Question

A galvanic cell is prepared using 0.150 M Co2+ and a cobalt electrode as one half-cell and 0.0350 M Ag+ and a silver electrode as the other half-cell.

1.) Write the reduction forms of the equation for the anodic half-reaction and the cathodic half-reaction

2.) Write the net ionic equation for the overall cell reaction. Ensure that your equation shows the spontaneous reaction direction.

3.) Calculate Ecell

Answer 1

The reduction potentials are given as follows:

Co²⁺ + 2e^- ---> Co   ; E = -0.29 V

Ag⁺ + e^- ---> Ag : E = +0.22 V

Since the reduction potential of Ag is more it acts as cathode

At anode ( oxidation ) :   Co   ----> Co²⁺ + 2e-   ; E = -0.29 V    -----(1)

At cathode( reduction ) : Ag⁺ + e^- ---> Ag : E = +0.22 V     ----(2)

(2) Net reaction is obtained by combining above equations so that the charge must be cancel each other.

This can be done as follows : Eqn(1) + [2xEqn(2)]    gives

Co + 2 Ag⁺ ---> Ag + Co²⁺

This is the balanced equation .

(3)

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

                                                      = E^o_Ag2+/Ag - E^o_Co2+/Co

                                                      = +0.22 - (-0.29) V

                                                      = +0.51 V

According to Nernst Equation ,

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

   = E^o - (0.059 / n) log ([Co²⁺] / [Ag⁺] )

Where

E = electrode potential of the cell = ?

E^o = standard electrode potential = +0.51 V

n = number of electrons involved in the reaction = 2

[Co²⁺] = 0.15 M

[Ag⁺] = 0.0350 M

Plug the values we get

E = 0.49 V