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Two students use same stock solution of `ZnSO_(4)` and a solution of `CuSO_(4)`. The `EMF` of one cell is `0.03` higher than the other. The concentration of `CuSO_(4)` in the cell with higher `EMF` value is `0.5M`. Find the concentration of `CuSO_(4)` in the other cell.
`(` Take `2.303 RT//F=0.06)`

Text Solution

Verified by Experts

The correct Answer is:
0.05
`E+0.03=E^(@)-(0.06)/(2)log(([Zn^(2+)])/(0.5))`
`E=E^(@)-(0.06)/(2)log(([Zn^(2+)])/(C))`
`C=0.05M`.
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