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Consider a concentration cell composed of a [tex]\text{Ni}^{2+}/\text{Ni}[/tex] compartment with a concentration of 0.0500 M and a [tex]\text{Ni}^{2+}/\text{Ni}[/tex] compartment with a concentration of 1.55 M. Determine the potential for this concentration cell at 298 K.

Answer :

Final answer:

To determine the potential of the concentration cell, use the Nernst equation. Substitute the given concentrations into the equation to calculate Q. Use the standard reduction potential for the half-reaction to determine E°cell. Finally, substitute all values into the Nernst equation to find the potential of the concentration cell at 298 K.

Explanation:

A concentration cell is constructed by connecting two half-cells with the same electrode and the same half-reaction, but differing concentrations of a redox species. The potential of a concentration cell is determined by the difference in concentration of the chosen redox species. To calculate the potential of the concentration cell, we can use the Nernst equation:

Ecell = E°cell - (0.0592V/n) log Q

In this case, the half-reaction is Ni²+(aq) + 2e- → Ni(s). The given concentrations are 0.0500 M and 1.55 M. The concentration ratio Q = [Ni²+(0.0500 M)]/[Ni²+(1.55 M)] = 0.0323. The standard potential E°cell is determined by the standard reduction potential of the half-reaction, which can be found in a table. Substituting these values into the Nernst equation will give you the potential for the concentration cell at 298 K.

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