1.

On passing C ampere of current for time t sec through 1 litre of 2(M) CuSO_(4) solution (atomic weight of Cu=63.5), the amount of m of Cu (in g) deposited on cathode will be:-

Answer»

`m=(Ct)/((63.5xx96500))`
`m=(Ct)/((31.25xx96500))`
`m(Cxx96500)/((31.25xxt))`
`m=(31.75xxCxxt)/(96500)`

Solution :According to faraday's LAW of ELECTROLYSIS
`mpropCt` or `m=ZCt`
where C=current, t=time
`Z=("Equivalent weight of metal")/(96500)`
EQ. WT of `Cu=(63.5)/(2)""(becauseCu^(2+)toCu)`
`Z=(63.5)/(2xx96500)""thereforem=(63.5xxCxxt)/(2xx96500)=(31.75xxCxxt)/(96500)`


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