Question

a) A Hittorf cell fitted with silver-silver chloride electrodes is filled with HCI solution that contains 0.3856 x 10 g HCL per g of water a current of 2 mA is passed for exactly three hours. The solutions are withdrawn, weighed and analysed. Total weight of the anode solution weighs 52.0461 g and contains 0.0133 g of HCI. Calculate the transference number of the H.​

Answer 1

Answer:

Explanation:

As per the question,

Given That:

When 0.3856 x 10 g of HCL per g of water is added to a Hittorf cell with silver-silver chloride electrodes, a current of 2 mA is transmitted through the cell for exactly three hours. The remedies are taken out, weighed, and examined. The anode solution weighs 52.0461 g in total and has an HCI content of 0.0133 g.

To Calculate:

the transference number of the H

Solution:

First we will calculate the mass of HCl present initially, Electricity $= 2^{-3} A \times 3\times60\times60s= 21.6 As = 21.6 C$ passed Amount of electrons passed = 21.6 C / 96500 C/mol = 0.000224 mol Mass of HCl present initially in 51.7169g (51.7436g - 0.0267g) of water $=(0.3856\times10^-3g / 1g H2O) \times(51.7169g H2O) = 0.01994 g$ Mass of HCl present after the electrolysis in 51.7169g of water = 0.0267 g (as given) So, mass of HCl gained = 0.0267g - 0.01994g = 0.00676 g therefore, amount of HCl gained = 0.00676g / 36.5g/mol = 0.0001853 mol If there is no migration, then gain of Cl- in the cathodic compartment would have been 0.000224 mol

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