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+]= 0.173 M , [Co2+]= 0.433 M , and [Cl−]= 0.306 M , and the pressure of Cl2 is PCl2= 9.00 atm ?

Respuesta :

Explanation:

The given data is as follows.

     [tex]E^{o}[/tex] = 0.483,     [tex][Co^{3+}][/tex] = 0.173 M,

     [tex][Co^{2+}][/tex] = 0.433 M,     [tex][Cl^{-}][/tex] = 0.306 M,

     [tex]P_{Cl_{2}}[/tex] = 9.0 atm

According to the ideal gas equation, PV = nRT

or,             P = [tex]\frac{n}{V}RT[/tex]    

Also, we know that

                Density = [tex]\frac{mass}{volume}[/tex]

So,         P = MRT

and,          M = [tex]\frac{P}{RT}[/tex]

                    = [tex]\frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}[/tex]

                    = [tex]\frac{9.0}{24.436}[/tex]

                    = 0.368 mol/L

Now, we will calculate the cell potential as follows.

          E = [tex]E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}[/tex]

             = [tex]0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}[/tex]

             = [tex]0.483 - 0.02955 log \frac{0.0689}{0.0162}[/tex]

             = [tex]0.483 - 0.02955 \times 0.628[/tex]

             =  0.483 - 0.0185

             = 0.4645 V

Thus, we can conclude that the cell potential of given cell at [tex]25^{o}C[/tex] is 0.4645 V.