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_((Ce^(4+)|Ce^(3+)))^(@)=1.44V, e_((Fe^(3+)|Fe^(3+)))^(@)=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) hArr Fe^(2+)(aq) + Cd(s) [ Given, E_(Cd^(2+) //Cd)^(@) = - 0.40V , E_(Fe^(2+)//Fe)^(@) = - 0.44 V ]

Calculate at 25^(@) C , the equilibrium constant for the reaction : 2 Fe^(3+) + Sn^(2+) hArr 2 Fe^(2+) + Sn^(4+) Given that E_((Fe^(3+)|Fe^(2+)))^(Theta) = 0.771 V , E_((Sn^(4+) |Sn^(2+)))^(Theta) = 0.150 V

If E_(Fe^(2+)//Fe)^(@)=-0.440 V and E_(Fe^(3+)//Fe^(2+))^(@)=0.770 V , then E_(Fe^(3+)//Fe)^(@) is -