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Question

In passing 3 F of electricity through three electrolytic cells connected in series containing Ag,Ca2+, and Al3+ ions, respectively. The molar ratio in which the three metal ions are liberated at the electrodes is:

A
1:2:3
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B
2:3:1
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C
6:3:2
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D
3:4:2
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Solution

The correct option is C 6:3:2
For all the solution Q=i×t is same,
nAg=QnF=QF(Ag++eAg)
nCa=Q2F(Ca2++2eCa)
nAl=Q3F(Al3++3eAl)
nAg:nCa:nAl=1:12:13=6:3:2
Molar ratio=6:3:2.


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