The overall formation constant for the reaction of 6 mole of `CN^(-)` with cobalt (II) is `1 xx 10^(19)`. The standard reduction potential for the reaction `[Co(CN)_(6)]^(3-) +e^(-) rarr Co(CN)_(6)^(4-)` is `-0.83V`. Calculate the formation constant of `[Co(CN)_(6)]^(3-)`
Given `Co^(3+) +e^(-) rarr Co^(2+) , E^(@) = 1.82V`

Answer

`K_(f) = 8.28 xx 10^(63)`

Explanation

Use `Co(CN)_(6)^(4-) rarr Co(CN)_(6)^(3-) +e, E_(OP)^(@) = + 0.83V`
`{:(Co^(3+)+e rarr Co^(2+),,,E_(RP)^(@) = 1.82V):}/(Co(CN)_(6)^(4-)+Co^(3+)hArrCo^(2+)+Co(CN)_(6)^(3-))`
and `E_(cell) = E_(cell)^(@) +(0.059)/(1) log_(10). ([Co^(3+)][Co(CN)_(6)^(4-)])/([Co^(2+)][Co(CN)_(6)^(3-)])`
or `E_(cell) = E_(cell)^(@) +(0.059)/(1)log_(10).([Co^(3+)][Co(CN)_(6)^(4-)][CN^(-)]^(6))/([Co^(2+)][Co(CN)_(6)^(3-)][CN^(-)]^(6))`
Also `6 CN^(-) +Co^(2+) hArr Co(CN)_(6)^(4-)`
and `K_(f_(1)) = ([Co(CN)_(6)^(4-)])/([Co^(2+)][CN^(-)]^(6))`
and `6CN^(-) +Co^(3+) hArr Co(CN)_(6)^(3-)`
and `K_(f_(2)) =([Co(CN)_(6)^(3-)])/([Co^(3+)][CN^(-)]^(6))`
`:. E_(cell) = E_(cell)^(@) +(0.059)/(1) log_(10).(K_(f_(1)))/(K_(f_(2)))`
`0 = 0.83 +1.82 +(0.059)/(1) log_(10). (10^(19))/(K_(f_(2)))`
(At equilibrium `E_(cell) = 0)`
`:. (K_(f_(2)))/(10^(19)) = 8.23 xx 10^(44)`
`:. K_(f_(2)) = 8.23 xx 10^(63)`