Question

For the reaction

2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g).  E∘=0.71 V

what is the cell potential at 25 ∘C if the concentrations are [Co3+]= 0.634 M , [Co2+]= 0.385 M , and [Cl−]= 0.491 M and the pressure of Cl2 is PCl2= 6.10 atm ?

Answer 1

Calculation of concentration of Cl₂ gas :

We know that PV = nRT

                       P = (n/V) RT

                       P = CRT               C - concentration

                        C = P / (RT)

Where

P = pressure of Cl2 gas = 6.10 atm

R = gas constant = 0.0821 L atm / (mol-K)

T = temperature =25 ^oC = 25+273 = 298 K

Plug the values we get C = 6.10 / (0.0821x298)

                                      = 0.250 M

2Co³⁺(aq)+2Cl⁻(aq) → 2Co²⁺(aq)+Cl₂(g)

Reaction quotient , Q = ([Co²⁺]²x[Cl₂]) / ([Co³⁺]² x[Cl^-]² )

                               = (0.385 ² x 0.250)/(0.634 ² x 0.491)

                               = 0.187

We know that E_cell = E^o -(0.059/n) log Q

Where

E^o = standard reduction potential = 0.71 V

E_cell = reduction potential of the cell = ?

n = number of electrons transferred = 1

Plug the values we get

E_cell = 0.71 -(0.059/1) log 0.187

       = 0.75 V

Therefore the cell potential is 0.75 V