Question

calculate E cell at 298 K (25 c) for Zn (s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu (s) when [Zn 2+]=.0010 M and [Cu 2+]=.10 M

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^o_anode

                                                      = 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⁺] = 0.10 M

[Zn²⁺] = 0.0010 M

Plug the values we get

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

   = +1.103 - (0.059 / 2 ) x log (0.0010 / 0.10 )

   = +1.162 V