[" 5.Calculate equilibrium constant for the following "],[" reaction "],[qquad [Fe(s)+Cd^(2+)(aq)longrightarrow Fe^(2+)(aq)+Cd(s)],[E_(Cd^(2*)/Cd)=-0.40V;E_(Fe^(2+)/Fe)=-0.44V]]

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 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 for the following reaction. Fe(s)+2Ag^+(aq) rarr 2Ag(s)+Fe^(2+)(aq) E_(Fe^(2+)//Fe)^@=-0.44V,E_(Ag^+//Ag)^=+0.80V .

(a) Express the relationship amongst cell constant , resistance of the solution in the cell and conductivity of the solution . How is molar conductivity of a solute related to conductivity of its solution ? (b) Calculate the equilibrium constant for the reaction Fe_((s)) + Cd_((aq))^(2+) hArr Fe_((aq))^(2+) + Cd (s) (Given : E_(Cd^(2+)|Cd)^(@) = 0.40 V , E_(Fe^(2+) |Fe)^(@) = -0.44 V ).

Calculate the equilibrium constant for the reaction: Cd^(2+) (aq) +Zn(s) hArr Zn^(2+) (aq) +Cd (s) If E_(Cd^(2+)//Cd)^(Theta) = -0.403V E_(Zn^(2+)//Zn)^(Theta) = -0.763 V

(a) Corrosion is essentially an electrochemical phenomenon. Explain the reactions occurring during corrosion of iron kept in an open atmosphere. (b) Calculate the equilibrium constant for the equilibrium reaction Fe_((s)) + Cd_((aq))^(2+) hArr Fe_((aq))^(2+) + Cd_((s)) (Given : E_(Cd^(2+)|Cd)^(@) = -0.40 V , E_(Fe^(2+)|Fe)^(@) = -0.44V ).

Corrosion is essentially an electrochemical phenomenon. Explain the reactions occurring during corrosion of iron kept in an open atmosphere. (b) Calculate the equilibrium constant for the equilibrium reaction Fe_((s)) + Cd_((aq))^(2+) hArr Fe_((aq))^(2+) + Cd_((s)) (Given : E_(Cd^(2+)|Cd)^(@) = -0.40 V , E_(Fe^(2+)|Fe)^(@) = -0.44V ).

Calculate the equilibrium constant for the reaction: Cd^(2+)+Zn(s)rarrZn^(2+)(aq)+Cd(s) If E_(Cd^(2+)|Cd)=-0.403V and E_(Zn^(2+)|Zn)=-.763V

Calculate the equilibrium constant for the reaction Fe(s)+Cd2+(aq)⇔Fe2+(aq)+Cd(s) (Given : E∘Cd2+∣Cd=0.40V,E∘Fe2+∣Fe=−0.44V).