Solution of Kohlrausch Law & Electrolysis Test 04 - Electrochemistry (Class 12, CBSE) Class 12 Notes | EduRev

Class 12 : Solution of Kohlrausch Law & Electrolysis Test 04 - Electrochemistry (Class 12, CBSE) Class 12 Notes | EduRev

 Page 1


CBSE TEST PAPER-04 
CLASS - XII CHEMISTRY (Electrochemistry) 
[ANSWERS] 
Topic: -Kohlrausch Law & Electrolysis 
1. An electrolyte that ionises completely in solution is a strong electrolyte eg. NaCl , CaCl 2
etc and an electrolyte that ionizes partially in solution is weak electrolyte eg CH 3
COOH , NH 4OH etc.
2. Kohlrausch Law of independent migration 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.
3. Relation between limiting molar conductance and degree of dissociation –
 
o
m
m
a
O
=
O
where  = degree of dissociation a
 = molar conductance m ?
0
 Limiting molar conductance m ?
Relation between dissociation constant and limiting molar conductance – 
2
0 0
 m
 =  where c = concentration
 ( m - m)
C
Ka
m
?
? ? ?
4. A = strong Electrolyte
B = weak Electrolyte
In case of B , it is not possible to get an exact value of limiting molar conductance.
5. 
0 o o o
4 m 4 m m
 (NH OH) =  (Na Cl) +  (NaOH)- (NaCl)
m
? ? ? ?
 = 129.8 +217.4 – 108.9    = 237.3  5 cm
2 
/mol 
Degree of dissociation , 
0
 = 
m
m
?
a
?
 = 
2
2
9.33 5 cm / mol
237.3 5 cm / mol
= 0.039   or  3.9 %. 
6. Faraday’s Laws of electrolysis
First Law: The amount of chemical reaction which occurs at any electrode during 
electrolysis by a current is proportional to the quantity of electricity 
passed through the electrolyte. 
Page 2


CBSE TEST PAPER-04 
CLASS - XII CHEMISTRY (Electrochemistry) 
[ANSWERS] 
Topic: -Kohlrausch Law & Electrolysis 
1. An electrolyte that ionises completely in solution is a strong electrolyte eg. NaCl , CaCl 2
etc and an electrolyte that ionizes partially in solution is weak electrolyte eg CH 3
COOH , NH 4OH etc.
2. Kohlrausch Law of independent migration 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.
3. Relation between limiting molar conductance and degree of dissociation –
 
o
m
m
a
O
=
O
where  = degree of dissociation a
 = molar conductance m ?
0
 Limiting molar conductance m ?
Relation between dissociation constant and limiting molar conductance – 
2
0 0
 m
 =  where c = concentration
 ( m - m)
C
Ka
m
?
? ? ?
4. A = strong Electrolyte
B = weak Electrolyte
In case of B , it is not possible to get an exact value of limiting molar conductance.
5. 
0 o o o
4 m 4 m m
 (NH OH) =  (Na Cl) +  (NaOH)- (NaCl)
m
? ? ? ?
 = 129.8 +217.4 – 108.9    = 237.3  5 cm
2 
/mol 
Degree of dissociation , 
0
 = 
m
m
?
a
?
 = 
2
2
9.33 5 cm / mol
237.3 5 cm / mol
= 0.039   or  3.9 %. 
6. Faraday’s Laws of electrolysis
First Law: The amount of chemical reaction which occurs at any electrode during 
electrolysis by a current is proportional to the quantity of electricity 
passed through the electrolyte. 
Second Law: The amount of different substances liberated by the same quantity of 
electricity passing through the electrolytic solution is proportional to 
their chemical equivalent weights. 
7. Faraday’s constant is the quantity of electricity carried by one mole of electrons.
1 F = 96500 C/mol
8. Cu
2+
 + 2e
- 
? Cu
Two faradays are needed to reduce 1g mole Cu
2+ 
 ?6 Faradays will be needed to
reduce 3g mole of Cu
2+
.
9. Q = It    = 1× 15 × 60 = 900C
The reaction is 2Cl
-
 ? Cl 2 + 2e
- 
      2mol  1mol  2mol 
2
 2F  . produces 1 mol of Cl ?
1mol of Cl 2 = 71g  
2
 2 96500 C produces 71g of Cl ? ×
900 C  will produce 
71 900
 g
2 96500
×  = 0.331 g of Cl 2. 
10. Q = It   = 5X 193 = 965 C
96500C = 1 mol of electrons = 6.022X10
23
 mol
-1
965 C = 6.022 X 10
23 
 965
96500
×
= 6.022 × 10
21
 electrons. 
11. Zn
2+
 has higher reduction potential (-0.76v) Than H 2O (- 0.83v) and therefore Zn
2+
 is
reduced to Zn preferentially at cathode.
12. 2.417 g of silver.
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