Two students use same stock solution of `ZnSO_(4)` and a solution of `CuSO_(4)` The e.m.f of one cell is `0.03V` higher than the other. The conc. Of `CuSO_(4)` in the cell with higher e.m.f., value is `0.5M`. Find outthe conc. Of `CuSO_(4)` in the other cell `[(2.303 RT//F)=0.06]` .

Explanation

As given:
`"Cell I: Zn"|{:(ZnSO_(4)),(C_(1)):}||{:(CuSO_(4)),(C_(2)):}|Cu`
`E_(cell)=E_(cell)^(0)+(0.06)/(2)log.(C_(2))/(C_(1))`....(1)
`"Cell II: Zn"|{:(ZnSO_(4)),(C_(1)):}||{:(CuSO_(4)),(C _(2)):}|Cu`
`E _(cell)=E_(cell)^(0)+(0.06)/(2)log.(C _(2))/(C_(1))`...(2)
If `E_(cell)gtE _(cell)`, then `E_(cell)-E _(cell)=0.03V` and
`C_(2)=0.5M`
By Eqs. (1) and (2),
`E_(cell)-E _(cell)=(0.06)/(2)log.(C_(2))/(C _(1))`
`thereforeC _(2)=0.05M`