1.

When 6 faraday electricity is pass through aqueous solution of silver nitrate, copper sulphate and gold chloride (AuCl_(3)), then what ratio of mole of metals obtained at cathode ?

Answer»

`1:1:1`
`3:2:1`
`1:2:3`
`6:3:2`

Solution :`AgNO_(3) to Ag^(+)+NO_(3)^(-)`
`Ag^(+)+E^(-) to Ag`
`therefore 6Ag^(+)+6e^(-) to 6Ag` (6 moles of Ag obtained by 6F)
`CuSO_(4) to CU^(2+)+SO_(4)^(2-)`
`Cu^(2+)+2E^(-) to Cu`
`therefore 3Cu^(2+)+6e^(-) to 3Cu` (3 moles of Cu obtained by 6F)
`AuCl_(3) to Au^(3+)+3CL^(-)`
`Au^(3+)+3e^(-) to Au`
`therefore 2Au^(3+)+6e^(-)to2Au` (2moles of Au obtained by 6F)
So, mole reaction of Ag, Cu and Au metal is 6:3:2.


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