How does the addition of NaCl to the cathode side effect the potential of a voltaic cell: Zn(s) | Zn2+(aq) 0.1M || Cu2+(aq) 0.1M | Cu(s)
@JR19. If you modify the anode compartment, and if you then observe that this change produces a decrease in the cell voltage, it means that [$\ce{Zn^{2+}}$] has increased. It is probably an effect of &quot;activity&quot;. At high concentration, concentration must be replaced by &quot;activity&quot;. The voltage delivered by the electrodes is not really proportional to the log of the concentration, but it is proportional to the log of the so called &quot;activity&quot;. I explain.
Let's consider the case of $\ce{HCl}$, which is best known. If you measure the pH of $0.01$ M HCl, you find pH = $2.0$. If now you add and dissolve some $\ce{NaCl}$ to this solution, the pH goes down to $1.5$ and even to $1.2$, as if the solution gets more and more concentrated. It is hard to believe, but easy to check. Do it in the lab, if you can. Here is an explanation taken from an old issue of J. Chem. Educ. (I don't remember which one). The measured &quot;concentration&quot; given by an electrode is the number of moles divided NOT by the volume of the solution, but divided by the volume of the &quot;free water&quot;, namely the volume of water not engaged in adsorption around the ions. This sort of &quot;new concentration&quot; is called activity. By adding more $\ce{NaCl}$, the volume of free water decreases, so that the activity of $\ce{HCl}$ increases. In $1$ liter concentrated solution of $\ce{NaCl}$, there is only about $0.2$ liter of free water. The activity is much higher than the concentration. It must be the same phenomena in your cell.