Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.887 M and [Sn2 ] = 0.0150 M. Mg(s)+Sn2+(aq)------->Mg2+ (aq)+Sn(s)

Answer 1

The cell reaction is Mg + Sn²⁺ -----> Mg²⁺+ Sn

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

                                                      = E^o_Sn2+/Sn - E^o_Mg2+/Mg

= -0.136 - (-2.363)

= +2.227 V

According to Nernst Equation ,

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

   = E^o - (0.059 / n) log ([Mg²⁺] / [Sn²⁺] )

Where

E = electrode potential of the cell = ?

E^o = standard electrode potential = +2.227 V

n = number of electrons involved in the reaction = 2

[Mg²⁺] = 0.887 M

[Sn²⁺] = 0.0150 M

Plug the values we get

E = 2.227- (0.059 / 2) xlog (0.887 / 0.0150)

= +2.175 V