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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

Answer»

1 MOLE: 2 mole : 3 mole
3 mole : 2 mole : 1 mole
1 mole : 1.5 mole : 3 mole
1.5 mole : 2 mole : 3 mole

Solution :`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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