Answer: The cell potential of the cell is +0.118 V
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): [tex]Ni(s)\rightarrow Ni^{2+}+2e^-[/tex]
Reduction half reaction (cathode): [tex]Ni^{2+}+2e^-\rightarrow Ni(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{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = ?
[tex][Ni^{2+}_{diluted}][/tex] = [tex]1.00\times 10^{-4}M[/tex]
[tex][Ni^{2+}_{concentrated}][/tex] = 1.0 M
Putting values in above equation, we get:
[tex]E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}[/tex]
[tex]E_{cell}=0.118V[/tex]
Hence, the cell potential of the cell is +0.118 V
The cell potential for the cell as calculated is 0.118 V.
The Nernst equation can be used to obtain the cell potential of a cell under non- standard conditions. The standard cell potential in this case is zero owing to the fact that both cathode and anode are made of nickel.
Hence;
Ecell = E°cell - 0.0592/nlog Q
Ecell = 0 - 0.0592/2 log (1 00 * 10^-4/1)
Ecell = 0.118 V
The cell potential for the cell as calculated is 0.118 V.
Learn more about Nernst equation: https://brainly.com/question/721749
