Discussion Forum

Two electrolytic cells are connected in series containing  CuSO₄ solution and molten AlCl₃ . If in electrolysis  0.4 moles of 'Cu'  are deposited on the cathode of the first cell. The number of moles of 'Al'  deposited on the cathode of the second cell is 

Answer:

Explanation:

 Key Idea  First find the weight of copper deposited, by using  the formula number of moles=  Weight/ molecular weight

and then calculate the weight of Al deposited and the number of moles of Al by using the second law of Faraday's.

 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 = $\frac {weight}{molecular weight}$

 $\therefore$    Weight of Cu = 0.4 x 63.5

 Now, from Eq.(i),

 = 0.4 x 6.5 / weight of Al deposited  = $\frac{\frac{63.5}{2}}{\frac{27}{3}}$

 $\therefore$  Weight of Al deposited=   $\frac{0.4 \times 63.5\times9}{31.75}=7.2 g$

 Now, number of moles of Al deposited = $\frac{7.2}{27}=0.27 $   moles