Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.799 M and [Sn2 ] = 0.0160 M. Standard reduction potentials can be found here.

Mg(s)+ Sn^2+ ---> Mg^2+ +Sn(s)

E= V units

Answer 1

Mg(s)+ Sn²⁺ Mg²⁺ +Sn(s)

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

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

                                                 = -0.14 - (-2.37) V

                                                 = +2.23 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 = +2.23 V

n = number of electrons involved in the reaction = 2

[Mg²⁺] = 0.799 M

[Sn²⁺] = 0.0160 M

Plug the values we get

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

   = +2.23 - (0.059 / 2 ) x log ( 0.799/0.0160 )

   = +2.18 V