Engineering Chemistry S Datta

222 ENGINEERING CHEMISTRY
Sol. According to Kohlrausch’s law,
λ α
= λ +
+ λ –
i.e., λ α
(AgCl) = λ(Ag + ) + λ(Cu – )
= 58.3 + 65.3 = 119.1 mhos eq –1 cm 2
λ v
= K v
V
∴ V = λ ∞ v 1191 .
=
–6
K v 124 . × 10
= 9.6 × 105 cm 3 eq –1
= 9.6 × 10 2 l eq –1
1
∴ Solubility =
9 . 6 × 10 2 eq l –1 = 0.00104 eq l –1
= (0.0104 × 143.5) g l –1
= 0.1495 g l –1 .
Example 11. The speed ratio of Ag + and NO 3
–
ions in a solution of AgNO 3
electrolysed
between Ag-electrodes is 0.916. Find the transport number of Ag + and NO 3
–
ions.
Sol.
Speed of Ag
Speed of NO
+
3 –
u
= =
v
0.916
1.000
∴
u
t Ag
+ =
u+ v
0.916
0.916 + 1
= 0.916
1.916 = 0.478
Hence, t = 1 – t – + NO3
= 1 – 0.478 = 0.522.
Ag
Example 12. A dilute solution of CuSO 4
was electrolysed, using Pt-electrodes. The amount
of Cu in the anodic solution was found to be 0.6350 g, and 0.6236 g before and after electrolysis,
respectively. The weight of Ag deposited in a silver coulometer, placed in series, was found to be
0.1351 g. Calculate the transport number of Cu +2 and SO
–2
4
ions.
[At. wts. = Ag(107.88), Cu(63.6)]
Sol. Wt. of Cu in anodic solution after electrolysis is = 0.6236 g.
Wt. of Cu in anodic solution before electrolysis = 0.6350 g.
∴ Loss in wt. of Cu from anodic solution (anolyte) = (0.6350 – 0.6236) g = 0.0114 g
Wt. of Ag deposited in voltameter = 0.1351 g.
∴ Amt. of Cu deposited equivalent to 0.1351 g of Ag.
=
=
wt. of Cu( x ) wt. of Ag
=
Eq. wt. of Cu Eq. wt. of Ag
x
63.6
2
0.1351
=
107.88
0.1351 × 31.8
∴ x = = 0.0398 g
107.88
Loss at anode 0.114
∴ t Cu
+ 2 = = = 0.286
Total loss 0.0398
∴ t ∞ SO4 = 1 – t Cu
+ 2 = 1 – 0.286 = 0.714.