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The overall formaion constant for the reaction of 6 mole of `CN^(-)` with cobalt (II) is `1xx10^(19)` the standard reduction potential constant of `[Co(CN)_(6)]^(3-)+e^(-)toCo(CN)_(6)^(4-)` is `-0.83V` Calcualte the formation constant of `[Co(CN)_(6)]^(3-)`. Given `Co^(3+)+e^(-)toCo^(2+),E^(@)=1.82V` |
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Answer» Correct Answer - `K_(f)=10^(63.915)` Anode: `[Co(CN)_(6)]^(3-)+e^(-)to[Co(CN)_(6)]^(4-)" "E_(SRP)^(0)=-0.83V` cathode: `Co^(3+)+e^(-)toCO^(2+)" "E_(SRP)^(0)=1.82V` so overall cell reaction is `Co^(3+)+[Co(CN)_(6)]^(4-)toCO^(2+)+[Co(CN)_(6)]^(3-)` `E_(cell)^(0)=E_(c)^(0)-E_(a)^(0)=1.82-(-0.83)=2.65V` `E_(cell)=E_(cell)^(0)-(0.059)/(1)log(([Co^(2+)][Co(CN)_(6)]^(3-))/([Co^(3+)][Co(CN)_(6)]^(4-)))` Now, `Co^(2+)+6CN^(-)hArr[Co(CN)_(6)]^(4-)" "K_(f_(1))=1xx10^(19)` `Co^(3+)+6CN^(-)hArr[Co(CN)_(6)]^(3-)" "K_(f_(2))` At equilibrium `E_(cell)=0` `E_(cell)^(0)=(0.059)/(1)log((k_(f_(2)))/(K_(f_(1)))`, solving we get `K_(1)=10^(63.915)` |
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