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)`

Explanation

`Zn(s)ZnSO_(4)(C_(1)) || CuSO_(4) (C_(2)|Cu(s)`
`E_(1) = E^(@)-(0.06)/(2)log(Cl)/(0.5) ……… (1)`
and 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)
`Rightarrow E_(1)- E_(2) = (0.06)/(2)["log"(C_(1))/(C_(2)) - log (C_(1))/(0.5)]` `Rightarrow 0.03 =(0.06/(2)[log ((0.5)/(C_(2))] Rightarrow C_(2) = 0.05`