Four faradays of electricity were passed through AgNO3, CdSO4, AICI3 and PbCI4 kept in four vessels using insert electrodes. The ratio of moles of Ag, Cd, AI and Pb deposited will be ?
A
12 : 4 : 6 : 3
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B
1 : 2 : 3 : 4
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C
12 : 6 : 4 : 3
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D
4 : 3 : 2 : 1
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Solution
The correct option is D 12 : 6 : 4 : 3 AgNO3⟶Ag++NO−3CdSO4⟶Cd2++(SO4)2−AlCl3⟶Al3++3Cl−PbCl4⟶Pb4++4Cl−
It requires 1F of current to deposit 1 mol of Ag.
Similarly 2F of current to deposit 1 mol of Cd.
3F of current to deposit 1 mol of Al.
4F of current to deposit 1 mol of Pb.
When 4 faradays of electricity were passed,
moles of Ag diposited=41=4mol.
moles of Cd diposited=42=2mol.
moles of Al diposited=43mol.
moles of lead diposited=44=1mol.
therefore Ratio of moles of Ag,Cd,Al,Pb will be 4:2:43:1