Respuesta :
Answer: The concentration of other (diluted) [tex]Fe^{2+}[/tex] solution is 0.0026 M
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): [tex]Fe(s)\rightarrow Fe^{2+}+2e^-[/tex]
Reduction half reaction (cathode): [tex]Fe^{2+}+2e^-\rightarrow Fe(s)[/tex]
In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}_{(diluted)}]}{[Fe^{2+}_{(concentrated)}]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = 0.047 V
[tex][Fe^{2+}_{(diluted)}][/tex] = ? M
[tex][Fe^{2+}_{(concentrated)}][/tex] = 0.10 M
Putting values in above equation, we get:
[tex]0.047=0-\frac{0.0592}{2}\log \frac{[Fe^{2+}_{(diluted)}]}{0.10M}[/tex]
[tex][Fe^{2+}_{(diluted)}]=0.0026M[/tex]
Hence, the concentration of other (diluted) [tex]Fe^{2+}[/tex] solution is 0.0026 M
