 or S m2 mol-1 (in SI unit).
1 S m2 mol-1 = 102 S cm2 mol-1

6. # We know that conductance of a solution of 1 cm3 is G
& conductance of 1 mole of V cm3 solution is Vk which is the definition of
molar conductivity.
Hence Λm = Vk.
Let the concentration of solution is M molar,
so, M mole of electrolyte is dissolved in 1 L = 1000 cm3 solution.
Hence, 1 mole of electrolyte is dissolved in
1000
M
cm3 of solution = V cm3 of
solution.
So, Λm = V.k or, Λm =
1000 k
M
# Unit of Λm=
𝐒𝐦−𝟏
𝐦𝐨𝐥 𝐋−𝟏 =
𝐒𝐦−𝟏
𝐦𝐨𝐥 𝐦−𝟑 𝟏𝟎 𝟑 So, S.I. unit of 𝚲 𝐦 is S m2 mol-1.
# With the increase of temperature, G, k & Λm increases.
# With the decrease in concentration on dilution, G increases, k decreases but
Molar
conductivity

7. Numerical Problems:
Examples: (1) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is
the cell constant f conductivity of 0.001 M KCl solution at 298 K is 0.146 x 10-3 S cm-1?
Solution: Given k = 0.146 x 10-3 S cm-1 & R = 1500 Ω We know, G = 1/R & k = GG* =
𝐺∗
𝑅
or,
G* = k.R = 0.146 x 10-3 x 1500 Hence G* = 0.219 cm-1 (Ans)
(2) A 0.05 M NaOH solution : a resistance of 31.6 ohm in a conductivity cell at 298 K. If area of plates of
conductivity cell is 3.8 cm2 & distance between them is 1.4 cm, calculate the molar conductivity of NaOH
solution.
Solution: Cell constant G* =
𝑙
𝑎
=
1.4 𝑐𝑚
3.8 𝑐𝑚2 = 0.368 cm-1. Given concentration = 0.05M, R = 31.6 ohm,
So, conductivity k = GG* =
𝐺∗
𝑅
=
0.368
31.6
– 0.0116 S cm-1. Hence, 𝛬 𝑚 =
1000 k
M
=
1000 x 0.0116
0.05
= 232 S cm2 mol-1 (Ans)
(3) A conductivity cell when filled with 0.01 M KCl has a resistance of 747.5 ohm at 298K. When the same cell
was filled with an aqueous solution of 0.005 M CaCl2 solution, the resistance was 876 ohm. Calculate (i)
conductivity and (ii) molar conductivity of CaCl2 solution. ( conductivity of 0.01 M KCl solution is 0.14114 S/m.
Solution: (a) For KCl solution: R = 747.5 ohm k = 0.14114 S/m
Hence Cell constant G* = R. k = 747.5 x 0.14114 = 105.5 m-1.
(b) For CaCl2 solution, conductivity cell is same. So cell constant (G*) is same.
Hence, conductivity k =
cell constant
R
=
105.5
876
= 0.1204 S/m
(c) For CaCl2 solution, molar concentration = 0.005 mol dm-3. 𝛬 𝑚 =
1000 𝑘
𝑀
=
1000 𝑥 0.1204
0.005
= 0.0241 S m2 mol-1 (Ans)

8. Numerical problems for practice:
(1) The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2mol-1.
Calculate the conductivity of this solution.
[Ans: 0.2083 S/cm]
(2) The conductivity of a solution containing 1.0 g of anhydrous BaCl2 in 200 mL of the solution
has found to 0.0058 Scm-1. Calculate the molar conductivity of the solution. (At. No. Ba = 137,
Cl = 35.5)
[ Ans241.67
Scm2mol-1]
(3) The resistance of a conductivity cell with 0.1 M KCl solution is found to be 200 ohm at 298
K. When the same cell was filled with 0.02 M NaCl solution, the resistance at the at the same
temperature is found to be 1100 ohm. Calculate (i) cell constant of the cell in m-1.
. (ii) the molar
conductivity of 0.02 M NaCl solution in S m2 mol-1. (k for 0.1 M KCl solution at 298 K = 1.29
S/m)
[Ans: 258 m-1, 1.175 x 10-2 S
m2 mol-1]
(4) The molar conductance of 0.05 M solution of MgCl2 is 194.5 S cm2 mol-1 at 298 K. A cell
with electrodes having 1.50 cm2 surface area and 0.50 cm apart is filled with 0.05 M solution of

9. Variation of molar conductivity with concentration for strong electrolytes.
In case of a strong electrolyte, molar conductivity increases slowly on dilution as ion-
ion interaction decreases. On further dilution till infinite dilution when concentration
tends to zero , molar conductivity value achieves a constant value for a particular
electrolyte.
The molar conductivity of an electrolyte when the concentration approaches zero is
called molar conductivity at infinite dilution (𝛬 𝑚
𝑜 )
We can say, 𝛬 𝑚 = 𝛬 𝑚
𝑜 when molar concentration C tends to zero. (infinite dilution)
Debye-Huckel Onsager equation :
It gives a relation between molar conductivity, Λm at a particular concentration and
molar conductivity at infinite dilution 𝛬 𝑚
𝑜 .
Λm = Λ0
m – A√C where, A is a constant. It depends upon the
nature of solvent and temperature.

10. 𝛬 𝑚
𝑜
Debye-Huckel Onsager equation:
Here, 𝛬 𝑚 = -A√C - 𝛬 𝑚
𝑜
(compare with
straight line eqn. y = mx + c
where Slope = -A and
y-intercept = Λm
o ,
The limiting value, Λ0
m or Λ∞
m.
(the molar conductivity at zero
concentration (or at infinite dilution)
can be obtained extrapolating the
graph. (Λ0
m is called limiting molar
conductivity).

11. Factors Affecting Conductivity :
(i) Nature of electrolyte:
The strong electrolytes like KNO3 KCl, NaOH, etc. are completely ionised in aqueous
solution and have high values of molar conductivity.
The weak electrolytes are ionised to a lesser extent in aqueous solution and have lower
values of molar conductivity.
(ii) Concentration of the solution:
The concentrated solutions of strong electrolytes have significant interionic attractions.
which reduce the speed of ions and lower the value of Λm.
The dilution decreases such attractions and increase the value of Λm.

12. Variation of molar conductivity (Λ0
m) with concentration of solution of weak electrolytes:
In case of weak electrolytes, the degree of
ionisation increases dilution which increases
the value of Λ m. The limiting value
Λ0
m (limiting molar conductivity) cannot be
obtained by extrapolating the graph. The
limiting value, Λ0
m for weak electrolytes is
obtained by Kohlrausch law of independent
migration of ions:
“At infinite dilution, the limiting molar
conductivity of an electrolyte is the sum of
the limiting ionic conductivities of all the
cations and anions.”
e.g., for AxBy  x Ay+ + y Bx-
Here 𝚲 𝐦
𝐨 𝐀 𝐱 𝐁 𝐲 = 𝐱𝛌 𝐀 𝐲+
𝐨
+ 𝐲𝛌 𝐁 𝐲−
𝐨
-

13. Variation of molar conductivity withTemperature:
The increase of temperature decreases inter-ionic
attractions of ions in the solution of an electrolyte and
increases kinetic energy of ions and their speed.Thus,
molar conductivity (Λm ) increase with temperature.

14. Applications of Kohlrausch law of independent migration of ions:
(i) We can determine the molar conductivities of weak electrolytes at infinite dilution,
e.g.,
𝛬 𝑚
𝑜
𝐶𝐻3 𝐶𝑂𝑂𝐻 = 𝛬 𝑚
𝑜
𝐶𝐻3 𝐶𝑂𝑁𝑎 + 𝛬 𝐻𝐶𝑙
𝑜
− 𝛬 𝑁𝑎𝐶𝑙
𝑜
𝛬 𝑚
𝑜
(𝑁𝐻4 𝑂𝐻) = 𝛬 𝑚
𝑜
𝑁𝐻4 𝐶𝑙 + 𝛬 𝑁𝑎𝑂𝐻
𝑜
− 𝛬 𝑁𝑎𝐶𝑙
𝑜
(ii) Determination of degree of dissociation (α) of an electrolyte at a given dilution.
𝝰 =
𝒎𝒐𝒍𝒂𝒓 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒗𝒊𝒕𝒚 𝒂𝒕 𝒂 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝑪
𝒎𝒐𝒍𝒂𝒓 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒗𝒊𝒕𝒚 𝒂𝒕 𝒊𝒏𝒇𝒊𝒏𝒊𝒕𝒆 𝒅𝒊𝒍𝒖𝒕𝒊𝒐𝒏 (𝒍𝒊𝒎𝒊𝒕𝒊𝒏𝒈 𝒎𝒐𝒍𝒂𝒓 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒗𝒊𝒕𝒚)
or 𝝰 =
𝜦 𝒎
𝜦 𝒎
𝒐
The dissociation constant (Kc) of the weak electrolyte (of type AB) at concentration C
of the solution can be calculated by using the formula Kc =
𝑪𝜶 𝟐
𝟏− 𝜶
where, α is the degree of dissociation of the electrolyte.

15. Applications of Kohlrausch law of independent migration
of ions:
(iii) Salts like BaSO4, PbSO4, AgCl, AgBr, AgI etc which do
not dissolve to a large extent in water are called sparingly
soluble salts.
The solubility of a sparingly soluble salt can be calculated
as 𝜦 𝒎
𝒐 =
𝟏𝟎𝟎𝟎 𝒌
𝑺
where S is the solubility of a salt in mol/L.Example: (1) The limiting molar conductivities of NaCl, NaAc & HCl are 126.4,
425.9 and 91.0 S cm2 mol-1 respectively. Calculate the limiting molar conductivity
of AcH.
Solution: ΛAcH
o
= λAc−
o
+ λH+
o
= λAc−
o
+ λNa+
o
+ λH+
o
+ λCl−
o
− λNa+
o
− λCl−
o
= ΛNaAc
0
+
ΛHCl
0
− ΛNaCl
0
= 425.9 + 91.0 – 126.4 = 226.0 S cm2 mol-1 (Ans)
Example (2) The conductivity (k) of 0.001028 M acetic acid is 4.95 x 10-5 S cm-1.
Calculate its dissociation constant if limiting molar conductivity is acetic acid is
390.5 S cm2 mol-1.
Solution: We know, Λm =
1000 𝑘
𝑀
=
1000 𝑥 4.95 𝑥 10−5
0.001028
= 48.15 S cm2mol-1.
Degree of dissociation 𝝰 =
𝛬 𝑚
𝛬 𝑚
𝑜 =
48.15
390.5
= 0.1233
𝐶𝛼2 0.001028 𝑥 (0.1233)2
-5 -1

16. Numerical problems for practice: (from Kohlrausch’s law)
(1) Suggest a way to determine limiting molar conductivity of water.
[Ans: 𝛬 𝐻2 𝑂
0
= 𝛬 𝑁𝑎𝑂𝐻
0
+ 𝛬 𝐻𝐶𝑙
0
− 𝛬 𝑁𝑎𝐶𝑙
0
]
(2) The molar conductivity of of 0.025 M HCOOH is 46.1 S cm2 mol-1.
Calculate its degree of dissociation & dissociation constant.
(Given: 𝜆 𝐻+
𝑜
= 349.6 S cm2 mol-1 & 𝜆 𝐻𝐶𝑂𝑂−
𝑜
= 54.6 S cm2 mol-1.
[Ans: 0.114, 3.67 x 10-4]
(3) The molar conductivity at infinity dilution of aluminium sulphate
is 858 S cm2 mol-1. Calculate the limiting molar ionic conductivity of
Al3+ ion.
(Given 𝜆 𝑆𝑂4
2−
0
= 160 S cm2 mol-1. [Ans. 189 Scm2 mol-1]
(4) The limiting molar conductivity of NaOH, NaCl and BaCl2 at 298K
are
2.481 x 10-2, 1.265 x 10-2 and 2.80 x 10-2 S cm2 mol-1 respectively.
Calculate 𝛬 𝑜 for Ba(OH) . [Ans 5.23 x 10-2 S cm2 mol-1]

17. In an electrolytic cell, an external source of voltage (electrical energy) is used
to bring about a non-spontaneous chemical reaction.
Electrolysis is the process of decomposition of an electrolyte when electric
current is passed through either its aqueous solution or molten state.
1. In electrolytic cell both oxidation and reduction takes place in the same cell.
2. Anode is positively charged where oxidation takes place and cathode is
negatively charged where reduction takes place in electrolytic cell.
[Anode (oxidation)  + ve and Cathode (reduction)  - ve]
3. During electrolysis of molten electrolyte, cations are liberated at cathode
(negatively charged) while anions at the anode (positively charged).
4. When two or more ions compete at the electrodes, the ion with higher
reduction potential gets liberated at the cathode while the ion with lower
reduction potential at the anode.
Electrolytic cells &
Electrolysis:

18. How to Predict the Products of Electrolysis?
(i) When an aqueous solution of an electrolyte is electrolysed, if the cation has higher reduction potential
than water (standard reduction potential H2O/ H2,OH- = -0.83 V), cation is liberated at the cathode (e.g.. in
the electrolysis of copper and silver salts , standard reduction potential of Cu2+/Cu & Ag+/Ag are +0.34 &
0.80V respectively) otherwise H2 gas is liberated due to reduction of water (e.g., in the electrolysis of K, Na,
Ca salts, etc, standard reduction potential of Na+/Na, K+/K, Ca2+/Ca are -2.71, -2.93 and -2.87V respectively).
(ii) Similarly if anion has higher oxidation potential than water (Standard oxidation potential:
H2O/ O2, H+ = - 1.23 V), anion is liberated (e.g., Br2/Br- = -1.09V), otherwise O2 gas is liberated due to
oxidation of water (e.g. in case of F- /F2 = -2.87 V, aqueous solution of Na2SO4 as standard oxidation potential
of SO4
2- is – 0.2 V).
(iii) For metals to be deposited on the cathode during electrolysis, the voltage required is almost the same
as the standard electrode potential. However for liberation of gases, some extra voltage is required than the
theoretical value of the standard electrode potential. The extra voltage thus required is called over voltage
or bubble voltage or over potential.

19. (1) Electrolysis of molten NaCl using graphite electrodes
NaCl (s) →
∆
[ Na+ + Cl- ] (molten state)
At Cathode (reduction, -ve electrode): Na+ + 1e  Na
At Anode (oxidation, +ve electrode): Cl-  ½Cl2(g) +1e
Overall reaction due to electrolysis:
2NaCl (molten)
𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑙𝑦𝑠𝑖𝑠 (𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑑𝑒𝑠)
2Na(at cathode) + Cl2 (at anode)
Products formed by electrolysis of electrolytes in molten or aqueous solution

20. (2) Electrolysis of aqueous NaCl solution using graphite electrodes:
NaCl + aq  Na+
(aq) + Cl-
(aq) and water dissociates very slightly as H2O ⇌ H+ + OH-
At cathode (reduction) H+ + 1e  ½H2
[Std reduction potential of H+/H2 (=0.00V) > Na+/Na
(= -2.71V)]
At anode (oxidation) Cl-  ½Cl2(g) + 1e
[Std reduction potential Cl2 / Cl- (= 1.36V < H2O/O2,H+ (= 1.23V), so O2 should be liberated
at anode but dissociation of H2O into O2 is kinetically slow. To enhance the rate , extra potential
is applied called over voltage or over potential, the oxidation potential of Cl-/Cl2 is achieved thus
Cl2 is preferentially produces at anode in place of O2]
& Na+ + OH-  NaOH ; thus, pH of the solution increases after the electrolysis.
Overall reaction:
[NaCl (aq) + H2O(l)
𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑙𝑦𝑠𝑖𝑠
[Na+(aq) + Cl-(aq) ] + ½H2 (at cathode) + ½Cl2(g) (at anode)

21. 3) Electrolysis of molten PbBr2 using Pt electrodes.
PbBr2 →
∆
Pb2+ + 2Br-(molten)
At Cathode(reduction): Pb2+ + 2e  Pb
At anode (oxidation) 2Br-  Br2 + 2e
Overall reaction: PbBr2 (molten)
𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑙𝑦𝑠𝑖𝑠
Pb (at cathode) + Br2 (at anode)
(4) Electrolysis of acidified water using Pt electrodes: (generally acidified by a
few drops of dilute sulphuric acid)
H2O ⇌ H+ + OH-
At cathode (reduction): H+ + 1e  ½H2(g)
At anode (oxidation): OH-  OH + 1e & 4OH  2H2O + O2(g)
Overall reaction: 2H2O
𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑙𝑦𝑠𝑖𝑠
2H2 (at cathode) + O2 (at anode)]

22. (5) Electrolysis of H2SO4 using Pt electrodes:
H2SO4 + aq  2H+(aq) + SO4
2-(aq)
At cathode (reduction) H+ + 1e  ½H2 [or 2H2O + 2e  H2 + 2
OH-]
At anode (oxidation) There are two possible reactions:
(i) 2H2O  O2 + 4H+ +4e (EH2O/O2
θ
= +1.23 V]
(ii) 2SO4
2-(aq)  S2O8
2-(aq) + 2e (ESO4
2−/ S2O8
2−
θ
= 1.96V)
(a) for dilute H2SO4 solution: reaction (i) is preferred & O2 is
liberated at anode.
(b) for higher concentration of H2SO4, reaction (ii) is followed &
peroxodisulphuric acid H2S2O8 is formed.

23. (6) Electrolysis of CuSO4 solution using Pt electrodes:
CuSO4 + aq  Cu2+(aq) + SO4
2- (aq)
At cathode (reduction) Cu2+ + 2e  Cu(s)
[ Here 𝐸 𝐶𝑢2+/𝐶𝑢
𝜃
(=0.34V) > 𝐸 𝐻+/𝐻2
𝜃
(= 0.00V) ]
At anode (oxidation): 2H2O  O2 + 4H+ +4e
(SO4
2- ion will not oxidise as maximum oxidation state
of S is +6)
[2H+ + SO4
2-  H2SO4, solution becomes more acidic pH of the
solution decreases.)

24. (7) Electrolysis of CuSO4 solution using Cu electrodes:
CuSO4 + aq  Cu2+(aq) + SO4
2- (aq)
At cathode (reduction) : Cu2+ + 2e  Cu(s) [ Here 𝐸 𝐶𝑢2+/𝐶𝑢
𝜃
(=0.34V) > 𝐸 𝐻+/𝐻2
𝜃
(=
0.00V) ]
At Anode (oxidation) : Cu  Cu2+ + 2e
[Here, Standard oxidation potential Cu/Cu2+( = - 0.34V) > H2O/O2, H+ ( = -1.23V)
]
We can find that Cu is deposited at cathode and Cu2+ ion is dissolved at anion, so
the blue colour of the solution does not fade out, pH of the solution remains same.
Try:
(1) Using Pt electrodes, the blue colour of CuSO4 fades out on electrolysis, but using Cu
electrodes, the blue colour of solution does not fade out. Why?
(2) Can we suitably add dilute HCl to acidify water before electrolysis to get H2 & O2?
(3) Suggest a list of metals that are extracted electrolytically.

25. Faraday’s Laws of
Electrolysis:
1. First law of electrolysis:
The amount of a chemical substance which occurs at any electrode during
electrolysis by a current is directly proport ional to the quantity of electricity
passed through electrolyte (molten or solution).
W ∝ I x t or, w = Z.I.t = Z.Q where Q (in coulomb) = I (in ampere) x t (in second)
& w is the mass of chemical substance in gram.
Z is a constant known as electrochemical equivalent (ECE).
When current of 1 ampere in 1 sec or charge in Q = 1 coulomb charge flows
through the solution or molten state of an electrolyte, mass of the substance
deposited or liberated is called its electrochemical equivalent,
( If Q = 1C or I = 1A & t = 1s, w = Z ).
[ Charge of 1 mole of electron = 1 F (faraday)
1F (faraday) = 6.023 x 1023 x 1.6021 x 10-19 = 96487 C mol-1 ≈ 96500 C mol-
1.]
Al3+
(aq) + 3e  Al(s)
1 mol 3F 1 mol = 27 g Al
i,e. 3F or 3 x 96500 C of charge is required to form 1 mole or 27 g of Al.
S, for 1C charge =
27
3 𝑥 96500
= 𝑍 𝐴𝑙

26. Example: A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 A.
What is the mass of copper deposited at cathode? (Atomic mass of Cu = 63.5 u)
Solution:
I = 1.5 A, t = 10 x 60 s = 600 s Charge Q = I.t = 1.5 x 600 = 900 C
We have: Cu2+ + 2e  Cu
2F 1 mol = 63.5g ⇒ 2 F = 2 x 96500 C charge
2 x 96500 C charge deposits 63.5 g Cu
900 C charge deposits
63.5 𝑥 900
2 𝑥 96500
= 0.2961 g of Cu (Ans)

27. 2. Second law of electrolysis:
When the same quantity of electricity is passed through different electrolytes, in
series, the amounts of the substances deposited or liberated at the electrodes arc
directly proportional to their equivalent weights, Thus,
When Q amount of charge is passed in series through different electrolytes A & B
are deposited.
According to second law,
𝑴𝒂𝒔𝒔 𝒐𝒇 𝑨 (𝑾 𝑨 )
𝑴𝒂𝒔𝒔 𝒐𝒇 𝑩 (𝑾 𝑩 )
=
𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑨 (𝑬 𝑨)
𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑩 (𝑬 𝑩)
or
𝑾 𝑨
𝑾 𝑩
=
𝑬 𝑨
𝑬 𝑩
=
𝑸.𝒁 𝑨
𝑸,𝒁 𝒃
=
𝒁 𝑨
𝒁 𝑩
Hence, electrochemical equivalent (Z) ∝ equivalent weight (E).
𝑊𝐴𝑔
𝑊𝐶𝑢
=
𝐸𝐴𝑔
𝐸 𝐶𝑢
=
𝑍 𝐴𝑔
𝑍 𝐶𝑢
𝑊𝐴𝑔
𝑊𝐶𝑢
=
𝐸𝐴𝑔
𝐸 𝐶𝑢
=
𝑍 𝐴𝑔
𝑍 𝐶𝑢

28. Numerical Problems for practice:
(1) How many coulombs are required to deposit 40.5 g of Al when the cathode reaction is Al3+
(aq) + 3e  Al(s)
[Ans. 434250C]
(2) How many coulombs of electricity are required for
(i) oxidation of 1 mol of H2O to O2. (ii) oxidation of 2 mol of FeO to Fe2O3
(iii) reduction of 1 mol of 𝑀𝑛𝑂4
−
to Mn2+. [Ans. 1.93 x 105C, 96500 C, 4.825 X 105 C]
(3) Calculate the time in hours that reduces 3 mol of Fe3+ to Fe2+ with 2 A current / (1F = 96500 C).
[Ans. 40.21 hours]
(4) In the electrolysis of acidified water, it is desired to obtain H2 gas at the rate of 1mL per second at STP. What
should be the current passed? [Ans. 8.616 A]
(5) Two electrolytic cells containing silver nitrate solution & dilute sulphuric acid were connected in series. A
steady current of 2.5 A was passed through them till 1.078 g of silver was deposited. (i) How much electricity was
consumed? (ii) What was the weight of oxygen gas liberated? [Ag = 107.8 g mol-1, O = 16 g mol-1]
(6) How many moles of mercury will be produced by electrolysing 1.0M Hg(NO3)2 solution with a current of 2.0A
for 3 hours. [Ans. 0.112 mol]
HOTS** (7) A current of 1.5A is passed through 500 mL of 0.25 M ZnSO4 solution for 1 hour with a current
efficiency of 90%.Calculate the final molarity if Zn2+ ions assuming volume of solution to be constant. Ans. 0.2 M]
**(8) An aqueous solution of an unknown salt of palladium is electrolysed by a current of 3.0 A passing for 1 hour.
During electrolysis, 2.977g of palladium ions are reduced at the cathode. What is the charge on the palladium
ions in solutions? [Ans. Charge on Pd = +4]

29. Batteries:
These are source of electrical energy which may have more than cells connected
in series.
A good quality battery, should be reasonably light. compact and its voltage should
not vary appreciably during its use.
Types:
Primary Batteries
In the primary batteries. the cell reaction occurs only once and after use over a period of time
battery becomes dead and cannot be reused again. Examples: (i) Dry cell or Leclanche cell,
(ii) mercury cell etc
Secondary Batteries:
In secondary cell after ist use can be recharged by passing current through itin opposite
direction so that it can be reused again. A good secondary cell should have a large number of
discharging & charging cycles. Example: Lead Storage battery, (ii) Nickel-cadmium cell etc.

30. (i) Dry cell or Laclanche cell:
Anode - Zinc container; Cathode - Graphite rod surrounded by MnO2 powder
(depolariser)
Electrolyte-Paste of NH4Cl + ZnCl2
SOME COMMERCIAL CELLS / BATTERIES
Anode reaction Cathode reaction,
Zn(s) → Zn2+(aq) +
2e-
MnO2(s) + 𝐍𝐇 𝟒(𝐚𝐪)
+
+ 2e → MnO(OH)(s) +
NH3(g)
# At cathode Mn+4 is reduced to Mn+3.
# NH3 formed , forms soluble complex with Zn2+ as
[Zn(NH3)4]2+
# Cell potential 1.25 V to 1.5 V,
# Used in transistors & clocks.

31. (ii) Mercury cell
Anode-Zn-Hg amalgam, Cathode-Paste of
(HgO + C)
Electrolyte-Moist paste of KOH - ZnO
Cathode reaction:
HgO(s) + H2O(l) + 2e  Hg(l) +
2OH-(aq)
Anode reaction: Zn(Hg) + 2OH-(aq)  ZnO(s) +
H2O + 2e
Net reaction: Zn(Hg) + HgO(l)  ZnO(s) + Hg(l)
# Cell Potential = 1.35 V
(Cell potential is constant during its life as the
overall reaction
does not involve any ion in solution whose
concentration does not change during its life time.
# Used in hearing aids, watches, calculators etc.

32. (i) Lead Storage battery (Secondary Batteries)
Anode-Spongy lead, Cathode-Grid of lead packed with PbO2
Electrolyte-38% H2SO4 by mass.
The cell reaction during discharging (when cell is in use)
Anode reaction: Pb(s) + 𝐒𝐎 𝟒 (𝐚𝐪)
𝟐−
 PbSO4(s) + 2e
Cathode reaction: PbO2(s) + 4H+
(aq) + 𝐒𝐎 𝟒 (𝐚𝐪)
𝟐−
+ 𝟐𝒆  PbSO4(s) +
2H2O(l)
Net reaction:
Pb(s) +PbO2(s) + 4H+(aq) + 2𝐒𝐎 𝟒 (𝐚𝐪)
𝟐−
 PbSO4(s) + 2H2O(l)
When concentration of H2SO4 decreases to nearly 19% by mass,
the
battery is recharged.
Here, the cell reactions are reversed.
At anode: PbSO4(s) + 2H2O(l)  PbO2(s) + 4H+
(aq) + 𝐒𝐎 𝟒 (𝐚𝐪)
𝟐−
+ 𝟐𝒆
At cathode: PbSO4(s) + 2e  Pb(s) + 𝐒𝐎 𝟒 (𝐚𝐪)
𝟐−
Net reaction: PbSO4(s) + 2H2O(l)  Pb(s) +PbO2(s) + 4H+(aq) +
2𝐒𝐎 𝟒 (𝐚𝐪)
𝟐−

33. (ii) Nickel-cadmium storage cell
Anode – Cadmium, Cathode - Metal grid containing
NiO2 Electrolyte-KOH solution
Anode reaction:
Cd(s) + 2OH-(aq)  CdO(s) + H2O(l) + 2e
Cathode reaction:
NiO2(s) + 2H2O(l) + 2e  Ni(OH)2(s) + 2OH-(aq)
Net reaction:
Cd(s) + NiO2(s) + 2H2O(l)  CdO(s) + Ni(OH)2(s) +
H2O(l)
# Ni-Cd cell has longer life than lead storage battery,
but is costly.
# It is used in camera flash light etc.

34. Fuel Cells:
Galvanic cells use fuel energy of combustion of fuels like H2, CH4,
CH3OH, etc., directly as the source to produce electrical energy. Oxygen gas is
used as oxidiser.
The fuel cells are pollution free and have high efficiency > 70%, but
costly, heavy & has corrosive chemicals. Example: H2 – O2 fuel cells, CH4 – H2
fuel cell, CH3OH – H2 fuel cell etc.
The H2 – O2 fuel cell was used for providing electrical power in the
APOLLO Space programme.

35. HYDROGEN-OXYGEN FUEL
CELL
• Electrodes-made of porous graphite impregnated with catalyst (Pt,
Ag or a metal oxide).
• Electrolyte-aqueous solution of KOH or NaOH.
• Oxygen and hydrogen are continuously fed into the cell.
• Oxidation half reaction at anode:
2H2(g) + 4OH-(aq)  4H2O(l) + 4e
• Reduction half reaction at cathode:
O2(g) + 2H2O(l) + 4e  4OH-(aq)
• Net cell reaction: 2H2(g) + O2(g)  2 H2O(l)
• EMF of this cell is 1.0 V
• Thermodynamic efficiency of the fuel cell = ɳ =
𝜟𝑮
𝜟𝑯
=
− 𝒏𝑭𝑬 𝒄𝒆𝒍𝒍
𝜟𝑯

36. Some applications of
electrochemistry:
(1) Electro-refining of metals like Cu, Ag
etc
Example: Cathode: pure Cu, Anode:
impure Cu & Electrolyte: Cu2+(aq)
(2) Electroplating of metals with Ag, Au,
Cu, Cr etc.

37. (iii)
Corrosion:
Slow formation of undesirable oxidised compounds such as oxides,
sulphides or carbonates at the surface of metals by its reaction with moisture,
oxygen and other atmospheric gases is known as corrosion.
Example: (i) Fe : Rust (Fe2O3.xH2O) a reddish brown solid. (Rusting of iron)
(ii) Cu : Cu(OH)2. CuCO3 a green coating on copper metal.
(iii) Al : Al2O3, dirty white coating on aluminium metal
Factors Affecting Corrosion :
1. Reactivity of metals: More reactive metals are corroded faster than less reactive
metals
2. Presence of moisture and atmospheric gases like CO2, SO2, etc rate of corrosion
increases.
3. Presence of impurities: increases the tare of corrosion.
4. Strains in the metal surface helps water droplets to accumulate where an
electrochemical cell is formed.
Some applications of
electrochemistry:

38. Rusting is an electrochemical process
Corrosion of iron is called rusting of iron.
An electrochemical cell, also known as corrosion cell, is developed at the stains of iron
surface in presence of dissolved air (containing CO2, SO2 etc) in water droplets where a local
cell is formed as given below:-
Anode- Pure iron Cathode- Impure Fe surface (containing dissolved air in water
droplets).
Rusting of Iron- Electrochemical
Theory:
Electrolyte: CO2 + H2O ⇌ H2CO3 ⇌2H+ + CO3
2−
Anode reaction:
2 Fe(s)  Fe2+ + 4e (𝐸 𝐹𝑒2+/𝐹𝑒
𝜃
= -0.44V)
Cathode reaction:
O2(g) + 4H+(aq) + 4e  2H2O(l) (EO2. H+/H2 O
θ
= 1.23V)
Net reaction:
2Fe(s) + 4H+(aq) + O2(g)  2Fe2+(aq) 2H2O(l) (𝑬 𝒄𝒆𝒍𝒍
𝜽
= 1.67 V)
On the surface: 2Fe2+(aq) + 4H2O(l) + O2(g)  2Fe2O3(s) + 8H+(aq)
Fe2O3(s) + x.H2O(l)  Fe2O3.xH2O(s) (rust)
The further production of H+ ions continue the rusting process.

39. Rusting of iron can be prevented by the following methods :
1. Barrier protection through coating of paints or by chemicals like bisphenol
or electroplating with less reactive metals.
2. Through coating by metal zinc called galvanisation or coating of surface
with tin metal or chromium etc.
3. By the use of antirust solutions (alkaline sodium chromate or phosphate)
which reacts with iron to form insoluble, thin hard coatings of iron(III)
chromate or phosphate)
4. By cathodic or sacrificial protection in which a metal is protected from
corrosion by connecting it to another metal that is more easily oxidised e.g.
Mg, Al, Zn etc. These metals corrodes itself by oxidation, prevents Fe to
oxidise & save from rusting. These metals have standard reduction potential
values lower than that of iron.

40. Some concept based questions:
(1) What are the signs of ΔG, equilibrium constant K & 𝐸 𝐶𝑒𝑙𝑙
𝜃
for an electrochemical cell.
(2) Why is a salt bridge or a porous plate not needed in a lead storage battery?
(3) Blocks of magnesium are often hanged in chains to the ocean going ships. Why?
(4) Rusting is prevented in an alkaline solution. Why?
(5) On the basis of the following data: (i) Co3+ + e  Co2+ : E0 = +1.82 V,
(ii) 2H2O  O2 + 4H+ + 4e E0 = +1.23 V. Explain why Co3+ salts are unstable in water.
(6) Why a dry cell becomes dead after a long time even if it has not been used?
(7) Why does the cell potential of Mercury Cell remain constant throughout its life?
(8) How will the pH of brine (aqueous NaCl solution) be affected when it is electrolysed?
(8) Why is alternating current used for measuring resistance of an electrolytic solution?
(9) How will the pH of the solution be affected when acidified water (dil. H2SO4) is electrolysed?
(10) What advantage do the fuel cells have over primary and secondary batteries?
(11) Why on dilution, the molar conductivity of CH3COOH increases drastically, while that of CH3COONa
increases gradually?
**(12) Show that foe two half reactions having potentials E1 & E2 which are combined to give a third half
reaction having potential E3 is E3 =
𝑛1 𝐸1+𝑛2 𝐸2
𝑛3
where n1, n2 & n3 are no. of electrons involved in the half
reactions respectively.