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

`7.236xx10^(12)`

Explanation

For the cell reaction `:`

Half cell reaction are `:`
`a. Fe^(2+)(aq) rarr Fe^(3+)(aq) +e^(-)" "E^(c-)=-0.68V`
`b. Ce^(4+)(aq)+e^(-) rarr Ce^(3+)(aq) " "E^(c-)=+1.44V`
`overlineunderline(Fe^(2+)(aq)+Ce^(4+)hArrFe^(3+(aq)+Ce^(3+)E_("Cell")^(Θ)=+0.76V)`
At equilibrium,
`E^(c-)._(cell)=(0.0591)/(n)log K_(eq)(` because at equilibrium, `E_(cell)=0)`.
`+0.76=(0.0591)/(1)log K_(eq)`
`logK_(eq)=(0.76)/(0.0591)=12.895impliesK_(eq)=7.236xx10^(12)`