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

The two cells may represented as
`Zn|Zn^(2+)(conc=c)||Cu^(2+)(c=?)|Cu,EMF=E_(1)`
`Zn|Zn^(2+)(conc.=c)||Cu^(2+)(0.5M)|Cu,EMF=E_(2)`
The cell reaction is: `Zn+Cu^(2+)toZn^(2+)+Cu`
`thereforeE_(1)=E^(@)-(2.303RT)/(2F)"log"(c)/([Cu^(2+)])`
`E_(2)=E^(@)-(2.303RT)/(2F)"log"(c)/(0.5)`
`thereforeE_(2)-E_(1)=(2.303RT)/(2F)("log"(c)/([Cu^(2)])-"log"(c)/(0.5))=0.03V` (given)
`therefore(0.06)/(2)"log"(0.5)/([Cu^(2+)])=0.03` or `"log"(0.5)/([Cu^(2+)])=1` or `(0.5)/([Cu^(2+)])=10` or `[Cu^(2+)]=(0.5)/(10)=0.05M`