1.

One faraday of electricity is passed through molten Al_(2)O_(3), aqueous solution of CuSO_(4) and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of Al, Cu and Na deposited at the respective cathodes is

Answer»

`2:3:6`
`6:2:3`
`6:3:2`
`1:2:3`

SOLUTION :`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`


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