# (a) Represent the cell in which the following reaction takes place :$2Al(s)+3Ni^{2+}(0.1M)\rightarrow 2Al^{3+}(0.01M)+3Ni(s)$Calculate its emf if $E_{cell}^{o}=1.41V$.(b) How does molar conductivity vary with increase in concentration for strong electrolyte and weak electrolyte? How can you obtain limiting molar conductivity $\wedge _{m}^{o}$ for weak electrolyte?

(a) $2Al(s)+3Ni^{2+}(0.1M)\rightarrow 2Al^{3+}(0.001M)+3Ni(s)$

Cell Repn: $Al(s)\mid Al^{3+}(0.01M)\parallel Ni^{2+}(0.1M)\mid Ni(s)$

Nernst Eqn : $E_{cell}=E^0cell-\frac{0.0591}{n}\log \frac{\left [ Al^{3+} \right ]^2}{[Ni^{2+}]^3}$

$=1.41-\frac{0.0591}{6}\log \frac{\left [ 10^{-2} \right ]^2}{[10^{-1}]^3}$ $=1.41-0.00985\log10^{-1}$

$=1.41985V=1.42V$

(b) With increase in temperature, the greater inter-ionic attraction retard the motion of the ions and therefore the molar conductivity falls in case of both strong and weak electrolytes.

The limiting molar conductivity $\left ( \wedge _{m}^{o} \right )$ for weak electrolyte can be calculated by using Kohlrausch's Law.

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