(A) Define molar conductivity of a substance and describe how for weak and strong electrolytes, molar conductivity changes with concentration of solute. How is such change explained?

(B) A voltaic cell is set up at 25°C with the following half cells:
${\mathrm{Ag}}^{+}(0.001\mathrm{M})\left|\right|\mathrm{Ag}\text{ and }{\mathrm{Cu}}^{2+}(0.10\mathrm{M})\mid \mathrm{Cu}$ What would be the voltage of this cell?

(a) Molar conductivity: Molar conductivity of a solution at a given concentration is tire conductance of the volume ‘V’ of a solution containing one mole of electrolyte kept between two electrodes with area of cross section ‘A’ and distance of unit length. It is represented by ${\mathrm{\Lambda }}_{\mathrm{m}}$ (lamda).
$\begin{array}{l}{\mathrm{\Lambda }}_{\mathrm{m}}=\mathrm{KAl}\therefore 1=1\text{ and }\mathrm{A}=\mathrm{V}\\ \therefore {\mathrm{\Lambda }}_{\mathrm{m}}=\mathrm{KV}\text{ Unit }=\mathrm{S}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}\end{array}$
Effect of change of concentrations on molar conductivity. In case of strong electrolytes there is a small increase in conductance with dilution because a strong electrolyte is completely dissociated in solution and the number of ions remains constant. Moreover there will be greater inter-ionic attractions at higher concentrations which retards the motion of ions and conductance decreases. In case of weak electrolytes there is increase in conductance with decrease in concentration due to the increase in the number of ions in the solution. The graph between ${\mathrm{\Lambda }}_{\mathrm{m}}$ and concentration also rectifies the above statement.

(B) The reaction takes place in cell as
$\begin{array}{l}\mathrm{Cu}+2{\mathrm{Ag}}^{+}⟶{\mathrm{Cu}}^{+2}+2\mathrm{Ag}\\ \text{ (0.001 M) }(0.10)\\ \mathrm{Cu}/{\mathrm{Cu}}^{+2}(0.10)\text{ II }{\mathrm{Ag}}^{+}(0.001\mathrm{M})/\mathrm{Ag}\\ {\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}^{\circ }-\frac{0.059}{\mathrm{n}}\log \frac{\left[{\mathrm{Cu}}^{+2}\right]}{{\left[{\mathrm{Ag}}^{+}\right]}^2}\\ {\mathrm{E}}_{\text{cell }}=0.46-\frac{0.059}{2}\log \frac{0.1}{(0.001{)}^2}\\ =0.46-\frac{0.059}{2}\log {10}^5\\ =0.46-\frac{0.059}{2}\times 5\log 10\\ \end{array}$

$\begin{array}{l}=0.46-0.0592\times 5\\ =0.46-0.1475=0.3125\end{array}$