InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Copper sulphate solution `(250 ML)` was electrolyzed using a platinum anode and a copper cathode. A constant current of `2mA` was passed for `16 mi n`. It was found that after electrolysis the absorbance of the solution was reducted to `50%` of its original value . Calculate the concentration of copper sulphate in the solution to begin with. |
|
Answer» We known, Equivalent of `Cu^(2+)` lost during electrolysis `=(ixxt)/(96500)=(2xx10^(-3)xx16xx60)/(96500)=1.989xx10^(-5)` or mole of `Cu^(2+)` lost during electrolysis `= (1.989xx10^(-5))/(2)` This value is 50% of the intial concentration of solution. Thus, initial mole of `CuSO_(4)=(2xx1.989xx10^(-5))/(2)` `=1.989xx10^(-5)` Thus, initial concentration of `CuSO_(4)=(1.989xx10^(-5)xx1000)/(250) [CuSO_(4)]=7.95xx10^(-5)M` |
|