Three Faradays of electricity are passed through molten `Al_2O_3` aqueous solution of `CuSO_4` and molten NaCl taken in different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of

Options

A. I mole : 2 mole : 3 mole
B. 3 mole : 2 mole : 1 mole
C. I mole : 1.5 mole : 3 mole
D. 1.5 mole : 2 mole : 3 mole

Answer

C

Explanation

At cathode : `AI^(3+)+3e^(-)toAI`
`E_(AI)= "Atomic mass"/3`
At cathode : `Cu^(2+)+2e^(-)toCu`
`E_(Cu)=("Atomic mass")/2`
At cathode : `Na^(+)+e^(-)to Na`
`E_(Na)=("Atomic mass")/1`
For the passage of 3 F :
Mole atoms od AI deposite=1
Mole atoms of Cu deposite =1.5
Mole atoms of deposite= `1xx3xx=3`