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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`

Text Solution

Verified by Experts

`E_(cell)^(@)=0.059/1 log K_(c)`
`E_(cell)^(@)=E_(Fe^(2+)//Fe^(3+))^(@)+E_(Ce^(4+)//Ce^(3+))^(@)`
`=-0.68+1.44=0.76 V`
`log_(10) K_(c)=0.76/0.059=12.8814`
`K_(c)=7.6cc10^(12)`
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