Question

Calculate the transport no. of $H^{+}$ ion from the following data obtained by moving boundary method using $CdC{l}_2$ as the indicator electrolyte:

Concentration of HCl solution $=0.1N$

Mass of Ag deposited in coulometer $=0.1209g$
Movement of boundary $=7.50cm$
Cross-section of tube $=1.24c{m}^2$

Write an answer to the nearest integer after multiplying the transport number calculated with 10.

Answer 1

Transport no. of $H^{+}ions=\frac{l\times a\times c}{1000Q}$ (moving boundary method)
l is length of migration $=7.50cm$
a is area of cross-section $=1.24c{m}^2$
c is concentration of solution $=0.1N$
Q is charge passed $=$ Equivalent of Ag deposited
$=\frac{0.1209}{108}Faraday(Q=\frac{W}{E})$
$\therefore t_{H^{+}}=\frac{7.5\times 1.24\times 0.1\times 108}{0.1209\times 1000}=0.8308$
Transport no. of $H^{+}$ ions is 0.8308

Thus the correct answer is $10\times \left(0.8308\right)=8.308\approx 8$
Now, $t_{H^{+}}+t_{C{l}^{-}}=1$
$\therefore t_{C{l}^{-}}=1-0.8308=0.1692$
Transport no. of $C{l}^{-}$ ions is 0.1692.