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

Answer

C

Explanation

At cathode: `Al^(3+) (aq) + 3e^(-) to Al(s)`
`E_(Al) = ("Atomic mass")/3`
at cathode: `Cu^(2+) (aq) + 2e^(-) to Cu(s)`
at Cathode: `Na^(+) (aq) + e^(-) to Na(s)`
`E_(Na) = ("Atomic mass")/1`
For the passage of 3 Faraday,
mole atoms of Al deposited = 1
mole atoms of Cu deposited `= (1 xx 3)/2 = 1.5`
mole atoms of Na deposited = `1 xx 3 = 3`