Question

An electric current is passed through three cells connected in series containing `ZnSO_(4)`, acidulated water and `CuSO_(4)` respectively. What amount of Zn and `H_(2)` are liberated when 6.25 g of Cu is deposited? Eq. wt. of `Cu` and `Zn` are 31.70 and 32.6 respectively.

Answer 1

`because` Eq. of `Cu=Eq`. Of `Zn=Eq`. Of `H_(2)`
`(6.25)/(31.70)=(W_(Zn))/(32.6)=(W_(H_(2)))/(1)`