The overall formaion constant for the reaction of 6 mole of `CN^(-)` with cobalt (II) is `1xx10^(19)` the standard reduction potential constant of `[C

Question

The overall formaion constant for the reaction of 6 mole of `CN^(-)` with cobalt (II) is `1xx10^(19)` the standard reduction potential constant of `[C

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`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

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)`