1.

100 ml, 0.05 M CuSO_4 solution is electrolysed by using current of 0.965 Å for 100 min.Find the pH of solution at the end of electrolysis.

Answer»


SOLUTION :pH=1
After 1000 second when total `Cu^2+` discharge then `H^+` and `OH^(-)` discharge equally so pH will not change further.
`Cu^(+2)+2e^(-)to Cu`
`n_(Cu^(+2))/1=n_(E^(-))/2=Q/(2F)=(it)/(2F)`
`(0.925xxt)/(2xx96500)=(0.05xx100)/1000`
t=1000 sec
`(n_(H^+))_("excess")=(n_(OH^(-)))_("discharge")=n_(e^(-))=Q/F=(it)/(F)=(0.965xx1000)/96500=10^(-2)`
`n_(H^+)=(MV)/1000`
`10^(-2)=(Mxx100)/1000implies M=10^(-1)`
`[H^+]=10^(-1)`
`pH= -"log" 10^(-1)=1`


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