The electrochemical cell shown below is a concentration cell M|M^ 2(Mx2)||M^ 2(0.001m)|M The emf of cell depends on the difference in concentration M^ 2 ion at two electrode.The emf of cell at 298k is 0.059v...Find solubility product of Mx2 at 298k???

In this question the cell notation shown is not clear.................Please write clearly and specify the concentrations and all other things given in the queston.

Use Nernst Equation Ecell = E0cell - (0.059/n)*LogK Use the formula :- ∆G = -nEF Answer will be -11.4 (d)