Can I get help on this please

Question

Calculate the cell potential for the galvanic cell in which the reaction

Cd(s)+Co2+(aq)↽−−⇀ Cd2+(aq)+Co(s)

occurs at 25 °C, given that [Cd2 + ] =  0.00140M and [Co2 + ] = 0.801

Answer 1

Since this cell is NOT operating under standard conditions, we will have to use the Nernst equation. But before using it, we will calculate the standard cell potential (Eºcell).

Cd²⁺ + 2e- ==> Cd ... Eº = -0.40 V (anode)

Co²⁺ +2e- ==> Co ... Eº = -0.28 V (cathode)

Co²⁺ + Cd ==> Co + Cd²⁺ cell reaction

Eº_cell = -0.28 + 0.40 = 0.12 V

Nernst equation @ 25ºC:

E_cell = Eº_cell - 0.0592/n ln Q

n = 2 mols electrons

Q = [Cd²⁺] / [Co²⁺] = 0.00140 / 0.801 = 0.00175

E_cell = - 0.0592 / 2 ln 0.00175

E_cell = - 0.0296 x -6.35

E_cell = 0.188 V