Calculate e.m.f. of cell for the reaction :
`Mg_((s))+Cu^(2+)"(0.0001 M)"rarr Mg^(2+)"(0.001 M)"+Cu_((s))`
Given that : `E_(Mg^(2+)//Mg)^(@)=-2.37V`
`E_(Cu^(2+)//Cu)^(@)=+0.34V`

Explanation

Cell equatio :
Mg(s
) +` Cu^(2+) (aq) to Mg^(2+) (aq)+Cu(s) underset(n=2)`
Nernst equaton:
`E_(cell)=-0.00591/2log. [Mg^(2+)]/[Cu^(2+)]``E_(cell)=-0.00591/2log. [Mg^(2+)]/[Cu^(2+)]`
EMF of the cell,
the limiting molar productivity electroyt can be represent as the sum of limiting molar conductivities of the individual cation and anion .
pplication : kohlraush law is appliedv to calculate the limiting molar conductivityb of any electroyt from the limitinf molar conductivity lambda @ of individual ion.s.
when concentration approches zero,the molar conductance is known as limitiomng molar conductance.