Question

A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-cells. The reduction potential of Ni2+ is �0.23 V. Calculate the potential of the cell at 25�C if the more dilute Ni2+ solution is in the anode compartment.

Answer 1

Answer: The cell potential of the cell is +0.118 V

Explanation:

The half reactions for the cell is:

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

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

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:

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

where,

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

$E_{cell}$ = ?

$[Ni^{2+}_{diluted}]$ = $1.00\times 10^{-4}M$

$[Ni^{2+}_{concentrated}]$ = 1.0 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}$

$E_{cell}=0.118V$

Hence, the cell potential of the cell is +0.118 V