Three faradays of electricity are passed through molten `Al_2O_3` aqueous solution of `CuSO_4` and molten `NaCl` taken in deffernt electrolytic cells. The amout of `Al,Cu` and Na deposited at the cathodes will be in the ration of .

Options

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

Answer

C

Explanation

At cathode: `Al^(3+)+3e^(-)toCu`
`E_(Cu)=("Atomic mass")/(3)`
At cathode: `Cu^(2+)+2e^(-)toCu`
`E_(Cu)=("Atomic mass")/(2)`
At cathode: `Na^(+)+e^(-)toNa`
`E_(Na)=("Atomic mass")/(1)`
For the passage of 3 faraday,
mole atoms of Al deposited=1
mole atoms of Cu deposited`=(1xx3)/(2)=1.5`
mole atoms of Na deposited `=1xx3=3`.