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

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

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+)hArr Fe^(3+)+Ce^(3+) Given, E_(Ca^(4+)//Ce^(3+))^(@)=1.44V and E_(Fe^(3+)//Fe^(2+))^(@)=0.68V

Determine the equilibrium constant for the reaction : Fe^(2+)(aq)+Ce^(4+)(aq)rarr Fe^(3+)(aq)+Ce^(3+)(aq) Givne : E_(Ce^(4+)|Ce^(3+))^(@)=1.44V & E_(Fe^(3+)|Fe^(2+))^(@)=0.77V .

Calculate the equilibrium constant for the reaction : Fe(s)Cd^(2+)(aq)harrFe^(2+)(aq)+Cd(s) ("Given" E_(cd^(2+)//Cd)^(@)=-0.40 V, E_(Fe^(2+)//Fe)^(@)=-0.44 V)

Calculate the equilibrium constant, K_(C) for the reaction 3Sn^(4+)+2Cr to 3Sn^(2+)+2Ce^(3+) , at 298K given E_(cell)^(@)=0.83V