6N.15 7th Ed

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1K Kevin
Posts: 47
Joined: Fri Sep 28, 2018 12:24 am

6N.15 7th Ed

Postby 1K Kevin » Sun Mar 03, 2019 7:30 pm

"Calculate the potential of a cell constructed with two nickel electrodes. The electrolyte in one compartment is 1.0 M Ni(NO3)2. In the other compartment, NaOH has been added to a Ni(NO3)2 solution until the pH = 11 at 298 K. Use Table 6I.1"
The problem is uses the Nernst Equation but Im confused on how to start it. What can we do with the pH in order to get all the info we need?

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

Re: 6N.15 7th Ed

Postby Matthew Tran 1H » Mon Mar 04, 2019 12:07 am

If the pH is 11, you know that the solution is basic. You would get that by adding OH- to solution via NaOH. However, in Table 6I.1, Ni(OH)2 has a Ksp of 6.5x10^-18, which means that Ni(OH)2 (s) is heavily favored vs. Ni2+ and OH- ions in solution. Therefore, in the context of this problem, when you add OH- to the solution, most of it will bond with the Ni2+ ions and precipitate to Ni(OH)2 (s). A small amount will stay ionized in solution, and that will contribute to the pH. Using the expression for the solubility constant, Ksp=[Ni2+][OH-], and the pH (calculate pOH then [OH-]), you can figure out [Ni2+], which is the unknown concentration of Ni(NO3)2. Then you can use the Nernst Equation.

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

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