1.

Consider two half cells based on the reaction Ag_((aq))^(+)etoAg_((s)).t he left half cell contain Ag^(+) ions at concentration of Ag^(+) ions, but just enough NaCl_((aq)) has been added to completely precipitate the Ag_((aq))^(+) as AgCl. If the emf of the cell is 0.29V, then log_(10)K_(sp) would have been

Answer»

9.804
`-9.804`
`-4.902`
`10.004`

Solution :The electrode potential is:
`E_(Cl^(-)//AGCL//AG)=E_(Cl^(-)//AgCl//Ag)^(o)+(0.0591)/(2)"LOG"(1)/([Cl^(-)])` . . .(i)
`E_(Cl^(-)//AgCl//Ag)=E_(Ag^(+)//Ag)^(o)+(0.0591)/(1)" log "K_(sp//(AgCl))`
log `K_(sp(AgCl))=-(0.29)/(0.0591)=-0.907`


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