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

Answer