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Knowing that `K_(sp)` for `AgCl` is `1.0 xx 10^(-10)`, calculate `E` for a silver `//` silver chloride electrode immersed in `1.00 M KCl ` at `25^(@)C.E^(c-)._(Ag^(o+)|Ag)=0.799V`. |
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Answer» `K_(sp)(AgCl)=[Ag^(o+)][Cl^(c-)],KCl=1.00M` `:. [Cl^(c-)]=1.00M` `:. [Ag^(o+)]=(K_(sp))/([Cl^(c-)])=(1.0xx10^(-10))/(1.00)=1.0=10^(-10)` `Ag^(o+)+e^(-)rarr Ag" "E^(c-)._(Ag^(o+)|Ag)=0.799V` `:. E=E^(c-)-(0.059)/(1)log.(1)/([Ag^(o+)])` `=0.799-(0.059)/(1)log.(1)/((0.10xx10^(-10))` `=0.799=0.592=0.207V` |
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