Question

A concentration cell is constructed by placing identical iron electrodes in two Fe2 solutions. The potential of this cell is observed to be 0.047 V. If the more concentrated Fe2 solution is 0.10 M, what is the concentration of the other Fe2 solution

Answer 1

Answer: The concentration of other (diluted) $Fe^{2+}$ solution is 0.0026 M

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):  $Fe(s)\rightarrow Fe^{2+}+2e^-$

Reduction half reaction (cathode):  $Fe^{2+}+2e^-\rightarrow Fe(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}_{(diluted)}]}{[Fe^{2+}_{(concentrated)}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.047 V

$[Fe^{2+}_{(diluted)}]$ = ? M

$[Fe^{2+}_{(concentrated)}]$ = 0.10 M

Putting values in above equation, we get:

$0.047=0-\frac{0.0592}{2}\log \frac{[Fe^{2+}_{(diluted)}]}{0.10M}$

$[Fe^{2+}_{(diluted)}]=0.0026M$

Hence, the concentration of other (diluted) $Fe^{2+}$ solution is 0.0026 M