Answer:
The correct answer is 0.186 V
Explanation:
The two hemirreactions are:
Reduction: Fe²⁺ + 2 e- → Fe(s)
Oxidation : Co(s) → Co²⁺ + 2 e-
Thus, we calculate the standard cell potential (Eº) from the difference between the reduction potentials of cobalt and iron, respectively, as follows:
Eº = Eº(Fe²⁺/Fe(s)) - Eº(Co²⁺/Co(s)) = -0.28 V - (-0.447 V) = 0.167 V
Then, we use the Nernst equation to calculate the cell potential (E) at 298.15 K:
E= Eº - (0.0592 V/n) x log Q
Where:
n: number of electrons that are transferred in the reaction. In this case, n= 2.
Q: ratio between the concentrations of products over reactants, calculated as follows:
[tex]Q = \frac{ [Co^{2+} ]}{[Fe^{2+} ]} = \frac{0.33 M}{1.47 M} = 0.2244[/tex]
Finally, we introduce Eº= 0.167 V, n= 2, Q=0.2244, to obtain E:
E= 0.167 V - (0.0592 V/2) x log (0.2244) = 0.186 V
