a silver nitrate solution was electrolysed between silver electrodes. the ratio of velocities of silver and nitrate ions is 0.916 . calculate the transport numbers of silver and nitrate ions
Answers
Answer:
After electrolysis:
∵20.09g of anodic solution contained 0.06227 g of AgNO
3
∴ Mass of water in solution =20.09−0.06227=20.02773g
Thus, 20.02773gH
2
O has 0.06277 g AgNO
3
=
170
0.06227
equivalentAgNO
3
=0.0003663equivalentAgNO
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 gAgNO
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. ofCu
+
deposited in coulometer
Eq. ofAg
+
lost in anodic cell
=
0.0002981
0.0001428
t
Ag
+
=0.479≈5
Now, t
Ag
+
+l
NO
3
−
=1
∴t
NO
3
−
=1−0.4792=0.521≈5