1.

If three faradyas of electricity is passed through the solutions of AgNO_(3),CuSO_(4) and AuCl_(3). The molar ratio of the cations deposited at the cathodes will be

Answer»

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

SOLUTION :`Ag^(+)+e^(-)rarrAg`
3F of electricity will produce 3 moles of Ag.
`CU^(2+)+2e^(-)rarrCu`
3F of electricity will produce 3/2 moles of Cu
`Au^(3+)+3E^(-)rarrAu`
3F of electricity will produce 1 mole of Au `therefore` Molar ratio of
`Ag:Cu:Au`
or `6:3:2`


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