6.N.15 7th Ed


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Ahmed Mahmood 4D
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

6.N.15 7th Ed

Postby Ahmed Mahmood 4D » Fri Mar 15, 2019 12:42 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"

I’ve found the concentrations of both nickel ion. When setting up ln(Q), how do I find out which nickel is the product and which one is the reactant?

David S
Posts: 54
Joined: Fri Sep 28, 2018 12:15 am

Re: 6.N.15 7th Ed

Postby David S » Fri Mar 15, 2019 5:29 pm

Looks like a concentration cell to me. If we want our concentration cell to do work, Ecell has to be positive.

Since Ecell = -(RT/nF)lnQ (E° = 0 for conc. cell), and the ln of numbers less than 1 is negative, we know that we want a Q that is less than one. For Q to be less than one, [P]<[R]. Thus, make the side with lower conc the P, and the higher conc the R.


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