Q13 of 47 Page 1

a. Following reaction takes place in the cell: Zn(s)+Ag2O(s)+H2O(l)⟶Zn2+(aq)+2Ag(s)+2OH−(aq) Calculate ΔrG0 of the reaction.
[Given: E0(Zn2+/Zn)=−0⋅76VE0(Ag+/Ag)=0⋅80V,1F=96,500Cmol−1J]


b. How can you determine limiting molar conductivity, (∧0m) for strong electrolyte and weak electrolyte?


a. E0(Zn2+/Zn) =−0⋅76V


E0(Ag+/Ag) =0⋅80V


1F=96,500Cmol−1J


n= 2


E0cell = E0(Ag+ / Ag) – E0(Zn2+ / Zn)
= 0.80- (- 0.76)
= 1.56V
ΔG0 =−nFE0
=-2 × 96500 × 1.56
= -301080 joules / mol
= -301.080 K/ mol


b. ∧0m for strong electrolyte is obtained as intercept from plot of ∧m verses c whereas ∧0m for weak electrolyte is obtained from Kohlrausch's law :


Λ0m=λ0++λ0-


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