Question

A Hittorf cell fitted with Ag/AgCl electrodes is filled with 0.0106 molal aqueous HCl (molar mass: 36.458). A 2.00 mA current was passed for 10800 sec. The cathode solution was then found to weigh 51.7436 g and contain 0.0267 g HCl after analysis. What is the transport number of H+?

a. 0.177

b. 0355

c. 0.645

d. 0.823

Answer 1

Step-by-step explanation:

The given data is as follows.

Current =
`2 * 10^(-3)` A

Time = 10800 sec

Weight of cathode solution = 51.7436 g

Weight of HCl after analysis = 0.0267 g (in cathode solution)

Weight of anode solution = 52.0461 g

Weight of HCl in anode solution = 0.0133 g

Amount of electrolyte present = molality × molar mass

=
`0.0106 * 36.4 * 10^(-3)`

=
`0.385 * 10^(-3) g`

Calculate the amount of HCl initially present as follows.

Electricity passed =
`2 * 10^(-3) * 10800` = 21.6 C

Hence, amount of electrons passed will be as follows.

`(21.6 C)/(96500 C/mol)`

= 0.00024 mol

Therefore, mass of HCl initially present is as follows.

= (51.7436 - 0.0267)

= 51.7169 g of water

=
`(0.3856 * 10^(-3))/(1g H_(2)O) * 51.7169 g H_(2)O`

= 0.01994 g

Mass of HCl present after the electrolysis in 51.7169 g of
`H_(2)O` = 0.0267 g

Mass of HCl gained = 0.0267 - 0.01994

= 0.00676 g

Hence, the amount of HCl gained =
`(0.00676 g)/(36.5 g/mol)`

= 0.0001853 mol

Therefore, gain of
`Cl^(-)` in cathodic compartment = 0.000224 mol

So, transport number of
`H^(+)` =
`\frac{\text{Amount of HCl gained}}{\text{moles of chlorine ions gained}}`

=
`(0.000185)/(0.00024)`

= 0.823

Thus, we can conclude that the transport number of
`H^(+)` is 0.823.