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Find `K_(c)` for the complex`:` `[Ag(NH_(3))_(2)]^(o+)hArrAg^(o+)+2NH_(3)` `E^(c-)._((Ag^(o+)//Ag))=0.8V` and `E^(c-)._{[Ag(NH_(3))_(2)]^(o+)|Ag|NH_(3))=0.37V` |
Answer» Anode reaction `:` `cancel(Ag) hArrAg^(o+)+cancel(e^(_))` Cathode reacion `:[Ag(NH_(3))_(2)]^(o+)hArrcancel(Ag)+2NH_(3)` Cell reaction`ulbar( :[Ag(NH_(3))_(2)]^(o+)hArrAg^(o+)+2NH_(3))` `E^(c-)._(cell)=(E^(c-)._(reduction))_(c)=(E^(c-)._(reduction))_(a)` `=0.37V-0.8V=-0.43V` `:. log K_(c)=(n_(cell)xxE^(c-)._(cell))/(0.0059)` `=(1xx(-.43V))/(0.59)=-7.22` `K_(c)=Anti l og(-7.22)` `=Antil og(-7-022+1-1)` `=Antil og(bar(8).78)=6xx10^(-8)` |
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