One Faraday of electricity is pa ssed through molten `Al_(2)O_(3)`, aqeusous solution of `CuSO_(4)` and molten NaCl taken in three different electrolytic cells connected in seris. The mole ratio of Al, Cu,Na deposted at the respective cathode is

Options

A. `2:3:6`
B. `6:2:3`
C. `6:3:2`
D. `1:2:3`

Answer

A

Explanation

`Ag^(3+)+3e^(-)toAl`
`Cu^(2+)+2e^(-)toCu`
`Na^(+)+e^(-)toNa`
thus 1F will be deposit `(1)/(3)` mole of `Al,(1)/(2)` mole of Cu and 1 mole of Na. Hence, molar ratio `=(1)/(3):(1)/(2):1=2:3:6`