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)`

Answer

`0.05M`

Explanation

Daniell cell is
`Zn(s)|Zn^(2+)(aq)||Cu^(2+)(aq)|Cu(s)`
Let there be two Daniell cells with their `E_(cell)` as given below `:`
`Zn(s)|Zn^(2+)(c_(1))||Cu^(2+)(aq)(c=?)|Cu(s), E_(cell)=E_(1)`
`Zn(s)|Zn^(2+)(c_(2))||Cu^(2+)(c=0.5M)|Cu(s), E_(cell)=E_(2)`
where `E_(2)gtE_(1)`.
According to problem, `E_(2)-E_(1)=0.03` and `c_(2)=c_(1)`.
The cell reaction is `:`

So, `E_(cell)=E^(c-)._(cell)-(2.303RT)/(nF)log.([Zn^(2+)])/([Cu^(2+)])`
`=E^(c-)._(cell)-(0.06)/(n)log.([Zn^(2+)])/([Cu^(2+)])`
Thus, `E_(1)=E^(c-)._(cell)-(0.06)/(2)log .(c_(1))/(c)`
`E_(2)=E^(c-)._(cell)-(0.06)/(2)log.(c_(2))/(0.05)`
So, `E_(1)-E_(1)=(0.06)/(2)(log.(c_(2))/(c)xx(0.5)/(c_(1))) ( :. c_(1)=c_(2))`
or `0.03=(0.06)/(2) log .(0.05)/(c)`
Hence, `c=0.05M`