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  • (a) Following reaction takes place in the cell :     <img alt="Zn(s)+Ag_{2}O(s)+H_{2}O(l)\rightarrow Zn^{2+}(aq)+2Ag(s)+2OH^{-}(aq)" src="https://entrancecorner.codecogs.com/gif.latex?%5Cinline%20Zn%28s%29&plus;Ag_%7B2

(a) Following reaction takes place in the cell :

    Zn(s)+Ag_{2}O(s)+H_{2}O(l)\rightarrow Zn^{2+}(aq)+2Ag(s)+2OH^{-}(aq)

    Calculate \Delta _{r}G^{\circ} of the reaction . 

    [Given : E^{0}_{(Zn^{2+}/Zn)}=-0.76\; V,

    E^{0}_{(Ag^{+}/Ag)}=0.80\; V,\; 1\; F=96,500\; C\; mol^{-1}]

(b)  How can you determine limiting molar conductivity, \left (\wedge ^{\circ} _{m}\right ) for strong         electrolyte and weak electrolyte ?

 

 

 

 
 
 
 
 

Answers (1)

(a) Zn(s)+Ag_{2}O(s)+H_{2}O(l)\rightarrow Zn^{2+}(aq)+2Ag(s)+2OH^{-}(aq)

E^{\circ}cell=E^{\circ}cathode-E^{\circ}anode

                =E^{\circ}_{Ag^{+}/Ag}-E^{\circ}_{Zn^{2+}/Zn}

                =0.80V-(-0.76V)=+1.56V

\Delta _{r}G^{\circ}=-nFE^{\circ}cell

                =-2\times 96500\times 1.56=-301080Jmol^{-1}

                =-301.08\; kJmol^{-1}

(b) The limiting molar conductivity \left (\wedge ^{\circ} _{m}\right ) for strong and weak electrolytes can be determined by Kohlrauschs law which states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contribution of the cation and anion of the electrolyte.

 

Posted by

Sumit Saini

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