Question

What is the potential of a cell made up of Zn|Zn²⁺ and Cu|Cu²⁺ half-cells at 25^o C if [Zn²⁺] = 0.25 M and [Cu²⁺] = 0.15 M?

(Answer should be E = +1.09 V but I need to know how to get to that)

Answer 1

The standard electrode potential for Zn system is Zn²⁺ + 2e^- Zn ; E^o red = -0.7628 V ---(1)

The standard electrode potential for Cu system is Cu²⁺ + 2e^- Cu ; E^o red = +0.340 V ---(2)

Since the reduction potential of Zn system is less it undergoes oxidation at anode & Cu undergoes reduction at cathode.

The cell reaction is Zn + Cu²⁺ Zn²⁺ + Cu

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

                                                      = E^o_Cu2+/Cu - E^o_Zn2+/Zn

                                                      = +0.340 - (-0.7628) V

                                                      = +1.103 V

According to Nernst Equation ,

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

   = E^o - (0.059 / n) log ([Zn²⁺] / [Cu²⁺] )

Where

E = electrode potential of the cell = ?

E^o = standard electrode potential = +1.103 V

n = number of electrons involved in the reaction = 2

[Cu²⁺] = 0.15 M

[Zn²⁺] = 0.25 M

Plug the values we get

E = E^o - (0.059 / n) xlog ([Zn²⁺] / [Cu²⁺] )

   = +1.103 - (0.059 / 2 ) x log ( 0.25 / 0.15 )

   = +1.09 V