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`

Explanation

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)`