Question

In: Chemistry

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2...

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

Solutions

Expert Solution

Mg(s)+ Sn2+ Mg2+ +Sn(s)

standard potential of the cell , Eo = Eocathode - Eoanode

                                                 = EoSn2+/Sn - EoMg2+/Mg

                                                 = -0.14 - (-2.37) V

                                                 = +2.23 V

                                 

According to Nernst Equation ,

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

   = Eo - (0.059 / n) log ([Mg2+] / [Sn2+] )

Where

E = electrode potential of the cell = ?

Eo = 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 = Eo - (0.059 / n) xlog ([Mg2+] / [Sn2+] )

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

   = +2.18 V


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