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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

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`Zn(s)|ZnSO_(4) (C_(1))||0.5 M CuCO_(4)|Cu(s)`
`E_(1)=E^(@)-0.06/2"log" C_(1)/0.5` ...(i)
student II. `Zn(s)|ZnSO_(4) (C_(1))||CuSO_(4) (C_(2))|Cu(s)`
`E_(2)=E^(@)-0.06/2"log" C_(1)/C_(2)` ...(ii)
`E_(1)-E_(2)=0.06/2 ["log"C_(1)/C_(2)-"log"C_(1)/0.5]`
`0.03=0.06/2[log (0.5/C_(2))]`
`1="log" 0.5/C_(2)`
`C_(2)=0.05 M`
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