Question

Two electrolytic cells are connected in series containing CuSO₄ solution and molten AlCl₃. 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

• 0.6 moles

• 0.27 moles

• 0.18 moles

• 0.4 moles

Answer 1

Two electrolytic cells are connected in series containing CuSO₄ solution and molten AlCl₃. 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 0.27 moles.

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"` ....(i)

∵ No. of moles = `"weight"/"molecular weight"`

∴ Weight of Cu = 0.4 × 63.5

Now, from Eq. (i),

`= (0.4 xx 63.5)/"weight of Al deposited" = (63.5/2)/(27/3)`

∴ Weight of Al deposited = `(0.4 xx 63.5 xx 9)/31.75` = 7.2 g

Now, number of moles of Al deposited = `7.2/27`

= 0.27 moles.