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.