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.
Answers
Answer: The cell potential of the cell is +0.118 V
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode):
Reduction half reaction (cathode):
In this case, the cathode and anode both are same. So, will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
=
= 1.0 M
Putting values in above equation, we get:
Hence, the cell potential of the cell is +0.118 V
The cell potential for the cell as calculated is 0.118 V.
What is the Nernst equation?
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: brainly.com/question/721749
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