Calculate the equilibrium constant for the reaction : `Fe^(2+)+Ce^(4+)hArr Fe^(3+)+Ce^(3+)` Given, `E_(Ca^(4+)//Ce^(3+))^(@)=1.44V` and `E_(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+)hArr Fe^(3+)+Ce^(3+)` Given, `E_(Ca^(4+)//Ce^(3+))^(@)=1.44V` and `E_(Fe^(3+)//Fe^(2+))^(@)=0.68V`
Answer» We know, `E_("cell")^(@)=(0.059)/(1)log_(10)K_(C )` `E_("cell")^(@)=E_(OP_(Fr^(2+)//Fe^(3+)))^(@)+E_(RP_(Ce^(4+)//Ce^(3+)))^(@)` `=-0.68+1.44=0.76V` `therefore log_(10)K_(C )=(0.76)/(0.059)=12.8814` `therefore K_(C )=7.6xx10^(12)`

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Calculate the equilibrium constant for the reaction : `Fe^(2+)+Ce^(4+)hArr Fe^(3+)+Ce^(3+)` Given, `E_(Ca^(4+)//Ce^(3+))^(@)=1.44V` and `E_(Fe^(3+)//Fe^(2+))^(@)=0.68V`

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