Answer :
Explanation:
The given data is as follows.
Current = [tex]2 \times 10^{-3}[/tex] 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
= [tex]0.0106 \times 36.4 \times 10^{-3}[/tex]
= [tex]0.385 \times 10^{-3} g[/tex]
Calculate the amount of HCl initially present as follows.
Electricity passed = [tex]2 \times 10^{-3} \times 10800[/tex] = 21.6 C
Hence, amount of electrons passed will be as follows.
[tex]\frac{21.6 C}{96500 C/mol}[/tex]
= 0.00024 mol
Therefore, mass of HCl initially present is as follows.
= (51.7436 - 0.0267)
= 51.7169 g of water
= [tex]\frac{0.3856 \times 10^{-3}}{1g H_{2}O} \times 51.7169 g H_{2}O[/tex]
= 0.01994 g
Mass of HCl present after the electrolysis in 51.7169 g of [tex]H_{2}O[/tex] = 0.0267 g
Mass of HCl gained = 0.0267 - 0.01994
= 0.00676 g
Hence, the amount of HCl gained = [tex]\frac{0.00676 g}{36.5 g/mol}[/tex]
= 0.0001853 mol
Therefore, gain of [tex]Cl^{-}[/tex] in cathodic compartment = 0.000224 mol
So, transport number of [tex]H^{+}[/tex] = [tex]\frac{\text{Amount of HCl gained}}{\text{moles of chlorine ions gained}}[/tex]
= [tex]\frac{0.000185}{0.00024}[/tex]
= 0.823
Thus, we can conclude that the transport number of [tex]H^{+}[/tex] is 0.823.