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One Faraday of electricity is passed thr...

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 cathode is

A

`2:3: 6`

B

`6:2:3`

C

` 6:3:2`

D

`1:2:3`

Text Solution

Verified by Experts

The correct Answer is:
A
`Al^(3+)+3e^(-)rarrAl`
`Cu^(2)+2e^(-)rarrCu`
`Na^(+)+e^(-)rarrNa`
Hence 1F will deposit 1/3 mole of Al, 1/2 mol of Cu and 1 mole of Na
`therefore` molar ratio `=(1)/(3):(1)/(2):1=2:3:6`.
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