CBSE Class 12 Chemistry 2019 Outside Delhi Set 2 Solved Paper

© examsnet.com
Question : 27
Total: 27
(a) The conductivity of 0.001molL1 acetic acid is 4.95×105Scm1. Calculate the dissociation constant if mo for acetic acid is 390.5Scm mol1.
(b) Write Nernst equation for the reaction at 25C :
2Al(S)+3Cu2+(aq)2Al3+(aq) +3Cu( s)
(c) What are secondary batteries? Give an example.
OR
(a) Represent the cell in which the following reaction takes place:
2Al( s)+3Ni2+(0.1M)2Al3+(0.01M) +3Ni( s)
Calculate its emf if E°cell =1.41V.
(b) How does molar conductivity vary with increase in concentration for strong electrolyte and weak electrolyte? How can you obtain limiting molar conductivity (mo) for weak electrolyte?
Solution:  
Λm=
k
c
=
4.95×105Scm1
0.001molL1
×
1000cm3
L

=49.5Scm2mol1
α=
Λm
Λα
=
49.5Scm2mol1
390.5Scm2mol1
=0.126
K=
cα2
(1α)
=
0.001molL1×(0.126)2
10.126

=1.8×105molL1
(If K=cα2, then K=1.6×105molL1 )
(b) E(cell )=E°(cell )
0.059
6
log
[Al3+]2
[cu2+]3

(c) Batteries which are rechargeable
Example- Lead storage, NiCd batteries (Or any other one example)
(a)
2Al( s)+3Ni2+(0.1M)2Al3+(0.01M) +3Ni( s)
Al( s)|Al3+||Ni2+|Ni( s) Cell reaction
Ecell =E
0.0591
6
log
[0.01]2
[0.1]3

=1.41
0.0591
6
log
0.1

=1.41+
0.0591
6
log
10

=1.41+
0.0591
6

=1.41+0.00985
=1.42V
(b) When the concentration of weak electrolyte becomes very low, its degree of ionization rises. This increase leads to increase in the number of ions in the solution. Thus, the molar conductivity rises sharply of a weak electrolyte at low concentration. The molar conductivity of strong electrolyte decreases a bit with an increase in concentration. This is due to increase in interionic attraction due to higher number of ions per unit volume. On dilution, ions move apart, weakening interionic attractions and thus conductance increases.
Limiting molar conductivity for weak electrolytes is obtained by using Kohlrausch law of independent migration of ions.
© examsnet.com
Go to Question: