One Faraday of electricity is pa ssed through molten `Al_(2)O_(3)`, aqeusous solution of `CuSO_(4)` and molten NaCl taken in three different electrolytic cells connected in seris. The mole ratio of Al, Cu,Na deposted at the respective cathode isA. `2:3: 6`B. `6:2:3`C. ` 6:3:2`D. `1:2:3`

One Faraday of electricity is pa ssed through molten `Al_(2)O_(3)`, aq

1.	One Faraday of electricity is pa ssed through molten `Al_(2)O_(3)`, aqeusous solution of `CuSO_(4)` and molten NaCl taken in three different electrolytic cells connected in seris. The mole ratio of Al, Cu,Na deposted at the respective cathode isA. `2:3: 6`B. `6:2:3`C. ` 6:3:2`D. `1:2:3`
Answer» Correct Answer - 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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One Faraday of electricity is pa ssed through molten `Al_(2)O_(3)`, aqeusous solution of `CuSO_(4)` and molten NaCl taken in three different electrolytic cells connected in seris. The mole ratio of Al, Cu,Na deposted at the respective cathode isA. `2:3: 6`B. `6:2:3`C. ` 6:3:2`D. `1:2:3`

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