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The standard potential of the following cell is 0.23 V at 15^@C and 0.21V at 35^@C (Pt) H_2(g) |HCl(aq)|AgCl(s)|Ag(s) Calculate the solubility of AgCl in water at 25^@C Given: E_(Ag^+,Ag)^@=0.80V at 25^@C |
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Answer» Solution :In the given cell `E_(cell)^@=E_(AgCl,Cl^-)^@-E_(H^+,H_2)^@` `0.22=E_(AgCl,Cl^-)-0`…..(AVERAGE VALUE of `E_(cell)^@` is considered ) or `E_(AgCl,Cl^-)=0.22` Let us now consider the following cell to calculate `K_(sp) (AgCl)` `Ag=Ag^+ +e, E^@=-0.8` (given) `AgCl+e=Ag+Cl^(-),E^@=0.22` (calculated) `therefore` for the cell REACTION `AgCl=Ag^+ +Cl^-` `E_(cell)=(E_(AgCl,Cl^-)^@-E_(Ag^+,Ag)^@)-0.0591/1 log[Ag^+][Cl^-]` At equilibrium `E_(cell)=0 and R.Q=K_(sp)` `therefore E_(AgCl,Cl^-)^@-E_(Ag^+,Ag)^@=0.0591 log K_(sp)` `0.22-0.8=0.0591 log K_(sp)` `therefore K_(sp)=1.535 times 10^-10` `therefore` solubility of `AgCl=sqrt(K_(sp))=sqrt(1.535 times 10^-10)` `=1.239 times 10^-5 mol e//L` |
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