Desscribe hittorf
method for the determination of transport number of AG positive and NO3 ions in AgNO3 solution when PT electrodes are used
Answers
Answer:
Correct option is
A
0.477, 0.523
After electrolysis:
∵ 20.09g of anodic solution contained 0.06227 g of AgNO
3
∴ Mass of water in solution = 20.09 − 0.06227 = 20.02773 g
Thus, 20.02773g H
2
O has 0.06277 g AgNO
3
.
170
0.0627
equivalent AgNO
3
= 0.0003663 equivalent AgNO
3
or Ag
+
Before Electrolysis:
∵ 10.0g of solution contained 0.01788 g AgNO
3
∴ Mass of water in solution =10−0.01788=9.98212g
Thus, 9.98212 g water has = 0.01788 g AgNO
3
=
170
0.01788
eq.AgNO
3
∴ 20.02773g water has =
170×9.98212
0.01788×20.02773
eq.AgNO
3
= 0.000211 equivalent of AgNO
3
or Ag
+
Thus, increase in concentration of Ag
+
during electrolysis
=0.0003663−0.000211
=0.0001553 equivalent
Also, Mass of Cu deposited in coulometer =0.009479 g
∴ Equivalent of Cu deposited in coulometer =
31.8
0.009479
∴ Equivalent of Cu deposited or actual increase around anodic solution
=0.0002981 eq.
(Since, equal equivalents are discharged at either electrode)
Since, Ag
+
had migrated from anode, which brings a fall in concentration around anode but due to attacked electrodes, (i.e., Ag in AgNO
3
) apparent increase is noticed.
Thus, fall in concentration of Ag
+
around anode.
= Actual increase which would have occur around anode - Apparent increase in Ag
+
around anode
= 0.0002981−0.0001553
= 0.0001428 equivalent of Ag
+
∴ Transport no. of t
Ag
+
=
Eq−of−Cu
+
Eq−of−Ag
+
=
0.0002981
0.0001428
= 0.479
Now , t
Ag
+
+t
NO
3
−
=1
∴t
NO
3
−
=1−0.4792=0.521
Explanation:
if it is helpful please mark as brainiest