Calculate the equilibrium constant for the reaction `:` `Fe^(2+)Ce^(4+)hArrFe^(3+)+Ce^(3+)` Given `:E^(c-)._((Ce^(4+)|Ce^(3+)))=1.44V` `E^(c-)._((Fe^(3+)|Fe^(2+)))=0.68V`

Calculate the equilibrium constant for the reaction `:` `Fe^(2+)Ce^(4+

1.	Calculate the equilibrium constant for the reaction `:` `Fe^(2+)Ce^(4+)hArrFe^(3+)+Ce^(3+)` Given `:E^(c-)._((Ce^(4+)\|Ce^(3+)))=1.44V` `E^(c-)._((Fe^(3+)\|Fe^(2+)))=0.68V`
Answer» We Know, `E_(cell)^(@) = (0.059)/(1)log_(10)K_(c)` `E_(cell)^(@) = E_(OP_(Fe^(2+)//Fe^(3+)))^(@)+E_(RP_(Ce^(4+)//Ce^(3+)))^(@)` `= -0.68+1.44 = 0.76V` `:. Log_(10)K_(c) = (0.76)/(0.059) = 12.8814` `:. K_(c) = 7.6 xx 10^(12)`

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Calculate the equilibrium constant for the reaction `:` `Fe^(2+)Ce^(4+)hArrFe^(3+)+Ce^(3+)` Given `:E^(c-)._((Ce^(4+)|Ce^(3+)))=1.44V` `E^(c-)._((Fe^(3+)|Fe^(2+)))=0.68V`

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