VIJETA SERIES CLASS-12th
PRACTICE SHEET
(CHEMISTRY)
Chapter: Electrochemistry
Multiple Choice Answer Type Questions
1.
In a galvanic cell, the salt bridge is used to;
(1) produce current at a constant strength.
(2) facilitate continuity of the cell reaction.
(3) prevent accumulation of charges around the
electrodes.
(4) All of these
2.
Three faradays of electricity was passed through an
aqueous solution of iron (II) bromide. The weight of
iron metal (At. wt. = 56) deposited at the cathode (in
gm) is;
(1) 56
(2) 84
(3) 112
(4) 168
3.
The molar conductivity is maximum for the solution
of concentration;
(1) 0.001 M
(2) 0.005 M
(3) 0.002 M
(4) 0.004 M
5.
Match the Column Answer Type Questions
6.
Assertion and Reason Answer Type Questions
Directions: These questions consist of two statements
each, printed as Assertion and Reason. While answering
these questions, you are required to choose any one of the
following four responses.
4.
Assertion (A): Conductivity of all electrolytes
decreases on dilution.
Reason (R): On dilution, number of ions per unit
volume decreases.
(1) Both Assertion (A) and Reason (R) are True,
and the Reason (R) is a correct explanation of
the Assertion (A).
(2) Both Assertion (A) and Reason (R) are True,
but Reason (R) is not a correct explanation of
the Assertion (A).
(3) Assertion (A) is True, but the Reason (R) is
False.
(4) Assertion (A) is False, but Reason (R) is True.
Assertion (A): Λm for weak electrolytes shows a
sharp increase when the electrolytic solution is
diluted.
Reason (R): For weak electrolytes, degree of
dissociation increases with dilution of solution.
(1) Both Assertion (A) and Reason (R) are True,
and the Reason (R) is a correct explanation of
the Assertion (A).
(2) Both Assertion (A) and Reason (R) are True,
but Reason (R) is not a correct explanation of
the Assertion (A).
(3) Assertion (A) is True, but the Reason (R) is
False.
(4) Assertion (A) is False, but Reason (R) is True.
Match the List-I with List-II.
List-I
A.
P.
m
B.
Ecell
Q.
C.
R.

D.
G°
S.
(1)
(2)
(3)
(4)
7.
List-II
Scm–1
m–1
S cm2mol–1
V
A-(Q); B-(R); C-(S); D-(P)
A-(Q); B-(S); C-(R); D-(P)
A-(R); B-(S); C-(P); D-(Q)
A-(Q); B-(P); C-(R); D-(S)
Match the List-I with List-II.
List-I
List-II
A.
P. Intensive property
m

B.
Q. Depends on number of
Ecell
ions/volume
C.
R. Extensive property

D.
r Gcell S. Increases with dilution
(1)
(2)
(3)
(4)
A-(P); B-(Q); C-(R); D-(S)
A-(S); B-(P); C-(Q); D-(R)
A-(P); B-(S); C-(R); D-(Q)
A-(R); B-(Q); C-(P); D-(S)
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Very Short Answer Type Questions
8.
conductance property of a solution containing one
mole of the electrolyte.
Both conductivity and molar conductivity change
with the concentration of the electrolyte. The molar
conductivity of a solution at infinite dilution, i.e.,
when concentration approaches zero is called
limiting molar conductivity. The Kohlrausch law
Independent migration of ions states that limiting
molar conductivity of an electrolyte can be
represented as the sum of the individual
contributions of the anion and cation of the
electrolyte.
Determine the values of equilibrium constant (Kc )
and ΔG° for the following reaction;
Ni(s) + 2Ag+ (aq) → Ni2+ (aq) + 2Ag(s)
E° = 1.05 V
(1F = 96500 C mol–1)
9.
Calculate the time to deposit 1.27 g of copper at
cathode when a current of 2A was passed through
the solution of CuSO4.
(Molar mass of Cu = 63.5 gmol–1, 1F = 96500 C mol–1)
I.
Which of the following is the SI unit of
conductivity?
(1) Sm–1
(2) Ohm–1
(3) S
(4) Ohm
Long Answer Type Question
10.
(I) What is limiting molar conductivity? Why is
there a steep rise in the molar conductivity of
weak electrolyte on dilution?
II. Which of the following equations represents
the correct relationship between conductivity
and molar conductivity of the solution?
(  m)

(1) m =
(2)  m =
c
c
m
(3) m =
(4) m =  + cm
(II) Calculate the emf of the following cell at 298 K:
Mg(s) | Mg2+ (0.1M) || Cu2+ (1.0 10−3 M) | Cu(s)
[Given = Ecell = 2.71V ].
11.
(I) What are fuel cells? Give an example of a fuel
cell.

(II) Calculate the equilibrium constant (log Kc) and
ΔrG° for the following reaction at 298 K.
Cu(s) + 2Ag+ (aq)
III. The unit of molar conductivity is;
(1) S
(2) Sm–1
(3) Sm2 mol–1
(4) Sm–1mol–2
Cu2+ (aq) + 2Ag(s)
Given E0cell = 0.46 V and IF = 96500 C mol−1
IV. Λm for NaCl, HCl, HCl and NaAc are 126.4,
425.9 and 91.0 Scm2mol–1 respectively.
Calculate Λ° for HAc.
(1) 284.1 S cm2 mol–1
(2) 390.5 S cm2 mol–1
(3) 162.7 S cm2 mol–1
(4) 132.8 S cm2 mol–1
Case Based Study Answer Type Questions
12.
Resistance is the property of a conductor due to
which it opposes the flow of current through it.
Electrical resistance of any object is directly
proportional to its length and inversely proportional
to its area of cross section. Inverse of resistance is
known as conductance. Inverse of resistivity is
known as conductivity. Molar conductivity is the
■■■
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Answer Key
1.
(4)
5.
(1)
2.
(2)
6.
(3)
3.
(1)
7.
(2)
4.
(1)
12.
I-(1); II-(2); III-(3); IV-(2)
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Hints & Solutions
Multiple Choice Answer Type Questions
1.
2.
3.
(4)
Salt bridge helps to facilitate the continuity of a cell
by maintaining the flow of ions of the solution. It
also helps to produce current at constant strength
by circulating electrons from the solution, thus,
preventing the accumulation of charges on or
around the electrodes.
Hence, option (4) is correct.
(2)
Fe2+ + 2e– → Fe
n = 2, as 2 electrons are exchanged in this reaction
3F charge is passed which implies 3 moles
electrons are passed. 1 mole of Fe is formed when
2 moles of electrons are passed.
For 3 moles of electrons passed 1.5 moles of Fe
will be formed. mass of iron deposited = 56 × 1.5
= 84g
5.
Match the Column Answer Type Questions
6.
(3)
7.
(2)
Very Short Answer Type Questions
8.
n = no of electrons gained or lost = 2
F = Faraday’s constant
E° = Standard E
G = −2  96500  1.05
G = −202650 J
The equilibrium constant is related to Gibbs free
energy by following relation:
G = −2.303RT  log K
G = Gibbs free energy
R = Gas constant = 8.314 J/Kmol
T = Temperature in kelvin = 298 K
G = −2.303RT  log K
Hence, the solution having maximum molar
conductivity has to be less concentrated.
Here, Option 1 is the one which is having less
concentration out of all other options and hence is
the correct answer
4.
(1)
Conductivity always decreases with the decrease in
concentration both for weak and strong
electrolytes. This can be explained by the fact that
the number of ions per unit volume that carry the
current in a solution decreases on dilution.
The standard emf of a cell is related to Gibbs free
energy by following relation
= −nFEcell
G = Gibbs free energy
Ni(s) + 2Ag+ (aq) →Ni2+ (aq) + 2Ag(aq)
(1)
Molar conductivity is defined as the conductivity
of an electrolyte solution divided by the molar
concentration of electrolyte.
We know the relation between Molar conductivity
and Molarity i.e.
1000
m =
M
 m is inversely proportional to Molarity
Assertion and Reason Answer Type Questions
(1)
Molar conductivity of weak electrolytic solution
increases on dilution because as we add excess
water to increase the dilution, degree of
dissociation increases which lead to increase in
number of ions in the solution. Thus, Λm shows a
very sharp increase.
–202650 J = –2.303 × 8.314 × 298 × log K
K = 3.16 × 1035
9.
From faraday’s law of electrolysis, we get
E
W = Zikt = it
F
where, W = Deposited amount
E = Equivalent weight;
F = Faraday
i = Electricity passed;
t = Time
63.5
 2  t  t = 965 sec
 1.27 =
96500
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Long Answer Type Question
10.
11.
(I) Limiting molar conductivity: It is the sum of
limiting ionic conductivities of cation and the
anions, each multiplied with the number of
ions present in one formula unit of electrolyte.
m = x+ + y−
+ and − → limiting molar conductivity of
(I) The galvanic cells in which the energy of
combustion of fuels is directly converted into
electrical energy are called fuel cells. Fuel cells
are different from ordinary galvanic cells.
The common example of a fuel cell is a
hydrogen-oxygen fuel cell.
(II) Cu(s) + 2Ag+ (aq)  Cu2+ (aq) + 2Ag(s)
cation and anion
There is steep rise in molar conductivity of
weak electrolyte on dilution because as the
concentration of weak electrolyte is reduced,
more of it ionizes. Thus, increase in
conductance with decrease in concentration is
due to increase in number of ions in the
solution.
Eocell = 0.46V,
G = −2nFEocell  n = 2
G = −2  96500  0.46
 G = −88780 J/mol
 G = −88780 J/mol
−G = 2.303RTlogkc
(II) Mg(s) | Mg2+ (0.1M) || Cu2+ (1.0 10−3 M) | Cu(s)
 88780 = 2.303 8.314  298logkc
Ecell = 2.71V .
log k c =
Reaction: Mg + Cu2+ → Mg2+ + Cu
2+
Mg → Mg
2+
+ 2e
Cu + 2e → Cu
n=2
Case Based Study Type Questions
2+ 
Mg
0.0591

log 
2
Cu 2+ 


0.0591
0.1
= 2.71 −
log
2
0.001
= 2.71 – 0.02955 log 102
= 2.71 – 0.02955 × 2 = 2.71 – 0.059
Ecell = 2.65 V
Ecell = Ecell −
88780
2.303  8.314  298
 logkc = 15.56
−
−
1F = 96500C mol–1
12.
I-(1), II-(2), III-(3), IV-(2)
I.
Conductivity (or specific conductance) of an
electrolyte solution is a measure of its ability to
conduct electricity. The SI unit of conductivity is
siemens per meter (S/m).
II.
(2)
III.
(3)
The unit of molar conductivity is Scm2mol–1.
IV.
(2)

m(HAc)= 
+ Ac =  + +  − +  − + + −  − −  +
H+
H
Cl
Ac
Na
Cl
Na
= m(HCl) + m(NaAc) − Am(NaCl)
= (425.9 + 91.0 – 126.4)Scm2mol–1
= 390.5 Scm2mol–1
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