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CE – 1
ELECTROCHEMISTRY
C1
Conductance :
The resistance of any conductor varies directly as its length (l) and inversely as its cross-sectional area (a).
R
l
a
R
l .
a
where  is a constant depending upon the nature of the material and is called specific resistance of
resistivity of the material.
Specific resistance is the resistance of one centrimetre cube of a material.
1
The reciprocal of the specific resistance,   is called specific conductance or conductivity..

Conductivity is the conductance of one cm cube of a material.
taking
l
1
1
 K (specific conductance),
 C (conductance) K = C ×  a  ....
 

R
l
  is called cell constant, l = distance between two electrodes , a = cross-sectional area of electrodes
a
C2A Equivalent Conductance :
It is defined as the conducting power of all the ions present in one gram equivalent of an electrolyte in a
given solution. At concentration N (in gm equivalent L–1) equivalent conductance is denoted by eq
(equivalent conductance at concentration c)
c =
k  1000
; units: ohm–1 cm2, eq–1
N
Equivalent conductance increases with dilution. When the solution is infinitely diluted the equivalent
conductance is denoted as . The  can be determined by extrapolation method, in which graph between
c and C is extended to zero concentration.
C2B
Molar Conductance :
It is defined as the conducting power of all the ions present in one mol of an electrolyte in a given solution.
m 
k  1000
unit : ohm–1 cm2 mol–1
M
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CE – 2
eq and m both increases with decrease in concentration.
C2C Kohlrausch’s Law :
At infinite dilution, when ionisation is complete, each ion makes its contribution towards equivalent
conductance of the electrolyte and at infinite dilution equivalent conductance is given by the sum of the
equivalent conductances of the contributing ions.
m = xa + yb
Thus for AxBy,
where, a is the molar conductance of A(say cation) and b that of B(say anion) at infinite dilution.
(degree of dissociation of weak electrolyte)  
cm
m
Practice Problems :
1.
Molar conductance at infinite dilution of BaCl2, H2SO4 and HCl aq. solution are x1, x2 and x3
respectively. Molar conductance of BaSO4 solution is :
(a)
x1 + x2 – x3
(b)
x1 – x2 – x3
(c)
x1 + x2 – 2x3
(d)
x1 – 2x2 + x3
[Answers : (1) c]
C3A Electrical Conductors :
All substances which can allow the flow of electricity are known as electrical conductors. Mainly we have
two types of electrical conductors which are given below :
1.
Electronic Conductor : They are those conductors in which the flow of electricity is due to the movement
of loosely bonded electrons in their own standard state. In this case, the movement of matter does not take
place during the flow of electricity. For example :
(a)
2.
all metals in their elemental state
(b)
graphite and
(c)
alloys
Electrolytical Conductors :
In this case of electrolytic conductors, the flow of electricity is due to the movement of ions i.e., here actual
transport of matter takes place.
C3B
Electrolytes :
Electrolytes may be pure substances (e.g., salts, acids or bases) in their fused states or more commonly they
are aqueous solutions of these compounds or they are sometimes pure liquid.
There are two types of electrolytes :
(a)
Strong Electrolyte : The compounds which are 100% ionised at any dilution, are treated as
strong electrolytes, for e.g., HClO4, HI, HBr, HCl, H2SO4, HNO3, NaOH, KOH, NaCl etc.
(b)
Weak Electrolyte : The compounds which are less or feebly ionised (at lower dilution) are
treated as weak electrolytes.
All weak acids, weak bases or the salts having less ionic character are treated as weak
electrolytes e.g. CH3COOH,
, H2CO3, Mg(OH)2, Zn(OH)2 etc.
C4A Electrolysis :
Electrolysis is a process which involves a chemical change at the electrodes when electricity is passes
through an electrolyte.
“Electrolysis is a process which involves a chemical change at the electrodes when electricity is passed
through an electrolyte”.
It was found experimentally that the positive ions from the electrolyte are attracted on the negative
electrode (cathode) and get reduced. Similarly, the negative ions are attracted on the positive electrode
(anode) and are oxidised.
Hence due to electrolysis, the species under consideration gets decomposed at cathode and anode.
Faraday has established a relationship between the amount of electricity passes through the electrolyte and
the amount of chemical change occurring at an electrode. This relationship is known as Faraday’s law of
electrolysis.
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 3
C4B
Faraday’s Laws of Electrolysis :
First Law of Electrolysis :
The amount of chemical change produced is proportional to the quantity of electric charge passing through
an electrolysis cell.
Suppose after passing Q coulombs of electricity (amount of electricity), W amount of a substance has
appeared (or disappeared).
Thus
WQ
W = ZQ (Where Z is a constant, which is known as electrochemical equivalent)
*
Also
W = Z i t.....
(Q = i t, where i is current passed for t seconds)
Q = nF.....
(F is a faraday constant i.e., it is charged carried by one mole of electron,
where n is no.of mole of electrons transfer takes place).
Second Law of Electrolysis :
When same amount of amount of electricity is passed through different electrolytes, the weight of the
substance appeared or disappeared by any electrode is always directly proportional to the equivalent weight
of the substance appeared or disappeated.
According to the second law of electrolysis
W  E, in general,
W1 W2

E1
E2
C4C Relation between Electrochemical Equivalent and Equivalent Weight :
By the use of above two laws
EZ
Thus,
E = FZ....
(where F is Faraday’s constant, E is equivalent
weight and Z is electrochemical equivalent).
If we combine first and second law of electrolysis then following expressions takes place :
n eq 
W Q
W It , (n )
= (neq)reduced

or mol  n.factor 

eq oxidised
E
F
E
F
Practice Problems :
1.
The density of Cu is 8.94 g cm–3. The quantity of electricity needed to plate an area 10 cm 10 cm to a
thickness of 10–2 cm using CuSO4 solution is
(a)
2.
(a)
3.
13586 C
(b)
27172 C
(c)
40758 C
(d)
20348 C
The same amount of electricity was passed through two cells containing molten Al2O3 and molten
NaCl. If 1.8 g of Al were liberated in one cell, the amount of Na liberated in the other cell is
4.6 g
(b)
2.3 g
(c)
6.4 g
(d)
3.2 g
Silver is removed electrolytically from 200 mL of a 0.1 N solution of AgNO3 by a current of 0.1
ampere. How long will it take to remove half of the silver from the solution (At. wt. of Ag = 108 g)
(a)
10 sec
Einstein Classes,
(b)
16 sec
(c)
100 sec
(d)
9650 sec
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 4
4.
Consider the following electrolysis
(1)
CuSO4
(2)
Fe2(SO4)3
(3)
AlCl3
(4)
AgNO3
The quantity of electricity needed to electrolyse completely 1 M solutions of these electrolytes will be
5.
(a)
2F, 6F, 3F and 1F respectively
(b)
6F, 2F, 3F and 1F respectively
(c)
2F, 6F, 1F and 3F respectively
(d)
6F, 2F, 1F and 3F respectively
Assume that during electrolysis of AgNO3, only H2O is electrolysed and O2 is formed as :
2H2O  4H+ + O2 + 4e—
O2 formed at N.T.P. due to passage of 2 amperes of current for 965 s :
(a)
0.112 L
(b)
0.224 L
(c)
11.2 L
(d)
22.4 L
[Answers : (1) b (2) a (3) d (4) a (5) a]
C5A Electrochemical Cell :
A devide that converts chemical energy into electrical energy. The cell is based on the principle of indirect
redox reactions, i.e, the oxidation and reduction reactions takes place in different container.
The electrochemical cells (galvanic or voltaic) consists of two half cells connected with each other by
means of an electric wire to allow an indirect redox reaction. Although the solutions of two half cells are in
different containers, but for the continuous flow of electricity, transfer of ions from one solution to another
solution are always allowed. Oxidation at anode and is a –ve electrode, reduction at cathode and is a +ve
electrode. For e.g., Zn-Cu galvanic cell is represented as follows :
C5B
Electrode Potential :
It is the potential difference established between the electrode and its electrolyte. It is of two types i.e.,
Reduction Potential and Oxidation Potential.
Reduction Potential is tandency of an electrode to receive electrons for the deposition of the solvated
positive ions from its own solution. Whereas oxidation potential is the tandency of an electron to get
oxidised in its own solution.
C5C E.M.F. (Electromotive Force) or Cell Potential of a Cell :
The difference between the reduction potentials of two half cells constituting the Galvanic cell is known as
its cell potential (or e.m.f.) i.e.,
Ecell(e.m.f.) = (Reduction Potential)Cathode – (Reduction Potential)Anode
or,
ECell = ECathode – EAnode
For the spontaneous flow of electricity from the Galvanic cell, the e.m.f. of the cell must be positive.
C5D Concentration Effect in Voltaic Cell – Nernst Equation
Nerst equation gives a quantitative relationship between the concentration of ions and electrode potential.
For a general electrode reaction : Mn+ + L(s)
Ln+ + M(s).
Emf(E) = E0 – 2.303
Einstein Classes,
RT
[Ln  ]
log
nF
[M n  ]
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 5
Practice Problems :
1.
2.
By how much will the potential of half-cell Cu2+ | Cu change if the solution is diluted to 100 times at
298 K ?
(a)
Increase by 59 mV
(b)
Decreases by 59 mV
(c)
Increase by 29.5 mV
(d)
Decreases by 29.5 mV
Which of the following changes will increase the EMF of the cell :
Co(s) | CoCl2(M1) || HCl (M2) | Pt (H2, g)
3.
(a)
increase in the volume of CoCl2 solution from 100 mL to 200 mL
(b)
increase M2 from 0.01 M to 0.50 M
(c)
increase the pressure of the H2(g) from 1.00 to 2.00 atm
(d)
increase M1 from 0.01 M to 0.50 M
The measured voltage of the following cell is 0.9 V at 250C. Pt, H2(1 atm) | H+(aq) || Ag+ (1.0 M) | Ag
If E 0 
 0.80V . The pH of the aqueous solution of H+ ion is
Ag / Ag
(a)
1.69
(b)
2.50
(c)
5.20
(d)
9.69
[Answers : (1) b (2) b (3) a]
C6
Application of Nernst Equation
1.
Electrical work :
G = –nFE (in a given state), G0 = –nFE0 (in a standard state), G0 = –RT In K aq
G10 + G20 = G30 (when different no. of element are involved, if equal no of electrons are involved
E10 + E20 = E30.
2.
At equilibrium
E0 
3.
E = 0 and Q = K (equilibrium constant)
RT
0.059
ln K 
log10 K
nF
n
For standard hydrogen electrods (SHE)
E0(SHE) = 0.00 V.
SHE Pt | H2(g) (1 atm) | HCl aq (1M)
SHE, colomel-electrode and silver, silver-chloride electrodes are used as reference half cells.
EH
4.
2
/ H
 0.059 p H
Concentration cells
Zn | Zn2+ (C1) || Zn2+ (C2) | Zn
E cell 
C
0.059
log 2 ........[C > C ]
1
2
n
C1
Pt (H2) (P1) | HCl | Pt (H2) (P2)
Ecell = 0.059 log
P1
......[P1 > P2]
P2
Practice Problems :
1.
E0 for the cell Zn | Zn2+(aq) || Cu2+ (aq) | Cu is 1.10 V at 250C. The equilibrium constant for the
reaction Zn + Cu2+ (aq)
(a)
10–37
Einstein Classes,
Cu + Zn2+ (aq) is of the order of
(b)
1037
(c)
10+18
(d)
1017
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 6
2.
Pt (H 2 ) | H  (1M) | Pt (H 2 ) (where p1 and p2 are pressures) cell reaction will be spontaneous if :
p1
3.
p2
(a)
p1 = p2
(c)
p2 > p1
(b)
p1 > p2
(d)
2+
2+
p1 = p2 = 1atm
+
The standard reduction potential of Cu | Cu and Cu | Cu are 0.337 and 0.153 V respectively. The
standard electrode potential of Cu+ | Cu half cell is
(a)
0.184 V
(b)
0.827 V
(c)
0.521 V
(d)
0.490 V
[Answers : (1) b (2) b (3) c]
C7A Batteries
Any battery (actually it may have one or more than one cell connected in series) or cell that we use as a
source of electrical energy is basically a galvanic cell where the chemical energy of the redox reaction is
converted into electrical energy.
1.
Primary Batteries : In primary batteries, the reaction occurs only once and battery then becomes dead
after use over a period of time and cannot be reused again, for e.g., :
(a)
Dry cell : Which is used commonly in our transistors and clocks. The cell consists of a
zinc container that also acts as anode and the cathode is a carbon (graphite) rod surrounded by
powdered magnese dioxide and carbon. The space between the electrodes is filled by a moist
paste of NH4Cl and ZnCl2. The electrode reactions are complex, but they can be written
approximately as follows :
Anode : Zn(s)  Zn2+ + 2e–
Cathode : MnO2 + NH4+ + e–  MnO(OH) + NH3
Ammonia produced in the reaction forms complex with Zn2+ to give [Zn(NH3)4]2+. The cell has
a potential of nearly 1.5 V.
(b)
Mercury cell : Suitable for the low current devides like hearing aids and camera etc. consists of
zinc-mercury amalgam as anode and a paste of HgO and carbon as the cathode. The electrolyte
is a paste of KOH and ZnO. The electrode reactions for the cell are given below :
Anode : Zn(Hg) + 2OH–  ZnO(s) + H2O + 2e–
Cathode : HgO + H2O + 2e–  Hg(l) + 2OH–
2.
Secondary Batteries : A secondary cell after use can be rechanged by passing current through it in
opposite direction so that it can be used again. A good secondary cell can undergo a large number of
discharging and charging cycles, for e.g. :
(a)
Lead storage battery : commonly used in automobiles and invertors. It consists of a lead anode
and a grid of lead packed with lead dioxide (PbO2) as cathode. A 38% solution of sulphuric acid
is used as an electrolyte.
The cell reactions when the battery is in use are given below :
Anode : Pb(s) + SO42–(aq)  PbSO4(s) + 2e–
Cathode : PbO2(s) + SO42–(aq) + 4H+(aq) + 2e–  PbSO4(s) + 2H2O (l)
i.e., overall cell reaction consisting of cathode and anode reaction is :
Pb(s) + PbO2(s) + 2H2SO4(aq)  2PbSO4(s) + 2H2O(l)
On charging the battery the reaction is reversed and PbSO4(s) on anode and cathode is converted
into Pb and PbO2, respectively.
(b)
Nickel-cadmium cell : Which has longer life than the lead storage cell but more expensive to
manufacture. We shall not go into detains of working of the cell and the electrode reactions
during charging and discharging. The overall reaction during discharge is :
Cd(s) + 2Ni(OH)3(s)  CdO (s) + 2Ni(OH)2 (s) + H2O(l)
3.
Fuel Cells : Production of electricity by thermal plants is not a very efficient method and is a major source
of pollution. In such plants, the chemical energy (heat of combustion) of fossil fuels (coal, gas or oil) is first
used for converting water into high pressure steam. This is then used to run a turbine to produce electricity.
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 7
It is now possible to make such cells in which reactants are fed continuously to the electrodes and products
are removed continuously from the electrolyte compartment. Galvanic cell that are designed to convert the
energy of combustion of fuels like hydrogen, methane, methanol etc. directly into electrical energy are
called fuel cells. One of the most successful fuel cells uses the reaction of hydrogen with oxygen to form
water. The cell was used for providing electrical power in the Apollo space programme. The water vapours
produced during the reaction were condensed and added to the drinking water supply for the astronauts.
Catode : O2(g) + 2H2O(l) + 4e–  4OH–(aq)
Anode : 2H2 + 4OH–(aq) 4H2O(l) + 4e–
Overall reaction being : 2H2(g) + O2(g)  2 H2O(l)
C7B
Corrosion
Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal. The rusting
of iron, tarnishing of silver, development of green coating on copper and bronze are some of the examples
of corrosion.
In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of iron
(commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite
complex but it may be considered essentially as an electrochemical phenomenon. At a particular spot of an
object made of iron, oxidation takes place and that spot behaves as an anode and we can write the
reaction :
Anode : 2 Fe(s)  2 Fe2+ + 4e–
E0(Fe2+, Fe) = –0.44 V
Cathode : O2(g) + 4H+(aq) + 4e–  2 H2O (l)
[E0 = 1.23 V]
The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the
form of hydrated ferric oxide (Fe2O3 . xH2O) and with further production of hydrogen ions.
Prevention of corrosion : one of the simplest method of preventing corrosion is to prevent the surface of
the metallic object to come incontact with atmosphere. This can be done by covering the surface by paint or
by some chemicals (e.g. bisphenol). Other simple method is to cover the surface by other metals (Sn, Zn
etc.) that are inert or react to save the object. An electrochemical method is to provide a sacrificial electrode
of another metal (like Mg, Zn etc.) which corrodes itself but saves the object.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 8
SINGLE CORRECT CHOICE TYPE
1.
2.
3.
The density of Cu is 8.94 g cm–3. The quantity of
electricity needed to plate an area 10 cm 10 cm to a
thickness of 10–2 cm using CuSO4 solution is
(a)
13586 C
(b)
27172 C
(c)
40758 C
(d)
20348 C
A gas X at 1 atm is bubbled through a solution
containing a mixture of 1 M Y— and 1 M Z— at 250C.
If the reduction potential of Z > Y > X, then
The Zn electrode had twice the surface
of the Cu electrode.
(c)
The [Zn2+] was larger than the [Cu2+]
(d)
The volume of the Zn2+ solution was
larger than the volume of the Cu2+
solution.
(b)
Y will oxidize Z and not X
(c)
Y will oxidize both X and Z
pressures) cell reaction will be spontaneous if :
(d)
Y will reduce both X and Z
(a)
p1 = p2
(b)
p1 > p2
(c)
p2 > p1
(d)
p1 = p2 = 1atm
0
cell
The E
7.
2+
+
2+
3+
(a)
50
(b)
10
(c)
1050
(d)
105
8.
Which of the following changes will increase the
EMF of the cell :
(a)
increase in the volume of CoCl2 solution
from 100 mL to 200 mL
(b)
increase M2 from 0.01 M to 0.50 M
(c)
increase the pressure of the H2(g) from
1.00 to 2.00 atm
(d)
increase M1 from 0.01 M to 0.50 M
Cu + Zn2+ (aq) is of the order
(a)
10–37
(b)
1037
(c)
10+18
(d)
1017
9.
Using the standard potential values given below,
decide which of the statements I, II, III, IV are
correct. Choose the right answer from (a), (b), (c)
and (d)
—
0
Fe + 2e = Fe,
E = 0.44 V
Cu2+ + 2e— = Cu,
E0 = +0.34 V
Ag+ + e— = Ag,
E0 = +0.80 V
I.
Copper can displace iron from FeSO4
solution
II.
Iron can displace copper from CuSO4
solution
III.
Silver can displace copper from CuSO4
solution
IV.
Iron can displace silver from AgNO3
solution
(a)
I and II
(b)
II and III
(c)
II and IV
(d)
I and IV
For the cell Zn(s) | Zn2+ || Cu2+ | Cu(s), the standard
cell voltage, E0cell is 1.10 V. When a cell using these
reagents was prepared in the lab, the measured cell
voltage was 0.98 V. One possible explanation for
the observed voltage is :
Einstein Classes,
p2
Co(s) | CoCl2(M1) || HCl (M2) | Pt (H2, g)
E0 for the cell Zn | Zn2+(aq) || Cu2+ (aq) | Cu is 1.10 V
at 250C. The equilibrium constant for the reaction
2+
Pt (H 2 ) | H  (1M) | Pt (H 2 ) (where p1 and p2 aree
p1
in which the reaction :
Zn + Cu2+ (aq)
of
6.
(b)
Y will oxidize X and not Z
MnO + Fe + H  Mn + Fe + H2O occurs is
0.59 V at 250C. The equilibrium constant for the
reaction is approximately of the order of
5.
There were 2.00 mol of Zn2+ but only 1.00
mol of Cu2+.
(a)
—
4
4.
(a)
10.
11.
12.
Cost of electricity for the production of x L H2 at
NTP at cathode is Rs. x, then cost of electricity for
the production of x L O2 gas at NTP at anode will
be (assume 1 mol of electrons as one unit of
electricity) :
(a)
2x
(b)
4x
(c)
16x
(d)
32x
Molar conductance at infinite dilution of BaCl2,
H 2SO 4 and HCl aq. solution are x 1, x 2 and x 3
respectively. Molar conductance of BaSO4 solution
is :
(a)
x1 + x2 – x3
(b)
x1 – x2 – x3
(c)
x1 + x2 – 2x3
(d)
x1 – 2x2 + x3
The same amount of electricity was passed through
two cells containing molten Al2O3 and molten NaCl.
If 1.8 g of Al were liberated in one cell, the amount
of Na liberated in the other cell is
(a)
4.6 g
(b)
2.3 g
(c)
6.4 g
(d)
3.2 g
Salts of A (atomic weight 7), B (atomic weight 27)
and C (atomic weight 48) were electrolysed under
identical conditions using the same quantity of
electricity. It was found that when 2.1 g of A was
deposited, the weights of B and C deposited were
2.7 g and 7.2 g. The valencies of A, B and C are
respectively
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 9
13.
14.
(a)
3, 1 and 2
(b)
1, 3 and 2
(c)
3, 1 and 3
(d)
2, 3 and 2
20.
Silver is removed electrolytically from 200 mL of a
0.1 N solution of AgNO 3 by a current of 0.1
ampere. How long will it take to remove half of the
silver from the solution (At. wt. of Ag = 108 g)
(a)
10 sec
(b)
16 sec
(c)
100 sec
(d)
9650 sec
21.
The cell reaction for the given cell is spontaneous
if :
The value of the standard potentials for reduction
reactions of A+/A, B2+/B, C2+/C and D2+/D at 250 are
0.80, 0.34, –0.76 and –1.66 volts respectively. The
correct sequence in which these metals will be
deposited on the cathode is
(a)
A, B, C, D
(b)
D, C, B, A
(c)
A, C, B, D
(d)
D, B, C, A
The standard reduction potential for half-reaction
for four different elements A, B, C and D are
PtCl2|Cl—(1M) || Cl— (1M) | PtCl2
P1
P2
15.
16.
17.
(a)
P1 > P2
(b)
P1 < P2
(c)
P1 = P2
(d)
P2 = 1 atm
B2 + 2e  2B—
E0 = +1.36 V
C2 + 2e—  2C—
E0 = +1.06 V
D2 + 2e—  2D—
E0 = +0.53 V
(a)
2.89 × 105
(b)
2.89
(a)
would be A and D respectively
(c)
1.89 × 102
(d)
5.89 × 109
(b)
would be D and A respectively
(c)
would be B and C respectively
(d)
cannot be ascertained from the given data
as the species being subjected to
oxidation or reduction have not been
indicated
The molar conductance of KCl, KNO3 and AgNO3
are 149.9, 145.0, and 133.4 ohm–1 cm2 mol–1. (all at
25 0C). The molar conductance of AgCl at this
temperature is
(a)
144.4
(b)
120.8
(c)
138.3
(d)
178.2
22.
Given the following half cell reaction and
corresponding reduction potentials :
The reaction ½ H2(g) + AgBr(s)  H+(aq) + Br—
(aq) + Ag(s) occurs in the galvanic cell
(a)
Ag | AgBr(s) | KBr (aq) || AgNO3 (aq) | Ag
(i)
A + e —  A—
E0 = – 0.24 V
(b)
Pt | H2(g) | HBr (aq) || AgNO3 (aq) | Ag
(ii)
B— + e—  B2—
E0 = 1.25 V
(c)
Pt | H2(g) | HBr (aq) | AgBr(s) | Ag
(iii)
C— + 2e—  C3—
E0 = 0.68 V
(d)
Pt | H2(g) | KBr (aq) || AgBr (s) | Ag
E + 4e—  E4—
E0 = 0.38 V
23.
The largest potential resulted by the combination
of two half cells is
In a electrolytic cell is which of the following
statement is correct :
(a)
oxidation occurs at cathode
(a)
E0 = 2.50 V
(b)
E0 = 1.50 V
(b)
reduction occurs at anode
(c)
E0 = 0.50 V
(d)
E0 = 1.20 V
(c)
anode acts as a negative terminal
(d)
flow of electron takes place from anode
to cathode
Consider the following electrolysis
(1)
CuSO4
(2)
Fe2(SO4)3
(3)
AlCl3
(4)
AgNO3
24.
The quantity of electricity needed to electrolyse
completely 1 M solutions of these electrolytes will
be
19.
–E0 = +2.85 V
The strongest oxidising and reducing agents among
these
The charge in coulomb on 1 g ion of N–3 is
(iv)
18.
A2 + 2e—  2A—
(a)
2F, 6F, 3F and 1F respectively
(b)
6F, 2F, 3F and 1F respectively
(c)
2F, 6F, 1F and 3F respectively
(d)
6F, 2F, 1F and 3F respectively
25.
The standard oxidation potentials of the electrodes
Ag|Ag+, Sn|Sn2+, Ca|Ca2+, Pb|Pb2+ are –0.8, 0.1.36,
2.866 and 10.126 V respectively. The most powerful oxidising agent among these metals is
(a)
Pb
(b)
Ca
(c)
Sn
(d)
Ag
By how much will the potential of half-cell
Cu2+ | Cu change if the solution is diluted to 100
times at 298 K ?
A solution of sodium sulphate in water is
electrolysed using inert electrodes. The products at
the cathode and anode are respectively
(a)
Increase by 59 mV
(b)
Decreases by 59 mV
(c)
Increase by 29.5 mV
(a)
H2, O2
(b)
O2, H2
(d)
Decreases by 29.5 mV
(c)
O2, Na
(d)
O2, SO2
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 10
26.
Given that
E 0Fe 3 |Fe and E 0Fe 2 |Fe are –0.36 V and
– 0.439 V, respectively. The value of
31.
E 0Fe3 ,Fe 2 |Pt
would be
27.
(a)
(–0.36 – 0.439)V
(b)
[3(–0.36) + 2(–0.439)]V
(c)
(–0.36 + 0.439)V
(b)
[3(–0.36) – 2(–0.439)]V
For the half cell
32.
A hydrogen electrode placed in a buffer solution of
CH3COONa and acetic in the ratio’s x : y and y : x
has electrode potential values E1 volt and E2 volt
respectively at 2500C. The pKa values of acetic acid
is (E1 and E2 are oxidation potential)
(a)
E1  E 2
0.118
(c)

E1  E 2
0.118
(b)
E 2  E1
0
0.118
(d)
E1  E 2
0
0.118
The measured voltage of the following cell is 0.9 V
at 250C. Pt, H2(1 atm) | H+(aq) || Ag+ (1.0 M) | Ag
If E 0 
 0.80V . The pH of the aqueous
Ag / Ag
+ 2H+ + 2e—, E0 = 1.30 V
solution of H+ ion is
At pH = 2, electrodes potential is :
(a)
1.36 V
(c)
28.
1.42 V
+
(b)
1.30 V
(d)
1.20 V
+
Ag | Ag (1M) || Ag (2M) | Ag
1 L solution
1 L solution
0.5 F of electricity in the LHS (anode) and 1F of
electricity in the RHS (cathode) is first passed
making them independent electrolytic cells at 298
K. EMF of the cell after electrolysis will be :
29.
30.
33.
34.
(a)
1.69
(b)
2.50
(c)
5.20
(d)
9.69
A 100 watt. 110 volt incandescent lamp in series with
an electrolytic cell containing cadmium sulphate
solution. The mass of cadmium will be deposited
by the current flowing for 10 hours is
(At. wt. Cd = 112.4).
(a)
16.02
(b)
19.06
(c)
20.22
(d)
25.22
The solubility product of AgI from the following
data
(a)
increased
(b)
decreased
E 0Ag  / Ag  0.80V and E 0I  |AgI|Ag  0.15V
(c)
no change
(a)
8.9 × 10–17
(b)
6.9 × 10–14
(d)
time is also required
(c)
2.3 × 10–10
(d)
5.2 × 10–9
During electrolysis of acidified water, O2 gas is
formed at the anode. To produce O2 gas at the
anode at the rate of 0.224 c.c. per second at STP,
current passed is
(a)
0.224 A
(b)
2.24 A
(c)
9.65 A
(d)
3.86 A
100 mL of a buffer of 1M NH3(aq) and 1 M NH4+
(aq) are placed in two voltaic cells separately. A
current of 1.5 A is passed through both cells for 20
minutes. If electrolysis of water only takes place :
2H2O + O2 + 4e—  4OH— (RHS)
+
—
2H2O  4H + O2 + 4e (LHS) then pH of the :
35.
36.
The standard reduction potential of Cu2+ | Cu and
Cu2+ | Cu+ are 0.337 and 0.153 V respectively. The
standard electrode potential of Cu+ | Cu half cell is
(a)
0.184 V
(b)
0.827 V
(c)
0.521 V
(d)
0.490 V
Zn + Cu2+ (aq)
Cu + Zn2+ (aq). Reaction
quotient is Q 
[ Zn 2 ]
. Variation of Ecell with log
[Cu 2 ]
Q is of the type with OA = 1.10 V. Ecell will be 1.1591
V when :
(a)
LHS half-cell will increase
(a)
[Cu2+]/[Zn2+] = 0.01
(b)
RHS half-cell will increase
(b)
[Zn2+]/[Cu2+] = 0.01
(c)
both half-cells will increase
(c)
[Zn2+]/[Cu2+] = 0.1
(d)
both half-cells will decrease
(d)
[Zn2+]/[Cu2+] = 1
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 11
41.
42.
37.
Assume that during electrolysis of AgNO3, only H2O
is electrolysed and O2 is formed as :
2H2O  4H+ + O2 + 4e—
38.
39.
108 g
59.0 g
52.0 g
22.4 L
(c)
108 g
108 g
108 g
(d)
108 g
117.5 g
166 g
ANSWERS (SINGLE CORRECT
CHOICE TYPE)
Chromium plating can involve the electrolysis of
an electrolyte of an acidified mixture of chromic
acid and chromium sulphate. If during electrolysis
the article being plated increases in mass by 2.6 g
and 0.6 dm3 of oxygen are evolved at an inert
anode, the oxidation state of chromium ions being
discharged must be
(a)
–1
(b)
Zero
(c)
+1
(d)
+2
Equal quantities of electricity are passed through
three voltmeters containing FeSO4, Fe2(SO4)3 and
Fe(NO3)3 consider the following statements in this
regard.
1.
The amount of iron deposited in FeSO4
and Fe2(SO4)3 is equal
2.
The amount of iron deposited in Fe(NO3)3
is two thirds of the amount of iron
deposited in FeSO4.
3.
The amount of iron deposited in Fe2(SO4)3
and Fe(NO3)3 is equal to these
statements
(a)
1 alone is correct
(b)
1 and 2 are correct
(c)
2 and 3 are correct
(d)
3 alone is correct
Einstein Classes,
One faraday of current was passed through the
electrolysis cells placed in series containing
solutions of Ag+, Ni++ and Cr+++ respectively. The
amount of Ag, Ni and Cr, atomic masses 108, 59,
and 52 g mol–1 respectively; deposited will be
(b)
(assuming Cr = 52 and 1 mole of gas at room
temperature and pressure occupies a volume of
24 dm3)
40.
4.8 L
0.224 L
How much will the reduction potential of a
hydrogen electrode change when its solution
initially at pH = 0 is neutralised to pH = 7
Decrease by 0.41 V
(d)
17.5 g
(d)
(d)
2.6 L
29.5 g
11.2 L
Increase by 0.41 V
(c)
108 g
(c)
(c)
2.4 L
(a)
(b)
Decrease by 0.059 V
(b)
Cr
0.112 L
(b)
1.2 L
Ni
(a)
Increase by 0.059 V
(a)
Ag
O2 formed at N.T.P. due to passage of 2 amperes of
current for 965 s :
(a)
In an electrolysis of an aqueous solution
containing sodium ions, 2.4 L of oxygen at STP was
liberated at anode. The volume of hydrogen at STP
liberated at cathode would be
1.
2.
b
a
22.
23.
c
d
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
c
b
d
c
b
b
a
c
a
b
d
b
a
c
b
a
a
a
a
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
d
b
d
c
c
d
b
a
a
b
a
c
b
a
d
d
c
d
a
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 12
EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
Comprehension-3
Comprehension-1
For the reaction MnO4– + 8H+ + 5Fe2+  Mn2+ +
4H2O + 5Fe3+, it is given that E0(MnO4–, Mn2+, H+|Pt)
= 1.51 V and E0(Fe3+, Fe2+|Pt) = 0.77 V.
Tollens reagent (ammonical solution of silver
nitrate) is used to test aldehydes. The following data
are available.
1.
Ag+ + e–  Ag :
E10 = 0.80 V
C6H12O7 + 2H+ + 2e– 
C6H12O6 + H2O;
[Ag(NH3)2]+ + e– 
Ag(s) + 2NH3
The value of log K
0
eq
7.
(a)
increasing [Mn2+]
E20 = 0.05 V
(b)
increasing [Fe3+]
(c)
decreasing [MnO4–]
E30 = 0.373 V
(d)
decreasing pH of the solution
8.
for the reaction
C6H12O6 + 2Ag+ + H2O = C6H12O7 + 2H+ + Ag is
2.
3.
(a)
12.7
(b)
25.4
(c)
29.27
(d)
58.54
The use of NH3 makes the pH of solution equal to
11. This causes
5.
(a)
–0.059 V
(b)
–0.0178 V
(c)
0.059 V
(d)
0.0178 V
–
(a)
decrease in the value of E2
(b)
increase in the value of E2
(a)
–0.018 V
(b)
0.0036 V
(c)
increase in the value of E1
(c)
0.018 V
(d)
–0.0036 V
(d)
increase in the value of E10
Comprehension-4
Ammonia is used in this reaction rather than any
other base. This is due to the fact that
The Edison storage cell is represented as
Fe(x) | FeO(s) | KOH (aq) | Ni2O3(s) | Ni(s)
(a)
[Ag(NH3)2]+ is a weaker oxidizing agent
than Ag+
(b)
ammonia prevents the decomposition of
gluconic acid
Ni2O3(s) + H2O(l) + 2e—
E0 = +0.40 V
(c)
silver precipitates gluconic acid as its
silver salt
FeO(s) + H2O(l) + 2e—
E0 = –0.87 V
(d)
the standard reduction potential of
[Ag(NH3)2]+ is changes.
The half-cell reaction are
10.
2NiO(s) + 2OH—
Fe(s) + 2OH—
The cell reaction is
Comprehension-2
(a)
Ni2O3(s) + Fe(s)
A saturated solution of silver bromide is made
10–7 M in silver nitrate. Given :
(b)
2NiO(s) + FeO(s)
Ksp0(AgBr) = 3.0 × 10–13, m(Ag+) = 6 × 10–3 S m2
mol–1,
(c)
Ni2O3(s) + Fe(s)
Ni(s) + Fe2+
(d)
Ni2O3(s) + Fe(s)
NiO + Fe2+
Calculate the specific conductance of AgNO3
11.
What is the cell e.m.f. ? How does it depend on the
concentration of KOH ?
14 × 10–7
(b)
Ecell = 1.27 V independent of
concentration of KOH
(b)
13 × 10–7
(c)
(d)
70 × 10
–7
Ecell = –1.27 V dependent of
concentration of KOH
(d)
(b)
13 × 10
–7
Ecell = 1.27 V dependent of
concentration of KOH
(d)
70 × 10–7
(b)
13 × 10–7
(c)
14 × 10–3
(d)
The specific conductor of AgBr is
(c)
14 × 10
–3
Specific conductance of solution is
–7
(a)
158 × 10
(c)
–3
14 × 10
Einstein Classes,
Ni2O3(s) + Fe(s)
Ecell = 0.27 V independent of
concentration of KOH
5 × 10–7
5 × 10–7
2NiO(s) + FeO(s)
(a)
(a)
(a)
6.
9.
Reducing [Fe 3+] to 0.50 M keeping all other
concentrations at unity, the emf of the cell will be
changed by
Reducing [MnO4 ] to 0.50 M keeping all other
concentrations at unity, the change in emf of the
cell will be changed by
m(NO3–) = 7 × 10–3 S m2 mol–1, m(Br–) = 8 × 10–3
S m2 mol–1, and K(water) = 7.5 × 10–6 S m–1.
4.
The cell emf could be increased above the standard
emf by
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 13
12.
The maximum amount of electrical energy that
can be obtained from one mole of Ni2O3
(a)
24.52 kJ
(b)
2.452 kJ
(c)
0.2452 kJ
(d)
245.11 kJ
(A)
Comprehension-5
The standard reduction potential for the following
half-cell is 0.78 V
–
3
+
(B)
—
NO (aq) + 2H (aq) + e  NO2(g) + H2O
13.
14.
15.
(A)
(B)
(C)
(D)
(E)
(A)
(B)
(C)
(D)
The reduction potential in 8 M H+ is
(a)
0.887 V
(b)
–0.887 V
(c)
0.32 V
(d)
–0.78 V
The reduction potential of the half-cell in a neutral
solution. Assume all the other species to be at unit
concentration.
(a)
0
(b)
0.887
(c)
0.0474 V
(d)
–0.0474 V
How much will the reduction potential of the above
mentioned half cell will change when its solution
initially at pH = 0 is neutralized to pH = 7 while
keeping the other ions concentration to be 1 molar.
(a)
decreases by 0.41 V
(b)
increases by 0.826 V
(c)
decreases by 0.826 V
(d)
increases by 0.41 V
MATRIX-MATCH TYPE
Matching-1
Column - A
Galvanic cell
Cathode of an
electrolytic cell
Electrode potential
Current, I
Faraday constant
Matching-2
Column-A
Faraday’s required
to reduce Cr2O7– to Cr3+
By passing one Faraday
of electricity
Zn | Zn2+ (C1 = 0.05) ||
Zn2+ (C2 = 0.5) | Zn
E0OCl–/Cl– = .94
and E0Cl–/Cl2 = –1.36 v.
The E0OCl–/Cl2 will be
Einstein Classes,
Column - B
(P)
Ecell = ER – EL
(Q)
reduction
potentials
(R)
Q/t
(S)
Symbol : F
(T)
negatively
charged
(C)
(D)
1.
2.
Column-B
(P)
1/3 mol of Al
deposited
(Q)
6F
(R)
–0.42 V
(S)
0.0295 V
(T)
3F
3.
Matching-3
Column-A
Column-B
The equivalent
(P)
189 ohm–1 cm2
conductivity of 1M
mol–1
H2SO4 solution whose
conductivity is 26 × 10–2
ohm–1 cm–1 is
Given that :
(Q)
130 ohm–1 cm2

m [Al2(SO4)3] =
eq–1
–1
2
–1
858 ohm cm mol
m(SO42–) = 160
ohm–1 cm2 mol–1
then m (Al3+) is
Given that :
(R)
1.66 cm–1
c
m (NH4OH) =
9.33 ohm–1 cm2
m(NH4OH) =
238.3 ohm–1 cm2 mol–1
the degree of dissociation
of NH4OH is
The conductivity of
(S)
5%
N/10 KCl solution at
200C is 0.0212 ohm–1
cm–1 and the resistance
to the cell containing this
solution at 200C is 55 ohm.
The cell constant is
(T)
3.92 %
MULTIPLE CORRECT CHOICE TYPE
Rusting of iron can be prevented
(a)
by electroplating the metal with silver
(b)
by electroplating the metal with Gold
(c)
by electropting the metal with Zn
(d)
by connecting the iron material with Mg
For the cell :
Tl|Tl+(0.0001M)||Cu2+(0.1 M)|Cu
Ecell = 0.83 at 298 k the cell potential can be increased
by
(a)
increasing [Cu2+]
(b)
increasing (Tl+)
(c)
decreasing [Cu2+]
(d)
decreasing [Tl+]
When net cell reaction is spontaneous which of the
following are correct
(a)
E0 cell is negative
(b)
Ecell > 0
(c)
Ecell = E0cell
(d)
G < 0
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 14
4.
5.
6.
7.
0
cell
In which case Ecell – E = zero
(a)
Cu|Cu2+ (0.01M) || Ag+(0.1M) | Ag
(b)
Pt(H2)|PH = 1 || Zn2+ (0.01M) | Zn
(c)
Pt(H2)|PH = 1||Zu2+(1M)|Zn
(d)
Pt(H2)|H+(0.01M) || Zn2+ (0.01M) | Zn
By passing IF of electricity
(a)
1 mol of Zn deposited
(b)
12 g of Mg deposited
(c)
1/3 mol of Al deposited
(d)
11.2 L of O2 at N.T.P. evolved
KI solution containing starch turns blue on
additionof Cl2. Which of the following statements
helps to explain this :
(a)
Reduction potential of Cl2 > I2
(b)
E0OX of Cl2 > E0OH of I2
(c)
the product of Cl2 and starch is blue in
colour
(d)
the product of I2 and starch is blue in
colour
The chemical reactions taken place in the process
of rusting of iron are listed below, pick up the correct chemical reactions
(a)
2Fe(s) + 4H++ O2 
 2Fe2+aq + 2H2O
(b)
4Fe2+aq + O2(g) + 4H2O 
 2Fe2O3 +
8H+
(c)
 Fe2O3 xH2O
Fe2O3 + x H2O 
(d)
3Fe + 4H2O 
 Fe2O3 + FeO + 4H2
8.
9.
10.
Which of the following are not reference electrode
(a)
Normal hydrogen electrode
(b)
Calomel electrode
(c)
Silver-Silver chloride electrode
(d)
Platinium electrode
For a galvanic cell,
(a)
anode is a negative terminal and cathode
is a positive terminal
(b)
oxidation takes place at anode and
reduction at cathode.
(c)
electrons in the external wire move from
anode to cathode.
(d)
Ecell = ER – EL
The emf of the cell
Pt|H2(g)||HCl(c1)||HCl(c2)|H2(g)|Pt can be increased
by
(a)
decreasing c2
(b)
decreasing c1
(c)
increasing c1
(d)
increasing c2
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
b
2.
a
3.
a
4.
b
5.
d
6.
a
7.
d
8.
d
9.
d
10.
a
11.
b
12.
d
13.
a
14.
d
15.
c
MATRIX-MATCH TYPE
1.
[A-P; B-T; C-P; D-R; E-S]
2.
[A-Q; B-P; C-S; D-R]
3.
[A-Q, B-P, C-T, D-R]
4.
b, c
6.
MULTIPLE CORRECT CHOICE TYPE
1.
c, d
2.
a, d
3.
b, d
7.
a, b,c
8.
a, c
9.
a, b, c, d
Einstein Classes,
a, b
5.
10.
b, d
a, d
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 15
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
2.
3.
4.
A current of 1.50 A flows through a cell containing
aqueous NiSO4 for 40 min. What mass of Ni is
deposited on the cathode and what volume at (STP)
of oxygen will be evolved at the anode ? (At. wt. of
Ni = 58.9).
In an electrolysis experiment electric current was
passed for 5 hrs through two cells connected in
series. The first cell contains gold salt and second
contain CuSO 4 solution. 9.83 g of gold was
deposited in the first cell. If the oxidation number
of Au is +3, find the amount of copper deposited in
the second cell. Also calculate the magnitude of the
current in amperes. (At wt. of Au = 197 and At. wt.
of Cu = 63.5)
An electric current is passed through two
electrolytic cells connected in series, one
containing a solution of silver nitrate and the other
solution of sulphuric acid. What volume of oxygen
measured at 250C and 750 mm of Hg would be
liberated from H2SO4 if (i) 1 mole and (ii) 8 × 1022
ions of Ag+ are deposited from the silver nitrate
solution ?
Calculate the e.m.f. of the following cell and
predict whether the given cell representation is
correct or wrong. If wrong, write the correct
representation and correct cell reaction
8.
The emf of the cell
Zn | ZnCl2(0.05 mol dm–3) | Ag+ | AgCl(s) | Ag
is 1.015 V at 298 K, the silver electrode being
positive, while the temperature coefficient of its emf
is –0.00492 VK–1. Write down the equation for the
reaction occuring when the cell is allowed to
discharge and calculate the changes in (a) free
energy (G), and (b) heat content (H) and (c)
entropy (S) accompanying this reaction, at 298 K.
9.
The standard reduction potential of Cu2+/Cu and
Ag +/Ag electrodes are 0.337 and 0.799 volt
respectively. Construct a galvanic cell using these
electrodes so that its standard emf is positive. For
what concentration of Ag+ will the emf of the cell at
25 0C be zero, if the concentration of Cu 2+ is
0.10 M ?
10.
The electrolysis of a metal salt solution was carried
out by passing a current of 4 amp for 45 min. It
resulted in deposition of 2.977 of a metal. If atomic
mass of the metal is 106.4 g mol–1, calculate the
charge on the metal cation.
11.
40 ml of 0.126 M NiSO4 solution is electrolysed by a
current of 0.05 amp for 40 min.
Cu(s) | Cu2+ (aq) || Zn2+(aq) | Zn(s)
(a)
Write equation for the reactions
occurring at each electrode
(b)
How many coulombs of electricity are
passed through the electrolyte
(c)
How many grams of product is deposited
at cathode
(d)
How long the same current will have to
be passed to remove completely the
metal ions from the solution.
Given :
E 0 Cu 2 / Cu = +0.34 V and E 0 Zn 2 / Zn  0.76 V
5.
Calculate emf of the cell
Pt(H2) | CH3COOH (0.1 M) || NH4OH (0.01 M) |
(H2) Pt
K a for CH 3 COOH = 1.8 × 10 —5 and K b for
NH4OH = 1.8 × 10–5
6.
7.
Consider the reaction 2Ag+ + Cd  2Ag + Cd2+.
The standard electrode potentials for Ag+/Ag and
Cd2+/Cd couples 0.80 V and 0.40 V respectively. (i)
What is the standard potential for this reaction ?
(ii) For the electrochemical cell in which the above
reaction takes place which electrode is negative
electrode ? (iii) Will the total emf of the given
reaction be more negative or positive, if the
concentration of Cd2+ ions is 0.1 M rather than
1.0 M ?
A zinc rod is placed in 0.1 M solution of ZnSO4 at
250C. Assuming that the salt is dissociated to the
extent of 95% at this dilution, calculate the
potential of the electrode at this temperature. Given
that
E
0
Zn
2
/ Zn
 0.76 V
Einstein Classes,
12.
Silver is electrodeposited on a metallic vessel of
surface area 800 cm2 by passing a current of 0.2
ampere for 3 hours. Calculate the thickness of the
silver deposited. Given the density of silver as
10.47 g/cc.
13.
Calculate standard electrode potential of Ni+2/Ni
electrode if the cell potential of a cell Ni | Ni+2 (0.01
M) || Cu 2+ (0.1 M) | Cu is 0.59 V. Given that
E 0Cu 2 / Cu  0.34V
14.
Calculate the maximum possible electrical work
that can be obtained from the following cell under
standard conditions at 250C.
Zn | Zn+2(aq) || Cu2+ (aq) | Cu
At 250C,
E 0Zn 2 / Zn  0.76V and E 0Cu 2 / Cu  0.34V
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 16
15.
Calculate the E.M.F. of the cell
2+
+
23.
Calculate pH of the following half cell, Pt H2 | H2SO4.
The oxidation electrode potential is + 0.3 V.
24.
(a)
–3
Mg | Mg (0.2 M) || Ag (1 × 10 M) | Ag
E 0Mg 2 / Mg  2.37 V, E 0Ag  / Ag  0.80 V
What will be the effect on EMF if concentrations of
Mg2+ ion is decreased to 0.1 M
16.
The measured voltage of the following cell is 0.9 V
at 250C.
+
volts
(b)
+
Pt, H2(1 atm) | H (aq) || Ag (1.0 M) | Ag
If E 0 
 0.80V . Calculate the pH of the
Ag / Ag
aqueous solution of H+.
17.
How many grams of silver could be plated out on a
serving tray by electrolysis of a solution containing
silver in + 1 oxidation state for a period of 8.0 hours
at a current of 8.46 Amperes ? What is the area of
the tray if the thickness of the silver plating is
0.00254 cm ? Density of silver is 10.5 g/cm3.
18.
A 100 watt. 110 volt incandescent lamp in series with
an electrolytic cell containing cadmium sulphate
solution. What mass of cadmium will be deposited
by the current flowing for 10 hours ? (At. wt.
Cd = 112.4).
19.
A current of 10.0 A is passed through 1.0 L of 1.0 M
HCl solution for 1.0 h. Calculate the pH of the
solution at the end of the experiment. What is the
volume of total gas evolved at STP ?
20.
Consider the cell
Zn | Zn2+ (aq. 1.0 M) || Cu2+ (aq 1.0 M) | Cu
21.
Write down the cell reaction
(ii)
Calculate the emf of the cell
(iii)
Is the cell reaction spontaneous or not ?
A current was passed through a series of cells
containing AgNO3, CuSO4 and H2SO4 solutions for
a period of 25 minutes. If the weight of silver
deposited was 0.5394 g, what would be (i) the weight
of copper and (ii) the volume of H 2 at N.T.P.
liberated by the current ? Also calculate the
magnitude (strength) of current assuming that it
remained constant. [At. wt. Ag = 108, Cu = 63.5].
26.
A solution of a salt of a metal of atomic weight 112
was electrolysed for 150 minutes with a current of
0.15 amperes. The weight of metal deposited was
0.783 mg. Find the equivalent weight and valency
of the metal in the salt.
27.
A current deposits 10–2 kg of Cu in 3.96 × 103 minute
from a solution of Cu++ ions. What is the strength
of current in amperes ? How many grams of Cu
will the same current deposit from a solution of
cuprous ions ?
28.
A 1.5276 g sample of CdCl 2 was converted to
metallic cadmium by an electrolytic process.
0.9367 g of cadmium was obtained. What is the
atomic mass of cadmium from this experiment if
the atomic mass of chlorine is taken as
35.453 g mol—1.
29.
Calculate heat of reaction inside the cell
Zn(s) + 2AgCl(s)
Given the following cell
ZnCl2(0.555 M) + 2Ag(s)
Al | Al3+ (0.1 M) || Fe2+ (0.2 M) | Fe
Given that E = 1.015 V at 00 C and
E 0Al3 / Al  1.66V and E 0Fe 2 / Fe  0.44V .
 E 
4
1
   4.02 10 VK
 T  P
Calculate the maximum work that can be obtained
by the cell.
22.
At what concentration of copper ions will
this electrode have a potential of zeo
volt ?
25.
The standard reduction potentials are 0.350 V for
Cu2+ (aq) + 2e—  Cu(s) and – 0.763 V for Zn2+ (aq)
+ 2e—  Zn(s)
(i)
Calculate the electrode potential at a
copper electrode dipped in a 0.1 M
solution of copper sulphate at 298 K;
assuming CuSO4 to be completely
dissociated. The standard electrode
potential of Cu2+ | Cu system is + 0.34
at 298 K.
30.
(i)
Determine the equilibrium constant of the
following reaction at 298 K
2Fe3+ + Sn2+
0
E 0Fe 2 / Fe 2  0.771V , E Sn
 0.150V
4
/ Sn 2
Einstein Classes,
Cl2(g) + SO2(g) + 2H2O(l)  2Cl—(aq) +
3H+(aq) + HSO4— (aq)
proceeds readily and rapidly in aqueous
acid solution. Write the half cell
reactions and construct the cell.
2Fe2+ + Sn4+
Also predict whether Sn2+ ions can reduce Fe3+ ions
to Fe2+ quantitatively or not
The chemical reaction :
(ii)
If the fully charged cell initially held
1.0 M of Cl2, for how many days could it
sustain a current of 0.05 A assuming that
the cell becomes inoperative when 90%
of the initial Cl2 has been consumed ?
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 17
31.
Zinc granules are added in excess to a 500 ml of 1.0
M nickel nitr ate solut ion at 250 C until the
equilibrium is reached. If the standard reduction
potential of Zn2+ | Zn and Ni2+ | Ni are –0.75 V and
–0.24 V respectively, find out the concentration of
Ni2+ in soluble at equilibrium.
32.
(a)
The standard reduction potential of
Cu++ / Cu and Ag+ / Ag electrodes are
0.337 and 0.799 volt respectively.
Construct a galvanic cell using these
electrodes so that its standard e.m.f. if
positive. For what concentration of Ag+
will the e.m.f. of the cell, at 250C, be zero
if the concentration of Cu++ is 0.01 M ?
(b)
33.
Calculate the quantity of electricity that
would be required to reduce 12.3 g of
nitrobenzene to aniline, if the current
efficiency for the process is 50 per cent.
If the potential drop across the cell is
3.0 volts, how much energy will be
consumed ?
A cell, Ag | Ag+ || Cu2+ | Cu, initially contains 1 M
Ag+ and 1 M Cu2+ ions. Calculate the change in the
cell potential after the passage of 9.65 A of current
for 1 h.
FINAL STEP EXERCISE
(SUBJECTIVE)
1.
2.
3.
4.
19 g fused SnCl 2 was electrolysed using inert
electrode 0.119 g Sn was deposited at cathode. If
nothing was given out during electrolysis, calculate
the ratio of weight of SnCl2 and SnCl4 in fused state
after electrolysis. (At. wt. of Sn = 119)
For the following galvanic cell, calculate the e.m.f.
at 25 0 C. Assign the correct polarity for the
spontaneous reaction to take place
Ag | AgCl(s), KCl (0.2 M) || KBr (0.001 M), AgBr(s)
| Ag
The solubility product of AgCl and AgBr are
2.8 × 1010 and 3.3 × 1013 respectively.
Calculate the e.m.f. of the following cell at 250C
H2(1 atm) | 0.5 M HCOOH || 1 M CH3COOH | H2
(1 atm)
The K a for HCOOH and CH 3 COOH are
1.77 × 10–4 and 1.80 × 10–5 respectively.
The following Galvanic cell Zn | Zn++ (1 M, 100 mL)
|| Cu++ (1 M, 100 mL) | Cu was operated as an
electrolytic cell as; Cu as anode and Zn as cathode.
0.48 A current was passed for 10 hours and then
the cell was allowed to function as galvanic cell.
Calculate e.m.f. of the cell at 250C, assuming that
only electrode reactions occuring were those
involving Cu/Cu++ and Zn/Zn++.
6.
7.
reaction, 2H2O + 2e—
H2 + 2OH— is – 0.8277
V. Calculate the equilibrium constant for the
reaction 2H2O
8.
9.
10.
An aqueous solution of NaCl on electrolysis gives
H2(g), Cl2(g) and NaOH according to the reaction :
2Cl–(aq) + 2H2O = 2OH— (aq) + H2(g) + Cl2(g).
A direct current of 25 amperes with a current
effeciency of 62% is passed through 20 litres of NaCl
solution (20% by weight). Write down the reactions
taking place at the anode and the cathode. How long
will it take to produce 1Kg of Cl2 ? What will be the
molarity of the solution with respect to hydroxide
ion ? (Assume no loss due to evaporation)
Einstein Classes,
H3O+ + OH— at 250C.
Find the solubility product of a saturated solution
of Ag2CrO4 in water at 298 K if the emf of the cell
Ag/Ag+ (satd. Ag2CrO4 soln.) // Ag+ (0.1 M)/Ag is
0.164 V at 298 K.
The standard reduction potential for Cu2+ / Cu is
+0.34 V. Calculate the reduction potential at
pH = 14 for the above couple. Ksp of Cu(OH)2 is
1.0 × 10–19.
An excess of liquid mercury is added to an
acidified solution of 1.0 × 10–3 M Fe3+. It is found
that 5% of Fe3+ remains at equilibrium at 250C.
0
Calculate E Hg 2 / Hg  , assuming that the only
reaction that occurs is
2Hg + 2Fe3+  Hg22+ + 2Fe2+
E 0Cu  / Cu  0.34V, E 0Zn  / Zn  0.76V, Zn  65.5 g mol 1 .
5.
An acidic solution of Cu2+ salt containing 0.4 g of
Cu 2+ is electrolysed until all the copper is
deposited. The electrolysis is continued for seven
more minutes with the volume of solution kept at
100 ml and the current at 1.2 amp. Calculate the
volume of gases evolved at NTP during the entire
electrolysis. (At. wt. Cu = 63.6)
The standard reduction potential at 250C of the
[ Given E 0 Fe3 / Fe 2   0.77V]
11.
12.
The standard reduction potential of the Ag+/Ag
electrode at 298 K is 0.799 V. Given that for AgI,
Ksp = 8.7 × 10 –17, evaluate the potential of the
Ag+/Ag electrode in a saturated solution of AgI. Also
calculate the standard reduction potential of the
I—/AgI/Ag electrode.
The E.M.F. of the following cell at 298 K is 1.0495V.
Pt/H2(1 atm) |LiOH (0.01 mol dm–3)| |LiCl (0.01)
mol/dms)| AgCl (s) / Ag. Determine the ionic
product of water.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CE – 18
ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)
1.
208.7 mL
2.
0.8026 amp.
4.
6.
6.
E0cell = – 1.10 V
Ecell = 0.4575 V
(i)
E0cell = 0.40 V
(iii)
Ecell = 0.4295 V
7.
E Zn / Zn 2 = – .79 volt
9.
11.
[Ag+] = 1.47 × 109M
(b)
120 c (c)
3.
(i)
(ii)
6.2 litre
0.823 litre
(ii)
Cd(s) | Cd2+ is negative electrone
8.
(a)
(b)
(c)
n=4
(d)
10.
.036 gm
0
Ni 2 / Ni
– 195.895 kJ mol–1
H298k = – 167.7 kJ mol–1
S298 = – 95.0 Jk–1 mol–1
5 hr 21 min.
12.
2.89 × 10–4 cm
14.
16.
18.
20.
212.3 kJ
15.
3.103 V, Ecell will increase to 3.02 V
1.69
17.
1.02 × 104 cm2
19.06 gm
19.
0.20, 8360.4 mL
(i)
Zn + Cu2+  Zn2+ + Cu
(ii)
1.113 V
(iii)
spontaneous
705.801 kJ
22.
1.035 × 1021, yes
5.08
24.
0.3105 V, 2.95 × 10–12 M
0.1589 gm, 56 mL, 0.31218 amperes
26.
55.97, 2
0.131 A, 20 gm
28.
110.88 gm mol–1
–2.171 × 102 kJ
Pt, SO2|H2SO4|| HCl | Cl2, Pt, 40.2 days
31.
[Ni2+] = 5.6 × 10–18 M
(a)
[Ag+] = 1.477 × 10–9 M
(b)
115800 coulomb, 347.4 kJ
0.01 V
21.
23.
25.
27.
29.
30.
32.
33.
13.
E
= .22 V
ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)
1.
2.
3.
5.
6.
8.
10.
SnCl2 : SnCl4 = 18.26 : 0.26
(s) Ag | AgBr(s), kBr || KCl, AgCl(s) | Ag(s), Ecell = –0.337 V
–.0204 V
4.
1.137
–
–
2Cl  Cl2 + 2e (Anode)
2e– + 2H2O  2OH– + H2 (cathode), 48.7 hr, 1.408 mol/lit
158.15 mL
7.
9.35 × 10–15
2.287 × 10–12
9.
–0.22 V
0.792 V
11.
E Ag  / Ag  0.325 V
E I  / AgI / Ag  0.1485 V
12.
10–14
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111