Question

What is the cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 83 ∘C when [Fe2+]= 3.30 M and [Mg2+]= 0.310 M .

Answer 1

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

So standard potential of the cell , E^o = E^o_cathode - E^o_anode

                                                      = E^o_Fe2+/Fe - E^o_Mg2+/Mg

= -0.44 V - (-2.363V)

= +1.923 V

According to Nernst Equation ,

E = E^o - (0.059 / n) log ([Products] / [reactants] )

   = E^o - (0.059 / n) log ([Mg²⁺] / [Fe²⁺] )

Where

E = electrode potential of the cell = ?

E^o = standard electrode potential = +1.923 V

n = number of electrons involved in the reaction = 2

[Mg²⁺] = 0.310 M

[Fe²⁺] = 3.30 M

Plug the values we get

E = +1.923 - (0.059 / 2 ) x log ( 0.310 / 3.30 )

   = +1.953 V