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Consider the cell Ag | AgBr | KBr || KCl | AgCl | Ag with EMF 0.059 V . Assume that [Br^(-)] = [Cl^(-)] . Here , conclusion inferred may be

Answer»

The ratio of the simultaneous solubilities of AgCl and AgBr in pure water is 1000
It is concentration cell
CHANGE in concentration of KCl will not EFFECT EMF
`K_(sp) (AgCl) gt K_(sp) (AgBr)`

SOLUTION :`E_(RP) = E_(SRP)^(0) -(0.0591)/(1) LOG""((Ag^(Ɵ))_a)/((Ag^(oplus))_e) ,0.059 = 0-(0.0591)/(1) log ""(K_(sp) AgBr[Br^(-)])/(K_(sp)AgCl[Cl^(-)]) = 10^(-1)`
` (K_(sp)AgBr)/([Br^(Ɵ)]) xx 10 = (K_(sp) AgCl)/([Cl^(Ɵ)]), K_(SP) [Br^(Ɵ)] = [Cl^(Ɵ)] ` Then ` K_(sp) AgCl to K_(sp) AgBr`


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