# Finding concentrations in a voltaic cell

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.

• Are we to assume that the temperature is $25^oC$? – ringo Apr 16 '15 at 6:20
• 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 – Jaroslav Kotowski 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 $E_\text{cell}=E_\text{cell}^\circ$

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

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

$$\ln Q=0$$

$$Q=1$$

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