Two electrolytic cells are connected in series containing `CuSO_(4)` solution and molten `AlCl_(3)`. If in electrolysis 0.4 moles of 'Cu' are deposited on cathode of first cell. The number of moles of 'Al' deposited on cathode of the second cell is

Options

A. 0.6 moles
B. 0.27 moles
C. 0.18 moles
D. 0.4 moles

Answer

B

Explanation

Given,
Number of moles of Cu deposited = 0.4 moles
According to Faraday's second law,
`("weight of Cu deposited")/("weight of Al deposited")=("Eq wt. of Cu")/("Eq wt. of Al")`
`because` No. of moles=`("weight")/("molecular weight")`
`therefore` Weight of Cu =`0.4xx63.5`
Now, from Eq. (i),
`=(0.4xx63.5)/("weight of Al deposited")=(63.5/2)/(27/3)`
`therefore` Weight of Al deposited`=(0.4xx63.5xx9)/(31.75)=7.2 g`
Now, number of moles of Al deposited`=7.2/27`=0.27
moles