A solution of $A g N O_{3}$ was electrolysed between silver electrodes. Before electrolysis, 10 g of solution contained 0.01788 g of $A g N O_{3}$ . After the experiment, 20.09 g of the anodic solution contained 0.06227 g of $A g N O_{3}$ . At the same time, 0.009479 g of Cu was deposited in the copper coulometer placed in series. Calculate transport no. of $A g^{+}$ . Write answer to the nearest integer after multiplying transport number calculated with 10.

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Solution

After electrolysis:
$∵ 20.09 g$ of anodic solution contained 0.06227 g of $A g N O_{3}$
$∴$ Mass of water in solution $= 20.09 - 0.06227 = 20.02773 g$
Thus, $20.02773 g H_{2} O$ has 0.06277 g $A g N O_{3}$
$= \frac{0.06227}{170} e q u i v a l e n t A g N O_{3}$
$= 0.0003663 e q u i v a l e n t A g N O_{3}$ or $A g^{+}$
Before Electrolysis:
$∵ 10.0 g$ of solution contained 0.01788 g $A g N O_{3}$
$∴$ Mass of water in solution $= 10 - 0.01788 = 9.98212 g$
Thus, 9.98212 g water has $= 0.01788 g A g N O_{3}$
$= \frac{0.01788}{170} e q . A g N O_{3}$
$∴ 20.02773 g$ water has $= \frac{0.01788 \times 20.02773}{170 \times 9.98212} e q . A g N O_{3}$
$= 0.000211$ equivalent of $A g N O_{3}$ or $A g^{+}$
Thus, increase in concentration of $A g^{+}$ during electrolysis
$= 0.0003663 - 0.000211$
$= 0.0001553 e q u i v a l e n t$
Also, Mass of Cu deposited in coulometer $= 0.009479 g$
$∴$ Equivalent of Cu deposited in coulometer $= \frac{0.009479}{31.8}$
$∴$ Equivalent of Cu deposited or actual increase around anodic solution
$= 0.0002981 e q .$
(Since, equal equivalents are discharged at either electrode)
Since, $A g^{+}$ had migrated from anode, which brings a fall in concentration around anode but due to attacked electrodes, (i.e., Ag in $A g N O_{3}$ ), apparent increase is noticed.
Thus, fall in concentration of $A g^{+}$ around anode
$=$ Actual increase which would have occur around anode - Apparent increase in $A g^{+}$ around anode
$= 0.0002981 - 0.0001553$
$= 0.0001428$ equivalent of $A g^{+}$
$∴$ Transport no. of $t_{A g^{+}} = \frac{Eq. of A g^{+} lost in anodic cell}{Eq. of C u^{+} deposited in coulometer}$
$= \frac{0.0001428}{0.0002981}$
$t_{A g^{+}} = 0.479 \approx 5$
Now, $t_{A g^{+}} + l_{N O_{3}^{-}} = 1$
$∴ t_{N O_{3}^{-}} = 1 - 0.4792 = 0.521 \approx 5$

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