On passing 3 faradays of electricity through three electrolytic cells connected in series containing Ag+, Ca2+ and Al3+ ion respectively, the molar ratio in which three metal ions are liberated at the electrode is 

On passing 3 faradays of electricity through three electrolytic cells connected in series containing Ag+, Ca2+ and Al3+ ion respectively, the molar ratio in which three metal ions are liberated at the electrode is 

  1. A

    1 : 2 : 3

  2. B

    6 : 3 : 2

  3. C

    3 : 2 : 1

  4. D

    2 : 1 : 3

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    Solution:

    Ag+ + e → Ag ; 

    Ca2++ 2e → Cl;

    Al3++ 3e → Al

    So, According to Faraday's second law:

    1 F produce = 1 mole of Ag   

    2F produce = 1 mole of Ca  or 1F produce = 12mole of Ca 

    3 F produce = 1 mole of Al   or  1 F produce = 13 mole of Al 

    Molar ratio is  =  1 : 12 : 13    or     6 : 3 : 2

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