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Three Faradays of electricity are passed...

Three Faradays of electricity are passed through molten `Al_(2)O_(3)` , aqueous solutio of `CuSO_(4)` and molten NaCl taken in different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of

A

1 mole : 2 mole : 3 mole

B

3 mole : 2 mole : 1 mole

C

1 mole : 1.5 mole : 3 mole

D

1.5 mole : 2 mole : 3 mole

Text Solution

Verified by Experts

The correct Answer is:
C
`Al^(3+) + 3e^(-) rarr Al, Cu^(2+) + 2e^(-) rarr Cu Na^(+) + e^(-) rarr Na.`
Three Faradays will deposit 1 mole of Al,1.5 moles of Cu and 3 moles of Na.
Hence, molar ratio `= 1 :1.5 : 3`
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