CBSE Class 12 Chemistry 2014 Delhi Set 1 Solved Paper

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Question : 28
Total: 30
(a) Define the following terms:
(b) Resistance of a conductivity cell filled with 0.1 mol L1KCl solution is 100W. If the resistance of the same cell when filled with 0.02molL1KCl solution is 520, calculate the conductivity and molar conductivity of 0.02molL1KCl solution. The conductivity of 0.1molL1KCl solution is 1.29×1021cm1.
OR
(a) State Faraday's first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1mol of Cu2+ to Cu.
(b) Calculate emf of the following cell at 298K :
Mg( s)|Mg2+(0.1M)||Cu2+(0.01M)|Cu( s)
[ Give Ecell 0=±+2.71V,F=96500Cmol1]
Solution:  
(a) (i) When the concentration of the electrolyte approaches zero, the molar conductivity is termed as limiting molar conductivity. It is represented by Λm.
(ii) Fuel cell: Fuel cells are the galvanic cells or electrochemical cells that transform the chemical energy into electrical energy from fuel combustion by redox reaction, such as hydrogen, methanol, etc.
Given,
For 0.1molL1KCl solution
Resistance (R)=100
Conductivity (k)=1.29×1021cm1
Cell constant (G*)=k×R
=1.29×1021cm1×100
=1.29cm1
For 0.02molL1KCl solution
Resistance (R)=520
Conductivity =
Cell constant (k)
Concentration (C)

K=
1.29cm1
520

K=2.48×1031cm1
(C)=0.02molL1
=0.02×103molcm3
Concentration (C)=0.02molL1
=0.02×103 mol cm3
Molar conductivity (Λm) =
Conductivity (k)
Concentration (C)

=
2.48×1031cm1
0.02×103molcm3

Λm=1241cm2mol1
OR
(a) Faraday's first law of electrolysis states that "The mass of a substance deposited at any electrode is directly proportional to the amount of charge passed."
m=Z×Q
where, m= mass of a substance deposited or liberated at an electrode.
Q= amount of charge passed through it and
Z= electrochemical equivalent
The reduction of one mol of Cu2+ to Cu can be represented as:
Cu2++2eCu
Since, 2mol of electrons are involved in the reduction, so the amount of charge required is 2F.
(b) The given cell reaction can be represented as
Mg( s)+Cu2+(aq)Mg2+(aq) +Cu( s)
Ecell =E
2.303RT
nF
log
Mg2+
Cu2+

Ecell =2.71
2.303×0.0831×298
2×96500
log
0.1
0.01

Ecell =2.71
0.0591
2
log
10

Ecell =2.68V
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