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

Explanation:

The given data is as follows.

            Current = $2 \times 10^{-3}$ 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

                                                   = $0.0106 \times 36.4 \times 10^{-3}$

                                                   = $0.385 \times 10^{-3} g$

Calculate the amount of HCl initially present as follows.

          Electricity passed = $2 \times 10^{-3} \times 10800$ = 21.6 C

Hence, amount of electrons passed will be as follows.

                     $\frac{21.6 C}{96500 C/mol}$

                    = 0.00024 mol

Therefore, mass of HCl initially present is as follows.

                     = (51.7436 - 0.0267)      

                     = 51.7169 g of water

               = $\frac{0.3856 \times 10^{-3}}{1g H_{2}O} \times 51.7169 g H_{2}O$    

               = 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 = $\frac{0.00676 g}{36.5 g/mol}$    

                                                        = 0.0001853 mol

Therefore, gain of $Cl^{-}$ in cathodic compartment = 0.000224 mol

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

                                                = $\frac{0.000185}{0.00024}$

                                                = 0.823

Thus, we can conclude that the transport number of $H^{+}$ is 0.823.