Electrochemistry

# Electrochemistry Notes - IIT JAM

## Document Description: Electrochemistry for IIT JAM 2022 is part of IIT JAM preparation. The notes and questions for Electrochemistry have been prepared according to the IIT JAM exam syllabus. Information about Electrochemistry covers topics like and Electrochemistry Example, for IIT JAM 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Electrochemistry.

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``` Page 1

1
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
Assignment    Physical Chemistry           IIT-JAM 2016
ELECTROCHEMISTRY
Ashoka Scientific Forum
Developing Scientific Temper Among Students
IIT-JAM | TIFR | DU | BHU | JNU | ISM | PU & All Other University M.Sc. Entrance Exam
CSIR-NET | GA TE | TIFR
Solved Problem
Subjective
1: During the discharge of a lead storage battery, the density of H
2
SO
4
falls from 1.294 to
1.139 g ml
-1
. H
2
SO
4
of density 1.294 g ml
-1
is 39% H
2
SO
4
by weight and that of density
1.139 g ml
-1
is 20% H
2
SO
4
by weight. The battery holds 3.5 litres of acid and volume
remains practically constant during the discharge. Calculate the number of ampere
hours for which the battery must have been used. The charging and discharging reactions
are:
Pb +
2
4
SO
?
? PbSO
4
+ 2e
-
(charging)
PbO
2
+ 4H
+
+ SO
4
2-
? PbSO
4
+ 2H
2
O (discharging)
Solution: The overall reaction is:
Pb + PbO
2
+ 2H
2
SO
4
? 2PbSO
4
+ 2H
2
O
It should be noted that 1 mole of Pb gets oxidised to 1 mole of Pb
2+
( in PbSO
4
). Therefore,
during the process, 2 moles of H
2
SO
4
are consumed with the utilisation of 2 moles of electrons. In
other words, the number of moles of H
2
SO
4
is equal to the number of Faradays consumed.
Mass of sample in 3.5 L  of acid before the discharge
= 3.5 ?1.294 ?10
3
= 4529 g
Mass of H
2
SO
4
= 4529 ?
100
39
=  1766.31 g
Moles of H
2
SO
4
=
98
31 . 1766
= 18.02
Mass of sample after discharge = 1.139 ? 10
3
? 3.5 = 3986.5 g
Mass of H
2
SO
4
=
20
100
? 3986.5 = 797.3 g
No. of moles of H
2
SO
4
=
797.3
98
= 8.135
Moles of H
2
SO
4
consumed = 18.02 – 8.135 =  9.88
? No of Faradays utilised = 9.88 F
?Charge = 9.88 ? 96500 = 953833.57 C
1 ampere hour = charge passed by one ampere in one hour
= 60 ? 60 = 3600 C
?No of ampere hours =
953833.57
3600
= 264.95
Page 2

1
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
Assignment    Physical Chemistry           IIT-JAM 2016
ELECTROCHEMISTRY
Ashoka Scientific Forum
Developing Scientific Temper Among Students
IIT-JAM | TIFR | DU | BHU | JNU | ISM | PU & All Other University M.Sc. Entrance Exam
CSIR-NET | GA TE | TIFR
Solved Problem
Subjective
1: During the discharge of a lead storage battery, the density of H
2
SO
4
falls from 1.294 to
1.139 g ml
-1
. H
2
SO
4
of density 1.294 g ml
-1
is 39% H
2
SO
4
by weight and that of density
1.139 g ml
-1
is 20% H
2
SO
4
by weight. The battery holds 3.5 litres of acid and volume
remains practically constant during the discharge. Calculate the number of ampere
hours for which the battery must have been used. The charging and discharging reactions
are:
Pb +
2
4
SO
?
? PbSO
4
+ 2e
-
(charging)
PbO
2
+ 4H
+
+ SO
4
2-
? PbSO
4
+ 2H
2
O (discharging)
Solution: The overall reaction is:
Pb + PbO
2
+ 2H
2
SO
4
? 2PbSO
4
+ 2H
2
O
It should be noted that 1 mole of Pb gets oxidised to 1 mole of Pb
2+
( in PbSO
4
). Therefore,
during the process, 2 moles of H
2
SO
4
are consumed with the utilisation of 2 moles of electrons. In
other words, the number of moles of H
2
SO
4
is equal to the number of Faradays consumed.
Mass of sample in 3.5 L  of acid before the discharge
= 3.5 ?1.294 ?10
3
= 4529 g
Mass of H
2
SO
4
= 4529 ?
100
39
=  1766.31 g
Moles of H
2
SO
4
=
98
31 . 1766
= 18.02
Mass of sample after discharge = 1.139 ? 10
3
? 3.5 = 3986.5 g
Mass of H
2
SO
4
=
20
100
? 3986.5 = 797.3 g
No. of moles of H
2
SO
4
=
797.3
98
= 8.135
Moles of H
2
SO
4
consumed = 18.02 – 8.135 =  9.88
? No of Faradays utilised = 9.88 F
?Charge = 9.88 ? 96500 = 953833.57 C
1 ampere hour = charge passed by one ampere in one hour
= 60 ? 60 = 3600 C
?No of ampere hours =
953833.57
3600
= 264.95
2
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
2: Cadmium amalgam is obtained by the electrolysis of a solution of cadmium chloride
using a mercury cathode. For how long the process of electrolysis should be carried out
to prepare a 12% (by wt.) of cadmium amalgam using 44 gms of mercury and a current of
5 amperes? (At. Wt. Of Cd = 112)
Solution: Since 88 gm Hg has 12 gm Cd
So 44 gm Hg require =
88
44 12 ?
gm Cd = 6 gm Cd
Now, 56 gm Cd is required to deposit 96500 C charge
So, 6 gm Cd required to deposit =
56
96500
? 6 C = 10339.286 C
Time  =
5
10339.286
=
I
Q
= 20678.57 Sec. =  34.4 min
3: Calculate the equilibrium constant for the reaction:
Fe
2+
+ Ce
4+
? Ce
3+
+ Fe
3+
Given that, Fe
3+
+ e
-
? Fe
2+
; E
0
= 0.771 V
Ce
4+
+ e
-
? Ce
3+
; E
0
= 1.61 V
Solution: The equilibrium reaction is Fe
2+
+ Ce
4+

Ce
3+
+ Fe
3+
Since the reaction is at equilibrium, E
cell
= 0
0
cell
E =
K log
1
0591 . 0
0
Ce / Ce
0
Fe / Fe
3 4 3 2
E E
? ? ? ?
? = 0.0591 log K
– 0.771 + 1.61 = 0.0591 log K
log K =
0591 . 0
839 . 0
= 14.1963
?K = 1.57 ? 10
14
4: Calculate the emf of the cell:
Pt H
2
| CH
3
COOH  || NH
4
OH | H
2
Pt
1 atm.   0.1 M            0.01 M   1 atm.
K
a
for CH
3
COOH = 1.8 x 10
-5
and K
b
for NH
4
OH = 1.8 x 10
-5
Solution: The reactions are : at anode : ½ H
2
?
?
a
H + e
–
at cathode :
?
c
H + e
–
? ½ H
2
———————————————————————
overall reaction
?
c
H ?
?
a
H
E = E
0
– 0.059 log
]
H
[
]
H
[
+
c
+
a
At anode:
From CH
3
COOH

CH
3
COO
–
+
?
a
H
? ] H [
a
?
= C ? = C C K
C
K
a
a
?
Page 3

1
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
Assignment    Physical Chemistry           IIT-JAM 2016
ELECTROCHEMISTRY
Ashoka Scientific Forum
Developing Scientific Temper Among Students
IIT-JAM | TIFR | DU | BHU | JNU | ISM | PU & All Other University M.Sc. Entrance Exam
CSIR-NET | GA TE | TIFR
Solved Problem
Subjective
1: During the discharge of a lead storage battery, the density of H
2
SO
4
falls from 1.294 to
1.139 g ml
-1
. H
2
SO
4
of density 1.294 g ml
-1
is 39% H
2
SO
4
by weight and that of density
1.139 g ml
-1
is 20% H
2
SO
4
by weight. The battery holds 3.5 litres of acid and volume
remains practically constant during the discharge. Calculate the number of ampere
hours for which the battery must have been used. The charging and discharging reactions
are:
Pb +
2
4
SO
?
? PbSO
4
+ 2e
-
(charging)
PbO
2
+ 4H
+
+ SO
4
2-
? PbSO
4
+ 2H
2
O (discharging)
Solution: The overall reaction is:
Pb + PbO
2
+ 2H
2
SO
4
? 2PbSO
4
+ 2H
2
O
It should be noted that 1 mole of Pb gets oxidised to 1 mole of Pb
2+
( in PbSO
4
). Therefore,
during the process, 2 moles of H
2
SO
4
are consumed with the utilisation of 2 moles of electrons. In
other words, the number of moles of H
2
SO
4
is equal to the number of Faradays consumed.
Mass of sample in 3.5 L  of acid before the discharge
= 3.5 ?1.294 ?10
3
= 4529 g
Mass of H
2
SO
4
= 4529 ?
100
39
=  1766.31 g
Moles of H
2
SO
4
=
98
31 . 1766
= 18.02
Mass of sample after discharge = 1.139 ? 10
3
? 3.5 = 3986.5 g
Mass of H
2
SO
4
=
20
100
? 3986.5 = 797.3 g
No. of moles of H
2
SO
4
=
797.3
98
= 8.135
Moles of H
2
SO
4
consumed = 18.02 – 8.135 =  9.88
? No of Faradays utilised = 9.88 F
?Charge = 9.88 ? 96500 = 953833.57 C
1 ampere hour = charge passed by one ampere in one hour
= 60 ? 60 = 3600 C
?No of ampere hours =
953833.57
3600
= 264.95
2
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
2: Cadmium amalgam is obtained by the electrolysis of a solution of cadmium chloride
using a mercury cathode. For how long the process of electrolysis should be carried out
to prepare a 12% (by wt.) of cadmium amalgam using 44 gms of mercury and a current of
5 amperes? (At. Wt. Of Cd = 112)
Solution: Since 88 gm Hg has 12 gm Cd
So 44 gm Hg require =
88
44 12 ?
gm Cd = 6 gm Cd
Now, 56 gm Cd is required to deposit 96500 C charge
So, 6 gm Cd required to deposit =
56
96500
? 6 C = 10339.286 C
Time  =
5
10339.286
=
I
Q
= 20678.57 Sec. =  34.4 min
3: Calculate the equilibrium constant for the reaction:
Fe
2+
+ Ce
4+
? Ce
3+
+ Fe
3+
Given that, Fe
3+
+ e
-
? Fe
2+
; E
0
= 0.771 V
Ce
4+
+ e
-
? Ce
3+
; E
0
= 1.61 V
Solution: The equilibrium reaction is Fe
2+
+ Ce
4+

Ce
3+
+ Fe
3+
Since the reaction is at equilibrium, E
cell
= 0
0
cell
E =
K log
1
0591 . 0
0
Ce / Ce
0
Fe / Fe
3 4 3 2
E E
? ? ? ?
? = 0.0591 log K
– 0.771 + 1.61 = 0.0591 log K
log K =
0591 . 0
839 . 0
= 14.1963
?K = 1.57 ? 10
14
4: Calculate the emf of the cell:
Pt H
2
| CH
3
COOH  || NH
4
OH | H
2
Pt
1 atm.   0.1 M            0.01 M   1 atm.
K
a
for CH
3
COOH = 1.8 x 10
-5
and K
b
for NH
4
OH = 1.8 x 10
-5
Solution: The reactions are : at anode : ½ H
2
?
?
a
H + e
–
at cathode :
?
c
H + e
–
? ½ H
2
———————————————————————
overall reaction
?
c
H ?
?
a
H
E = E
0
– 0.059 log
]
H
[
]
H
[
+
c
+
a
At anode:
From CH
3
COOH

CH
3
COO
–
+
?
a
H
? ] H [
a
?
= C ? = C C K
C
K
a
a
?
3
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
] H [
a
?
= 1 . 0 10 8 . 1 ] COOH CH [ K
5
3 a
? ? ? ?
?
= 1.34 ? 10
–3
M
At cathode:
From NH
4
OH

?
4
NH + OH
–
[OH
–
] = C ? = C K
C
K
b
b
?
? ] H [
c
?
=
C K
K
] OH [
K
b
w w
?
? =
] OH NH [ K
K
4 b
w
?
=
01 . 0 10 8 . 1
10
5
14
? ?
?
?
=  2.359 ? 10
-11
M
?E = 0  –  0.0591 log
11
3
10 359 . 2
10 34 . 1
?
?
?
?
= –0.4575 V
5: A current of 15 amperes is employed to plate nickel in a NiSO
4
bath. Both Ni and H
2
are
formed at cathode. The current efficiency with respect to the formation of Ni is 60%. (a)
How many g. of Ni are plated out in one hour? (b) What is the thickness of the plating
of the cathode consisting of a square sheet of metal of 4 cm. side which is coated on both
faces? The density of nickel is 8.9 g/cc. (c) What volume  of H
2
(STP) is formed per hour?
Solution: Total charge passed in one hour= (15 ? 60 ? 60) C
= 54000 C
But current efficiency with respect to Ni is 60%
Therefore, charge used up in deposition of Ni = 54000 ?
100
60
C
= 32400 C
And remaining charge will be used up to evolve H
2
i.e. equal to = (54000 – 32400) C = 21600 C
? Ni
+2
+ 2e
–
? Ni
Wt. Of Ni (plated) =
96500 2
32400 69 . 58
?
?
gm =  9.85 gm
Now, volume of square sheet of Ni metal required to coated on
both faces =
cc
8.9
9.85
so volume of square sheet of Ni metal required to coated on one face
=
cc
8.9 2
9.85
?
which should be equal to = (area of square) ? thickness of plate
= 16  cm
3
? thickness of plate
? thickness of plate =
16 2 9 . 8
85 . 9
? ?
= 0.035 cm
Now, 2H
+
+ 2e
–
? H
2
Page 4

1
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
Assignment    Physical Chemistry           IIT-JAM 2016
ELECTROCHEMISTRY
Ashoka Scientific Forum
Developing Scientific Temper Among Students
IIT-JAM | TIFR | DU | BHU | JNU | ISM | PU & All Other University M.Sc. Entrance Exam
CSIR-NET | GA TE | TIFR
Solved Problem
Subjective
1: During the discharge of a lead storage battery, the density of H
2
SO
4
falls from 1.294 to
1.139 g ml
-1
. H
2
SO
4
of density 1.294 g ml
-1
is 39% H
2
SO
4
by weight and that of density
1.139 g ml
-1
is 20% H
2
SO
4
by weight. The battery holds 3.5 litres of acid and volume
remains practically constant during the discharge. Calculate the number of ampere
hours for which the battery must have been used. The charging and discharging reactions
are:
Pb +
2
4
SO
?
? PbSO
4
+ 2e
-
(charging)
PbO
2
+ 4H
+
+ SO
4
2-
? PbSO
4
+ 2H
2
O (discharging)
Solution: The overall reaction is:
Pb + PbO
2
+ 2H
2
SO
4
? 2PbSO
4
+ 2H
2
O
It should be noted that 1 mole of Pb gets oxidised to 1 mole of Pb
2+
( in PbSO
4
). Therefore,
during the process, 2 moles of H
2
SO
4
are consumed with the utilisation of 2 moles of electrons. In
other words, the number of moles of H
2
SO
4
is equal to the number of Faradays consumed.
Mass of sample in 3.5 L  of acid before the discharge
= 3.5 ?1.294 ?10
3
= 4529 g
Mass of H
2
SO
4
= 4529 ?
100
39
=  1766.31 g
Moles of H
2
SO
4
=
98
31 . 1766
= 18.02
Mass of sample after discharge = 1.139 ? 10
3
? 3.5 = 3986.5 g
Mass of H
2
SO
4
=
20
100
? 3986.5 = 797.3 g
No. of moles of H
2
SO
4
=
797.3
98
= 8.135
Moles of H
2
SO
4
consumed = 18.02 – 8.135 =  9.88
? No of Faradays utilised = 9.88 F
?Charge = 9.88 ? 96500 = 953833.57 C
1 ampere hour = charge passed by one ampere in one hour
= 60 ? 60 = 3600 C
?No of ampere hours =
953833.57
3600
= 264.95
2
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
2: Cadmium amalgam is obtained by the electrolysis of a solution of cadmium chloride
using a mercury cathode. For how long the process of electrolysis should be carried out
to prepare a 12% (by wt.) of cadmium amalgam using 44 gms of mercury and a current of
5 amperes? (At. Wt. Of Cd = 112)
Solution: Since 88 gm Hg has 12 gm Cd
So 44 gm Hg require =
88
44 12 ?
gm Cd = 6 gm Cd
Now, 56 gm Cd is required to deposit 96500 C charge
So, 6 gm Cd required to deposit =
56
96500
? 6 C = 10339.286 C
Time  =
5
10339.286
=
I
Q
= 20678.57 Sec. =  34.4 min
3: Calculate the equilibrium constant for the reaction:
Fe
2+
+ Ce
4+
? Ce
3+
+ Fe
3+
Given that, Fe
3+
+ e
-
? Fe
2+
; E
0
= 0.771 V
Ce
4+
+ e
-
? Ce
3+
; E
0
= 1.61 V
Solution: The equilibrium reaction is Fe
2+
+ Ce
4+

Ce
3+
+ Fe
3+
Since the reaction is at equilibrium, E
cell
= 0
0
cell
E =
K log
1
0591 . 0
0
Ce / Ce
0
Fe / Fe
3 4 3 2
E E
? ? ? ?
? = 0.0591 log K
– 0.771 + 1.61 = 0.0591 log K
log K =
0591 . 0
839 . 0
= 14.1963
?K = 1.57 ? 10
14
4: Calculate the emf of the cell:
Pt H
2
| CH
3
COOH  || NH
4
OH | H
2
Pt
1 atm.   0.1 M            0.01 M   1 atm.
K
a
for CH
3
COOH = 1.8 x 10
-5
and K
b
for NH
4
OH = 1.8 x 10
-5
Solution: The reactions are : at anode : ½ H
2
?
?
a
H + e
–
at cathode :
?
c
H + e
–
? ½ H
2
———————————————————————
overall reaction
?
c
H ?
?
a
H
E = E
0
– 0.059 log
]
H
[
]
H
[
+
c
+
a
At anode:
From CH
3
COOH

CH
3
COO
–
+
?
a
H
? ] H [
a
?
= C ? = C C K
C
K
a
a
?
3
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
] H [
a
?
= 1 . 0 10 8 . 1 ] COOH CH [ K
5
3 a
? ? ? ?
?
= 1.34 ? 10
–3
M
At cathode:
From NH
4
OH

?
4
NH + OH
–
[OH
–
] = C ? = C K
C
K
b
b
?
? ] H [
c
?
=
C K
K
] OH [
K
b
w w
?
? =
] OH NH [ K
K
4 b
w
?
=
01 . 0 10 8 . 1
10
5
14
? ?
?
?
=  2.359 ? 10
-11
M
?E = 0  –  0.0591 log
11
3
10 359 . 2
10 34 . 1
?
?
?
?
= –0.4575 V
5: A current of 15 amperes is employed to plate nickel in a NiSO
4
bath. Both Ni and H
2
are
formed at cathode. The current efficiency with respect to the formation of Ni is 60%. (a)
How many g. of Ni are plated out in one hour? (b) What is the thickness of the plating
of the cathode consisting of a square sheet of metal of 4 cm. side which is coated on both
faces? The density of nickel is 8.9 g/cc. (c) What volume  of H
2
(STP) is formed per hour?
Solution: Total charge passed in one hour= (15 ? 60 ? 60) C
= 54000 C
But current efficiency with respect to Ni is 60%
Therefore, charge used up in deposition of Ni = 54000 ?
100
60
C
= 32400 C
And remaining charge will be used up to evolve H
2
i.e. equal to = (54000 – 32400) C = 21600 C
? Ni
+2
+ 2e
–
? Ni
Wt. Of Ni (plated) =
96500 2
32400 69 . 58
?
?
gm =  9.85 gm
Now, volume of square sheet of Ni metal required to coated on
both faces =
cc
8.9
9.85
so volume of square sheet of Ni metal required to coated on one face
=
cc
8.9 2
9.85
?
which should be equal to = (area of square) ? thickness of plate
= 16  cm
3
? thickness of plate
? thickness of plate =
16 2 9 . 8
85 . 9
? ?
= 0.035 cm
Now, 2H
+
+ 2e
–
? H
2
4
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Volume of H
2
evolved at STP =
96500 2
21600 4 . 22
?
?
= 2.5 litre e
6: How many moles of iron metal will be produced by passage of 4A of current through 1L
of 0.1 M Fe
3+
solution for 1 hour? Assume only iron Fe
+3
+ 3e
–
? Fe

is reduced.
Solution: Total charge passed = (4 ? 1 ? 60 ? 60) C
= 14400 C = 0.149 F
First of all ferric ion charges to ferrous and  than  ferrous changes to Fe.
Fe
+3
+ e
–
?? Fe
+2
Fe
+2
+ 2e
–
?? Fe
So, charge used up in formation of Fe(II) = 0.1 F
Therefore, remaining charge left = (0.149 – 0.1) F   =  0.049 F
Which is used in conversion of Fe
+2
to Fe
So, number of mole of Fe produced =
2
049 1 ?
=  0.0245
7: Per di-sulphuric acid (H-
2
­ S
2
O
8
) can be prepared by electrolytic oxidation of H
2
SO
4
as
2H
2
SO
4
? H
2
S
2
O
8
+ 2H
+
+ 2e
-
. O
2
and H
2
are byproducts. In such an electrolysis 0.87 g
of H
2
and 3.36 g of O
2
were generated at STP. Calculate the total quantity of current
passed through the solution to carry out electrolysis. Also report the weight of H
2
S
2
O
8
formed.
Solution: At anode: 2H
2
SO
4
? H
2
S
2
O
8

+

2H
+
+ 2e
–
2H
2
O ? O
2
+ 4H
+
+ 4e
–
At cathode:2H
+
+ 2e
–
? H
2
In electrolysis of H
2
SO
4
,
firstly H
2
SO
4
oxidises at anode untill all H
2
SO
4
converts into H
2
S
2
O
8
and
then O
2
is evolved by the electrolysis of H
2
O.
Simultaneously at that time H
2
is evolved at cathode which continue during whole electrolysis
process.
So, total charge required to produce 0.87 gm H
2
=
2
87 . 0 96500 2 ? ?
C
= 83955 C
Now, quantity of charge required to produce 3.36 gm
O
2
=
32
36 . 3 96500 4 ? ?
C = 40530 C
Thus, quantity of charge used up in the formation of H
2
S
2
O
8
= (83955 – 40530)=  43425 C
Therefore, weight of H
2
S
2
O
8
formed =
96500 2
43425 194
?
?
=  43.65 gm
8: Consider the disproportionation 2Cu
+

Cu
2+
+ Cu (s). At equilibrium the value of
? ?
2
Cu
2
Cu
?
?
is 1.8 ? 10
6
. If the standard potential for Cu
2+
| Cu
+
is +0.15V , calculate
o
Cu / Cu
E
?
o
Page 5

1
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Assignment    Physical Chemistry           IIT-JAM 2016
ELECTROCHEMISTRY
Ashoka Scientific Forum
Developing Scientific Temper Among Students
IIT-JAM | TIFR | DU | BHU | JNU | ISM | PU & All Other University M.Sc. Entrance Exam
CSIR-NET | GA TE | TIFR
Solved Problem
Subjective
1: During the discharge of a lead storage battery, the density of H
2
SO
4
falls from 1.294 to
1.139 g ml
-1
. H
2
SO
4
of density 1.294 g ml
-1
is 39% H
2
SO
4
by weight and that of density
1.139 g ml
-1
is 20% H
2
SO
4
by weight. The battery holds 3.5 litres of acid and volume
remains practically constant during the discharge. Calculate the number of ampere
hours for which the battery must have been used. The charging and discharging reactions
are:
Pb +
2
4
SO
?
? PbSO
4
+ 2e
-
(charging)
PbO
2
+ 4H
+
+ SO
4
2-
? PbSO
4
+ 2H
2
O (discharging)
Solution: The overall reaction is:
Pb + PbO
2
+ 2H
2
SO
4
? 2PbSO
4
+ 2H
2
O
It should be noted that 1 mole of Pb gets oxidised to 1 mole of Pb
2+
( in PbSO
4
). Therefore,
during the process, 2 moles of H
2
SO
4
are consumed with the utilisation of 2 moles of electrons. In
other words, the number of moles of H
2
SO
4
is equal to the number of Faradays consumed.
Mass of sample in 3.5 L  of acid before the discharge
= 3.5 ?1.294 ?10
3
= 4529 g
Mass of H
2
SO
4
= 4529 ?
100
39
=  1766.31 g
Moles of H
2
SO
4
=
98
31 . 1766
= 18.02
Mass of sample after discharge = 1.139 ? 10
3
? 3.5 = 3986.5 g
Mass of H
2
SO
4
=
20
100
? 3986.5 = 797.3 g
No. of moles of H
2
SO
4
=
797.3
98
= 8.135
Moles of H
2
SO
4
consumed = 18.02 – 8.135 =  9.88
? No of Faradays utilised = 9.88 F
?Charge = 9.88 ? 96500 = 953833.57 C
1 ampere hour = charge passed by one ampere in one hour
= 60 ? 60 = 3600 C
?No of ampere hours =
953833.57
3600
= 264.95
2
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2: Cadmium amalgam is obtained by the electrolysis of a solution of cadmium chloride
using a mercury cathode. For how long the process of electrolysis should be carried out
to prepare a 12% (by wt.) of cadmium amalgam using 44 gms of mercury and a current of
5 amperes? (At. Wt. Of Cd = 112)
Solution: Since 88 gm Hg has 12 gm Cd
So 44 gm Hg require =
88
44 12 ?
gm Cd = 6 gm Cd
Now, 56 gm Cd is required to deposit 96500 C charge
So, 6 gm Cd required to deposit =
56
96500
? 6 C = 10339.286 C
Time  =
5
10339.286
=
I
Q
= 20678.57 Sec. =  34.4 min
3: Calculate the equilibrium constant for the reaction:
Fe
2+
+ Ce
4+
? Ce
3+
+ Fe
3+
Given that, Fe
3+
+ e
-
? Fe
2+
; E
0
= 0.771 V
Ce
4+
+ e
-
? Ce
3+
; E
0
= 1.61 V
Solution: The equilibrium reaction is Fe
2+
+ Ce
4+

Ce
3+
+ Fe
3+
Since the reaction is at equilibrium, E
cell
= 0
0
cell
E =
K log
1
0591 . 0
0
Ce / Ce
0
Fe / Fe
3 4 3 2
E E
? ? ? ?
? = 0.0591 log K
– 0.771 + 1.61 = 0.0591 log K
log K =
0591 . 0
839 . 0
= 14.1963
?K = 1.57 ? 10
14
4: Calculate the emf of the cell:
Pt H
2
| CH
3
COOH  || NH
4
OH | H
2
Pt
1 atm.   0.1 M            0.01 M   1 atm.
K
a
for CH
3
COOH = 1.8 x 10
-5
and K
b
for NH
4
OH = 1.8 x 10
-5
Solution: The reactions are : at anode : ½ H
2
?
?
a
H + e
–
at cathode :
?
c
H + e
–
? ½ H
2
———————————————————————
overall reaction
?
c
H ?
?
a
H
E = E
0
– 0.059 log
]
H
[
]
H
[
+
c
+
a
At anode:
From CH
3
COOH

CH
3
COO
–
+
?
a
H
? ] H [
a
?
= C ? = C C K
C
K
a
a
?
3
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
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] H [
a
?
= 1 . 0 10 8 . 1 ] COOH CH [ K
5
3 a
? ? ? ?
?
= 1.34 ? 10
–3
M
At cathode:
From NH
4
OH

?
4
NH + OH
–
[OH
–
] = C ? = C K
C
K
b
b
?
? ] H [
c
?
=
C K
K
] OH [
K
b
w w
?
? =
] OH NH [ K
K
4 b
w
?
=
01 . 0 10 8 . 1
10
5
14
? ?
?
?
=  2.359 ? 10
-11
M
?E = 0  –  0.0591 log
11
3
10 359 . 2
10 34 . 1
?
?
?
?
= –0.4575 V
5: A current of 15 amperes is employed to plate nickel in a NiSO
4
bath. Both Ni and H
2
are
formed at cathode. The current efficiency with respect to the formation of Ni is 60%. (a)
How many g. of Ni are plated out in one hour? (b) What is the thickness of the plating
of the cathode consisting of a square sheet of metal of 4 cm. side which is coated on both
faces? The density of nickel is 8.9 g/cc. (c) What volume  of H
2
(STP) is formed per hour?
Solution: Total charge passed in one hour= (15 ? 60 ? 60) C
= 54000 C
But current efficiency with respect to Ni is 60%
Therefore, charge used up in deposition of Ni = 54000 ?
100
60
C
= 32400 C
And remaining charge will be used up to evolve H
2
i.e. equal to = (54000 – 32400) C = 21600 C
? Ni
+2
+ 2e
–
? Ni
Wt. Of Ni (plated) =
96500 2
32400 69 . 58
?
?
gm =  9.85 gm
Now, volume of square sheet of Ni metal required to coated on
both faces =
cc
8.9
9.85
so volume of square sheet of Ni metal required to coated on one face
=
cc
8.9 2
9.85
?
which should be equal to = (area of square) ? thickness of plate
= 16  cm
3
? thickness of plate
? thickness of plate =
16 2 9 . 8
85 . 9
? ?
= 0.035 cm
Now, 2H
+
+ 2e
–
? H
2
4
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
Volume of H
2
evolved at STP =
96500 2
21600 4 . 22
?
?
= 2.5 litre e
6: How many moles of iron metal will be produced by passage of 4A of current through 1L
of 0.1 M Fe
3+
solution for 1 hour? Assume only iron Fe
+3
+ 3e
–
? Fe

is reduced.
Solution: Total charge passed = (4 ? 1 ? 60 ? 60) C
= 14400 C = 0.149 F
First of all ferric ion charges to ferrous and  than  ferrous changes to Fe.
Fe
+3
+ e
–
?? Fe
+2
Fe
+2
+ 2e
–
?? Fe
So, charge used up in formation of Fe(II) = 0.1 F
Therefore, remaining charge left = (0.149 – 0.1) F   =  0.049 F
Which is used in conversion of Fe
+2
to Fe
So, number of mole of Fe produced =
2
049 1 ?
=  0.0245
7: Per di-sulphuric acid (H-
2
­ S
2
O
8
) can be prepared by electrolytic oxidation of H
2
SO
4
as
2H
2
SO
4
? H
2
S
2
O
8
+ 2H
+
+ 2e
-
. O
2
and H
2
are byproducts. In such an electrolysis 0.87 g
of H
2
and 3.36 g of O
2
were generated at STP. Calculate the total quantity of current
passed through the solution to carry out electrolysis. Also report the weight of H
2
S
2
O
8
formed.
Solution: At anode: 2H
2
SO
4
? H
2
S
2
O
8

+

2H
+
+ 2e
–
2H
2
O ? O
2
+ 4H
+
+ 4e
–
At cathode:2H
+
+ 2e
–
? H
2
In electrolysis of H
2
SO
4
,
firstly H
2
SO
4
oxidises at anode untill all H
2
SO
4
converts into H
2
S
2
O
8
and
then O
2
is evolved by the electrolysis of H
2
O.
Simultaneously at that time H
2
is evolved at cathode which continue during whole electrolysis
process.
So, total charge required to produce 0.87 gm H
2
=
2
87 . 0 96500 2 ? ?
C
= 83955 C
Now, quantity of charge required to produce 3.36 gm
O
2
=
32
36 . 3 96500 4 ? ?
C = 40530 C
Thus, quantity of charge used up in the formation of H
2
S
2
O
8
= (83955 – 40530)=  43425 C
Therefore, weight of H
2
S
2
O
8
formed =
96500 2
43425 194
?
?
=  43.65 gm
8: Consider the disproportionation 2Cu
+

Cu
2+
+ Cu (s). At equilibrium the value of
? ?
2
Cu
2
Cu
?
?
is 1.8 ? 10
6
. If the standard potential for Cu
2+
| Cu
+
is +0.15V , calculate
o
Cu / Cu
E
?
o
5
2637, Hudson Lane, Behind Khalsa College, Near G .T.B. Nagar Metro Station Gate No. 3 & 4, New Delhi - 110009
Mob. 09555785548, 08860929430, e-mail: info@asfinstitute.com, www .asfinstitute.com
Solution: 2Cu
+
Cu
++
+ Cu
Given:
? ?
?
?
?
2
2
Cu
Cu
1.8 ? 10
6
& ? ?
Cu /
0
Cu
2
E = + 0.15 V
At equilibrium EMF is zero
So,
0
cell
E =
2
2
] Cu [
] Cu [
log
n
0591 . 0
?
?
1
0591 . 0
E E
0
Cu / Cu
0
Cu / Cu
2
? ?
? ? ? log (1.8 ? 10
6
)
–0.15 +
0
Cu / Cu
E
? = 0.3697
0
Cu / Cu
E
? = + 0.5197 V
9: A current of 40 microamperes is passed through a solution of AgNO
3
for 32 minutes
using Pt electrodes. An uniform single atom thick layer of Ag is deposited covering 43%
cathode surface. What is the total surface area of cathode, if each Ag atom  covers 5.4 ?
10
-16
cm
2
?
Solution: Total charge passed = (4 ?10
–6
? 32 ? 60) C =  0.0768 C
Ag
+
+ e
–
? Ag
So, weight of Ag deposited at cathode surface =
96500
0768 . 0 108 ?
gm
= 8.6 ? 10
-5
gm
Now, total no. of Ag atoms deposited =
23
5
10 023 . 6
108
10 6 . 8
? ?
?
?
= 4.8 ? 10
17
So, surface area of cathode which is covered by Ag atoms
= (4.8 ? 10
17
? 5.4 ? 10
–16
)= 259.2 cm
2
?Total surface area =
43
100 2 . 259 ?
cm
2
=

602.8 cm
2
10: Calculate the potential of an indicator electrode versus the standard hydrogen electrode
which originally has 0.1M  MnO
4
–
and 0.8M H
+
and which has been treated with 90% of
the Fe
2+
necessary to reduce all the MnO
4
–
to Mn
2+
.
MnO
4
–
+ 8H
+
+ 5e
–
? ? Mn
2+
+ 4H
2
O E° = 1.51 V
Solution: Let us consider Galvanic cell is
H
+
(1M)  | H
2
(1atm), Pt || MnO
4
–
(H
+
) | Mn
+2
, Pt
Anode half cell : 2H
+
(1M) ?? H
2
(1atm) + 2e
–
Cathode half cell: MnO
4
–
+ 8H
+
+ 5e
–
??  Mn
+2
+ 4H
2
O
Initial Conc.: 0.1 0.8    0 0
Alter Complete
?
?
?
?
?
?
? ? 90
100
1 . 0
1 . 0 ?
?
?
?
?
?
?
?
? 8
100
90 1 . 0
8 . 0
100
90 1 . 0 ?
reaction with Fe
+2
(0.01) (0.08) (0.09)
So, electrode potential of indicator electrode
2
4
Mn / MnO
E
? ?
=
8 -
4
2
Mn / MnO
o
] [H ] [MnO
] Mn [
log
5
0591 . 0
E
2
4
?
?
?
? ?
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