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Three faradays of electricity are passed through molten `Al_2O_3` aqueous solution of `CuSO_4` and molten `NaCl` taken in deffernt electrolytic cells. The amout of `Al,Cu` and Na deposited at the cathodes will be in the ration of .A. ` 1` mole : 15 ,mole `1` molesB. 1 mole : 2 moles :3 molesC. 1 mole : 2 moles : 3 molesD. 1.5 moles : 2 moles : 3 moles |
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Answer» Correct Answer - A At cathode : `Al^(3+) +3e^- rarr Al` `E_(Al) = ("Atomic mass")/3` At cathode : ` Cu^(2+) +2 e^- rarr Cu` `E_(Cu) = ("Atomic mass")/2` At cathode , `Na^+ + e^- rarr N` ` E_(Na) = ("Atomic mass")/1` for the passage of ` 3F` mole atoms of Al deposited `=1` mole atoms of Cu deposited ` = ( 1xx 3)/2 = 1.5` mole atoms of Na deposited `= 1xx 3 =3`. |
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