`6.537 xx 10^(-2)f ` of metallic `Zn` was added to `100 mL` of saturated solution of `AgCl` . Calculate `log .([Zn^(2+)])/([Ag^(o+)])`.
Given `: E^(c-)._(Ag^(o+)|Ag)=0.80V, E^(c-)._(Zn^(2+)|Zn)=-0.763V`.
`K_(sp)` of `AgCl~~10^(-10),` atomic weight of `Zn=65.37`

Options

A. `26.5`
B. `13.24`
C. 53
D. 106

Answer

c

Explanation

At anode `:Zn rarr Zn^(2+)+2e^(-)" "E^(c-)._(red)=-0.763V`
At cathode `:2Ag^(o+)+2e^(-)rarr 2Ag" "E^(c-)._(red)=0.8V`
Cell reaction `ulbar( :2Ag^(o+)=Zn rarr 2Ag+Zn^(2+))`
`E^(c-)._(cell)=(E^(c-)._(c)-E^(c-)._(a))_(red)=0.8-(-0.763)=1.563V`
Given `K_(sp)` of `AgCl=[Ag^(o+)][Cl^(c-)]`
`10^(-10)=[Ag^(o+)][Cl^(c-)]`
Solubility `(S)` of `AgCl=sqrt(10^(-10))=10^(-5)M`
`:. [Ag^(o+)]=10^(-5)M`
`M of Zn =(W_(2)xx1000)/(Mw_(2)xx"Volume of solution in mL" )`
`=(6.537xx10^(-2)xx1000)/(65.37xx10mL)=0.01M`
`{:( :. ,2Ag^(o+),+,Zn,rarr,2Ag,+,Zn^(2+)),(Initial conc.,10^(-5)M,0.01M, ,- ,,-):}`
`(` Almost whole of `Ag^(o+)` is consumed. `)`
At equilibrium,
`log .([Zn^(2+)])/([Ag^(o+)])=(E^(c-)._(cell)xxn_(cell))/(0.059)=(1.563Vxx2)/(0.059)~~53`
`:. K_(eq)=([Zn^(2+)])/([Ag^(o+)]^(2))`