(a) What is limiting molar conductivity ? Why there is step rise in the molar conductivity of weak electrolyte on dilution ?
(b) Calculate the emf of the following cell at 298 K : Mg(s)`|Mg^(2+)(0.1 M)||Cu^(2+)(1.0 xx 10^(-3)M)|Cu(s)`
[Given : `E_("cell")^(@)` = 2.71 V]

Explanation

(a) Limiting molar conductivity : It is the sum of limiting ionic conductivities of cation and the anions each multiplied with the number of ions present in one formula unit of electrolyte.
`wedge^(@)m " for " Axx"By"=xlambda_(+)^(@)+ylambda_(-)^(@)`
limiting molar `" "lambda^(@+)` and `lambda^(@-)rarr` limiting molar
Conductivity `" Conductivity of cation and anion"`
There is steep rise in molar conductivity of weak electrolyte on dilution because as the concentration of weak electrolyte is reduced, more of it ionizes. Thus, increase in conductance with decrease in concentration is due to increase in number of ions in the solution.
(b) `Mg(s)|Mg^(2+)(0.1 M)||Cu^(2+)(1.0 xx 10^(-3)M)|Cu(s)`
`E^(@)`Cell = 2.71 v.
Reaction : `Mg+Cu^(2+)rarrMg^(2+)Cu`
`MgrarrMg^(2+)+2e^(-)`
`Cu^(2+) + 2e^(-) rarrCu`
n=2
`E" cell" = E^(@)cell - (0.0591)/(2) log. ([Mg^(2+)])/([Cu^(2+)])`
`=2.71 - (0.0591)/(2)log.(0.1)/(0.001)`
= 2.71 - 0.02955 log 10^(2)`
`=2.71 - 0.02955 xx 2`
= 2.71 - 0.059
E cell = 2.65 V