a. For the reaction
Given : `Ag^(o+)(aq)+Cl^(o+)(aq)rarr AgCl(s)`

Write the cell representation of above reaction and calculat `E_("cell")^(@)` at 298K.
Also find the solubility product of AgCl.
b. If `6.539xx10^(-2)g` of metallic zinc is added to 100ml saturated solution of AgCl. Find the value of
`log_(10).([Zn^(2+)])/([Ag^(+)]^(2))`
How many moles of Ag will be preciptated in the above reaction. Given that
`Ag^(+)+e^(-)rarrAg, E^(@)=0.80V`
`Zn^(2+)+2e^(-)rarr Zn,E^(@)=-0.76V`

Explanation

(a) 0.59V, `10^(-10)`
Cell reactions are `Ag+1/2 Cl_(2) to AgCl .......(i)`
`Ag to Ag^(+)+e .......(ii)`
`1/2 Cl_(2)+e to Cl^(-) .....(iii)`
`=Ag^(+)+Cl ^(-) to AgCl ..........(i-ii-iii)`
hence cell representation is `Ag|Ag^(+)|AgCl|Cl^(-)|Cl_(2),Pt`
`Ag^(+)(aq)+Cl^(-) (aq) to AgCl(s), sum triangleG_(p)^(@) -sum triangleG_(R)^(@)`
(i) `triangleG^(@)=-109-(-129+77)`
=109+129-77
`=20-77 =-57=-1 xx F xx E^(@)`
`-57=-1 xx 96500 xx E^(@)`
`rArr E^(@)=(57000)/(96500) =0.59"Volts"`
(ii) `-57=-2.303Rt log K_(0)`
`log K_(0)=(57 xx 1000)/(2.303 xx 8.314 xx 298)`
`log K_(0)=9.98 approx 10`
`K_(0)=10^(10)`
`K_(sp)=1//K_(0)`
`K_(sp)=10^(-10)`
`log K_(sp)=-10`
(iii) `AgCl(s) Leftrightarrow underset(S)(Ag^(+)(aq))+underset(S)(Cl^(-) (aq))`
`10^(-10)=S^(2)`
`S=10^(-5)m//L`
(b) `52.9, 10^(-6)`
When `(65.39 xx 10^(-2))/(65.39)=10^(-3)` moles of Zn has been added,
`2Ag^(+)+2e to 2Ag," "E^(0)=0.80V`
`Zn to Zn^(2+)+2e," "E^(0)=0.77V`
`=2Ag_((aq))^(+) +Zn_((s)) to Zn_((aq))^(2+)+2Ag_((s)), E^(0)=1.57V`
`10^(-6)" mole "10^(-3)" moles"`
`log_(10)K_((eq))=52.8`
Therefore, this reaction will move in formward direction completely. Hence moles of Ag formed will be `10^(-6)`
At equilibrium `(E_("cell")=0)`
`E_("cell")^(0)=(+0.0591)/(2) log_(10) ([Zn^(2+)])/([Ag^(+)]^(2))`
`=(1.56 xx 2)/(0.0591)=log ([Zn^(+2)])/([Au^(+2)]^(2)=52.8`