The students use same stock solution of `ZnSO_(4)` and a solution of `CuSO_(4)`. The EMF of one cell is 0.03V higher than the other. The concentration of `CuSO_(4)` in the cell with higher EMF value is 0.5 M. Find the concentration of `CuSO_(4)` in the other cell. (Take, 2.303 RT/F =0.06)

Explanation

The two given cells are represented as :
`Zn(s)|Zn^(2+)(C_(1))||Cu^(2+)(aq)(C=?)|Cu(s)," "E_("cell")=E_(1)`
`Zn(s)|Zn^(2+)(C_(2))||Cu^(2+)(aq)(C=0.5M)|Cu(s)," "E_("cell")=E_(2)`
Given, `E_(2)gtE_(1)" "E_(2)-E_(1)=0.03 and C_(1)=C_(2)`
[concentration of `Zn^(2+)` is the same in both the solutions]
`therefore" Cell reaction: "Zn(s)+Cu^(2+)(aq)hArr Zn^(2+)(aq)+Cu(s)`
`therefore" "E_("cell")=E_("cell")^(@)-(2.303)/(2F)log.([Zn^(2+)])/([Cu^(2+)])`
`"For cell 1, "E_(1)=E_("cell")^(@)-(0.06)/(2)log.(C_(1))/(C)" "["Given: "(2.303)/(F)=0.06]`
`"For cell 2, "E_(2)=E_("cell")^(@)-(2.303RT)/(nF)log.(C_(2))/(0.05)`
`"or, "E_(2)=E_("cell")^(@)-(0.06)/(2)log.(C_(2))/(0.05)`
`therefore" "E_(2)-E_(1)=(E_("cell")^(@)-(0.06)/(2)log.(C_(2))/(0.05))-(E_("cell")^(@)-(0.06)/(2)log.(C_(1))/(2))`
`"or, "0.03=(0.06)/(2)(log.(C_(2))/(C)xx(0.5)/(C_(1)))=(0.06)/(2)log.(0.5)/(C)" "[because C_(1)=C_(2)]`
`"or, "log.(0.5)/(C)=(0.03xx2)/(0.06)" or, "(0.5)/(C)=10" "therefore" "C=0.05M`