(a) Calculate e.m.f. of the following cell :

                Zn(s)/Zn^{2+}(0.1\; M)\left | \right |(0.01\; M)\; Ag^{+}/Ag(s)

Given : E^{\circ}_{Zn^{2+}/Zn}=-0.76\; V.E^{\circ}_{Ag^{+}/Ag}=+0.80\; V

        \left [ Given: log\; 10=1 \right ]

(b) X and Y are two electrolytes. On dilution molar conductivity of 'X' increases 2.5 times while that Y increases 25 times. Which of the two is a weak electrolyte and why ?

 

 
 
 
 
 

Answers (1)

(a)

            Zn(s)/Zn^{2+}(0.1\; M)\left | \right |(0.01\; M)\; Ag^{+}/Ag(s)

Given : E^{\circ}_{Zn^{2+}/Zn}=-0.76\; V.E^{\circ}_{Ag^{+}/Ag}=+0.80\; V

        \left [ Given: log\; 10=1 \right ]

E0cell = E0C - E0A

= 0.80 – (-0.76) = 1.56 V

Ecell = E0cell – 0.0591/n log [Zn+2]/[Ag+]2

        = 1.47 V

(b)  Y is weak electrolyte , as molar conductivity increases with dilution due to increase in degree of dissociation.

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