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 = 0.1209 g
Movement of boundary = 7.50 cm
Cross-section of tube= 1.24 $c{{m}^{2}}$
Write an answer to the nearest integer after multiplying the transport number calculated with 10.
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Hint:. As we know that the total current carried by the ion in the solution is called transport number. It is found that the transport number depends on the size of the ion. We will calculate the transport number of ${{H}^{+}}$ ions by the formula:
${{H}^{+}}$ ions = $\dfrac{l\times a\times c}{1000Q}$
Complete step by step answer:
- We are being provided with the information that Concentration of HCl solution= 0.1N,
Mass of Ag deposited in coulometer = 0.1209 g, Movement of boundary = 7.50 cm, and
Cross-section of tube= 1.24 $c{{m}^{2}}$.
- We will calculate the transport number of ${{H}^{+}}$ ions by moving the boundary method. We will calculate this by the formula:
${{H}^{+}}$ ions =$\dfrac{l\times a\times c}{1000Q}$
Where l is the length of migration that is equal to 7.50 cm. a is the area of cross – section is equal to 1.24 $c{{m}^{2}}$. And c is the concentration of solution equal to 0.1 N.
- As we know that Q is the charge passed which is equal to the equivalent of Ag deposited.
We can write the formula for this as: $Q =\dfrac{W}{E}$
- Now, by putting all the values given in this equation we get:
$\begin{align}
& =\dfrac{0.1209}{108}Faraday \\
& \therefore {{t}_{{{H}^{+}}}} =\dfrac{7.5\times 1.24\times 0.1\times 108}{0.1209\times 1000} \\
& = 0.8308 \\
\end{align}$
- Hence, we get the value of the transport number of ${{H}^{+}}$ ions as 0.8308.
Therefore, we get the correct as 10 X (0.8308) = 8.308 or 8
- Now, further as we know that ${{t}_{{{H}^{+}}}}+{{t}_{C{{l}^{-}}}}=1$
${{t}_{C{{l}^{-}}}}$ = 1 - 0.8308 = 0.1692
transport no. of $C{{l}^{-}}$ ion is 0.1692.
- Hence, we can conclude that the value of the transport number of ${{H}^{+}}$ ions is 0.8308.
Note: - It is found that transport number is affected by concentration of electrolyte, temperature and by the nature of co-ion. We should note that total transport of cation and anion of electrolyte is equal to one.
${{H}^{+}}$ ions = $\dfrac{l\times a\times c}{1000Q}$
Complete step by step answer:
- We are being provided with the information that Concentration of HCl solution= 0.1N,
Mass of Ag deposited in coulometer = 0.1209 g, Movement of boundary = 7.50 cm, and
Cross-section of tube= 1.24 $c{{m}^{2}}$.
- We will calculate the transport number of ${{H}^{+}}$ ions by moving the boundary method. We will calculate this by the formula:
${{H}^{+}}$ ions =$\dfrac{l\times a\times c}{1000Q}$
Where l is the length of migration that is equal to 7.50 cm. a is the area of cross – section is equal to 1.24 $c{{m}^{2}}$. And c is the concentration of solution equal to 0.1 N.
- As we know that Q is the charge passed which is equal to the equivalent of Ag deposited.
We can write the formula for this as: $Q =\dfrac{W}{E}$
- Now, by putting all the values given in this equation we get:
$\begin{align}
& =\dfrac{0.1209}{108}Faraday \\
& \therefore {{t}_{{{H}^{+}}}} =\dfrac{7.5\times 1.24\times 0.1\times 108}{0.1209\times 1000} \\
& = 0.8308 \\
\end{align}$
- Hence, we get the value of the transport number of ${{H}^{+}}$ ions as 0.8308.
Therefore, we get the correct as 10 X (0.8308) = 8.308 or 8
- Now, further as we know that ${{t}_{{{H}^{+}}}}+{{t}_{C{{l}^{-}}}}=1$
${{t}_{C{{l}^{-}}}}$ = 1 - 0.8308 = 0.1692
transport no. of $C{{l}^{-}}$ ion is 0.1692.
- Hence, we can conclude that the value of the transport number of ${{H}^{+}}$ ions is 0.8308.
Note: - It is found that transport number is affected by concentration of electrolyte, temperature and by the nature of co-ion. We should note that total transport of cation and anion of electrolyte is equal to one.