## 6N.15

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Rachel Yoo 1F
Posts: 65
Joined: Fri Sep 28, 2018 12:24 am

### 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 at 298 K. Use Table 6I.1
I got the [OH-} concentration but I'm not really sure how to approach the problem

EllerySchlingmann1E
Posts: 76
Joined: Fri Sep 28, 2018 12:24 am

### Re: 6N.15

For this problem, you want to recognize that it is a concentration cell reaction, so standard cell potential is zero and cell potential is equal to -(RT/nF) ln Q. In the 6I.1 table, you can find that Ksp = 6.5×10^−18 for Ni(OH)2 which means [Ni][OH]^2 = 6.5x10^-18. Use your concentration of OH to find the concentration of nickel in the one compartment and use that concentration compared to the 1.0 M nickel solution to find Q and subsequently cell potential. Hope that helps!

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