A voltaic cell with $\ce{Ni/Ni^2+}$ and $\ce{Co/Co^2+}$ half cells has the initial concentrations of $\ce{Ni^2+}$ $\pu{0.80 M}$ and $\ce{Co^2+}$ $\pu{0.20 M}$.
(a) Find initial cell potential;
(b) Find concentration of $\ce{Ni^2+}$ when $E_\mathrm{cell}$ reaches $\pu{0.03 V}$

In part (a), I correctly found the initial $E_\mathrm{cell}$ to be $\pu{0.05 V}$.

In part (b), the book doesn't have any other examples that are explained showing how to get one of the concentrations, it's supposed to be $\ce{0.50 M}$, but I'm not sure how they get to that. From my tables, I found $E$ half cell for $\ce{Ni}$ to be $\pu{-0.25 V}$ and $E$ half cell for $\ce{Co}$ to be $\pu{-0.28 V}$, and the overall then to be $\pu{0.03 V}$.

Tried plugging in everything I have into the Nernst equation, but that still leaves me coming up short.

  • 1
    $\begingroup$ Are we to assume that the temperature is $25^oC$? $\endgroup$
    – ringo
    Apr 16 '15 at 6:20
  • 2
    $\begingroup$ Idea: the Initial cell potential is 0.05V, the potential drops to 0.03V. Since the overal reaction is equimolar and half cells potential similar, Ni ions concentration drops by 0.03/0.05 $\endgroup$ Apr 16 '15 at 9:39

I think you were on the right track!

You were right that you need the Nernst equation, which must be how you determined the initial $E_\text{cell}$.

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

You correctly identified that your $\ce{Co|Co^2+||Ni|Ni^2+}$ cell has the cell standard potential

$$E_\text{cell}=E_\text{cathode}^\circ-E_\text{anode}^\circ=-0.25\ \mathrm V--0.28\ \mathrm V=0.03\ \mathrm V$$

So the question is essentially asking what concentrations are needed for your cell to have the standard potential. Again looking at the Nernst equation, you can solve for the reaction quotient $Q$ when $E_\text{cell}=E_\text{cell}^\circ$

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

$$E_\text{cell}-E_\text{cell}^\circ=0=\frac{RT}{nF}\ln Q$$

$$\ln Q=0$$


At this point, you should be able to solve for the concentration of $\ce{Ni^2+}$ with simple algebra.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.