In an experiment, `6.539xx10^(-2)g` of metallic Zn (M = 65.39) was added to 100 mL of saturated solution of `AgCl`. Calculate `log[Zn^(2+)]//[Ag^(+)]^(2)`, given that: `Ag^(+)+e rarr Ag," "E^(@)=0.80V,`
`Zn^(2+)+2e rarr Zn, E^(@)=-0.78`
Also calculate the number of moles of Ag formed.

Explanation

Half - cell reactions are -
`{:(" "Ag^(+)(aq)+e" "rarr" "Ag(s)),(" "Ag(s)+Cl^(-)(aq)" "rarr" "AgCl(s)+e),(bar("Cell reaction: "Ag^(+)(aq)+Cl^(-)(aq)rarrAgCl(s))):}`
Amount of zinc added `=(6.539xx10^(-2))/(65.39)=10^(-3)` mol
Therefore, the following reactions will occur :
`{:(" "2Ag^(+)(aq)+2e" "rarr" "2Ag(s),),( " "Zn(s)" "rarr" "Zn^(2+)(aq)+2e,),(bar(2Ag^(+)(aq)+Zn(s)" "rarr" "Zn^(2+)(aq)+2Ag(s),)):}`
`therefore" "E^(@)=(0.059)logK" or, "-0.59=(0.059)/(1)logK_(sp)`
`"or, logK_(sp)=-(0.59)/(0.059)" "therefore" "K_(sp)=10^(-10)`
Solubility of `AgCl` in `"0.1 M AgNO"_(3)=S(S+0.1)`
`therefore" "S(S+0.1)=10^(-10)" or, "S=10^(-9)M`
`therefore" Amount of AgCl dissolved in "10^(9)L" solution = 1 mol."`
`therefore" Amount of AgCl that dissolves in "10^(6)L" of the solution "`
`=(1)/(10^(9))xx10^(6)=10^(-3)"mol = 1.0 mmol"`