How long will it take for a uniform current of 6.0 ampere to deposite 78.0 g gold from asolution of `AuCl_(4)^(-)` ? What mass of chlorine gas will be formed simultaneously at the anode in the electrolytic cell ?

Options

A. t = 3010 sec, w = 35.50 g
B. t = 20306 sec, w = 45.54 g
C. t = 19500 sec, w = 54.5 g
D. t = 19139 sec, w = 42.24 g

Answer

D

Explanation

Reactions take place at electrodes are : At cathode : `AuCl_4^(-) + 3e^(-) to Au + 4Cl^(-)`
At anode : `Cl^(-) to 1/2Cl_2 + e^(-)`
For the deposition of 197 g ( 1 mole ) of Au = 3F of charge is required thus, for the deposition of 78 g of Au , Charge required =`3/197xx78`=1.19 F = 1.19 x 96500 coulombs
From Q=I x t
`t=Q/t=(1.19xx96500)/6`=19136.16 sec
By 1 F charge , 35.5 g of `Cl_2` gas is formed , thus from 1.19 F, mass of chlorine gas formed = 35.5 x 1.19 = 42.24 g