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Maya Gollamudi 1G
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Postby Maya Gollamudi 1G » Wed Mar 04, 2020 8:28 pm

How do you solve problem 6N.15:
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=11.0 at 298 K

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Re: 6N.15

Postby Chem_Mod » Thu Mar 05, 2020 12:24 pm

You know that the redox half reactions should relate to Ni2+ + 2e- --> Ni. When you add OH- to one of the compartments, Ni2+ + 2OH- --> Ni (OH)2. Because Ni2+ ions are being used in this compartment to form Ni(OH)2, your two compartments have a different concentration of Ni2+. You can use the Nernst equation to solve for Ecell.

Cooper Baddley 1F
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Re: 6N.15

Postby Cooper Baddley 1F » Thu Mar 05, 2020 3:19 pm

You set up the concentration cell and then use the Nernst equation to solve for potential.

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

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