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  • (a) The electrical resistance of a column of solution of length <img alt

(a) The electrical resistance of a column of 0.05\; M KOH solution of length 50\; cm and area of cross-section0.625\; cm^{2}  is 5\times 10^{3}\; ohm. Calculate its resistivity, conductivity and molar conductivity.

(b) Predict the products of electrolysis of an aqueous solution of CuCl_{2} with platinum electrodes.

        (Given : E^{\circ}_{Cu^{2+}/Cu}=+0.34\; V.E^{\circ}_{(\frac{1}{2}Cl_{2}/Cl^{-})}=+1.36\; V

                        E^{\circ}_{H^{+}/H_{2}(g),Pt}=0.00\; V.E^{\circ}_{(\frac{1}{2}O_{2}/H_{2}O)}=+1.23\; V )

 

 
 
 
 
 

Answers (1)

(a) 

R = \rho l / a = 5\times 10^{3} \times 0.625/50 = 62.5 \Omega \\ Conductivity = \kappa = 1/\rho = 1/62.5 = 0.016 \Omega^{-1}cm^{-1}\\ Molar -conductivity= \Lambda _{m} = \kappa\times 1000/c = 320 \Omega^{-1}cm^{2}mol^{-1}

(b) 

        Given : 

                          E^{\circ}_{Cu^{2+}/Cu}=+0.34\; V.E^{\circ}_{(\frac{1}{2}Cl_{2}/Cl^{-})}=+1.36\; V

                        E^{\circ}_{H^{+}/H_{2}(g),Pt}=0.00\; V.E^{\circ}_{(\frac{1}{2}O_{2}/H_{2}O)}=+1.23\; V 

At cathode :

Cu2+ + 2e– \rightarrow Cu

Because E0 Cu2+ /Cu > E0 H+/H2

At anode:

H2O \rightarrow 1/2 O2 + 2H+ + 2e–

This reaction should occur at anode but due to over-potential of O2, oxidation of Cl– is preferred.

Hence the reaction is

2Cl– \rightarrowCl2 + 2e–

Posted by

Safeer PP

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