Question

Consider the following liquid junction cell at room temperature:

Zn(s) / ZnSO4(aq) // CuSO4(aq) / Cu(s)

The left side of the cell contains an electrolyte solution at 9.26*10^-2 M and the right side of the cell contains an electrolyte solution at 7.32*10^-2 M.

Calculate the observed cell potential using the Nernst Equation, the Debye-Huckel Limiting Law, and ionic activities.

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)

= +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⁺] = 7.32x10^-2 M

[Zn²⁺] = 9.26x10^-2 M

Plug the values we get

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

   = +1.103 - (0.059 / 2 ) x log ( ( 9.26x10^-2 ) / ( 7.32x10^-2 ) )

   = +1.09 V