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

As given :
Cell I : `Zn|{:(ZnSO),(C_(1)):}|{:(CuSO_(4)),(C_(2)):}:|Cu`
`E_(cell) = E_(cell)^(@) + (0.06)/(2)log.(C_(2))/(C_(1))` ….(1)
Cell II : `.Zn|{:(ZnSO),(C_(1)):}|{:(CuSO_(4)),(C'_(2)):}:|Cu`
`E'_(cell) = E_(cell)^(@) + (0.06)/(2)log.(C'_(2))/(C_(1))` ....(2)
If `E_(cell)gtE'_(cell)`, then `E_(cell) - E'_(cell) = 0.03 V` and
`C_(2) = 0.5 M`
By Eqs. `(1)` and `(2)`,
`E_(cell) - E'_(cell) = (0.06)/(2)log.(C_(2))/(C'_(2))`
`0.03 = (0.06)/(2)log.(0.5)/(C'_(2))`
`:. C'_(2) = 0.05 M`