(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? 

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

(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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