Question

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [Co3+]= 0.228 M , [Co2+]= 0.732 M , and [Cl−]= 0.816 M and the pressure of Cl2 is PCl2= 6.70 atm ?

Answer 1

+3    -1 +2 0               

2Co³⁺(aq)+2Cl⁻(aq)→2Co²⁺(aq)+Cl₂(g) : E∘=0.483 V.

According to Nernst Equation ,

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

   = E^o - (0.059 / n) log (([Co²⁺]²xp_Cl2) /([Co³⁺]²[Cl^-]² ))

Where

E = electrode potential of the cell = ?

E^o = standard electrode potential = +0.483 V

n = number of electrons involved in the reaction = change in oxidation number = 1

[Co²⁺]= 0.732 M

p_Cl2= 6.70 atm

[Co³⁺] = 0.228M

[Cl^-] = 0.816 M

Plug the values we get

E = 0.483 - (0.059 / 1) xlog ((0.732 ² x 6.70) / ( 0.228² x 0.816² ) )

= +0.385 V

Therefore the cell poytential is +0.385 V

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

   = +1.09 V