If 2.7 gm aluminum metal is deposited on electrodes when two different electrolytic cell having molten `Cu(NO_(3))_(2)` and `Al(NO_(3))_(3)` respectively are arranged in series, then how much copper metal is produced ? `(Cu=63.5,Al=27.0" gm "mol^(-1))`.

Options

A. 9.525 gm
B. 31.75 gm
C. 63.5 gm
D. 190.5 gm

Answer

A

Explanation

`Al^(3+)+3e^(-) to Al`
So, 3 F electricity gives 27 gm of Al, but Al is produced only 2.7 gm
So, Faraday `=(2.7)/(27)xx3=0.3` Faraday
Reaction : `Cu^(2+)+2e^(-) to Cu`
So 2 F electricity gives 63.5 gm of Cu
So, 0.3 F electricity gives `=(0.3xx63.5)/(2)`
`=9.525gm` Cu