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

(A) `E_(cell)^0= E_(Fe//Fe^(2+))^0=E_(Fe^(3+)//Fe^(2+))^0`
=0.441 + 0.771=1.212 V