# 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)

By addition of $$\ce{NaCl}$$ to the $$\ce{Zn}$$ side of the cell, I observed precipitation of $$\ce{ZnCl2}$$.

I measured the voltage to $$\pu{1.05 V}$$ before adding $$\ce{NaCl}$$ to the cathode side, I then observed the voltage go down by addition of the salt.

According to the Nernst equation: as the $$\ce{Zn^2+}$$ ion concentration reduces with precipitation, there should be an increase in voltage.

$$E = 1.05 - \frac{RT}{nF} \ln Q$$

$$E= 1.05 - \frac{RT}{nF} \ln \left(\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}\right)$$

since $$\ln Q \lt 0$$ for $$[\ce{Zn^2+}] < [\ce{Cu^2+}]$$

"$$-\frac{RT}{nF} \ln Q$$" term would be positive. Therefore the voltage should go up. But the opposite was observed and I don't know how to explain it.

• Do you consider activity coefficients and chlorocomplexes ? – Poutnik Mar 18 at 20:15
• There are two mistakes here. First $\ce{ZnCl2}$ will not precipitate. It is one of the best soluble compounds of zinc. Second, you say that you add NaCl to the cathode side and the cathode of the cell is made of copper. Then you say later on that you have added NaCl in the zinc solution. So please explain what you have done and observed. – Maurice Mar 18 at 20:23
• @Maurice Thank you. Yes I mixed them up, apologies. I had an equal solution of ZnSO4 (on Zink side) and CuSO4 (on copper side). I then added som NaCl to the Zink side, anode side that is. I then observed some kind of precipitation on the anode side that fell to the bottom. I also observed the voltage go down on my voltmeter. The impression given by the instructor was that the Zn2+ would be precipitated with Cl-, thus explaining the shift in potential. However that doesn't seem correct by the calculation. Please let me know again if I'm beeing unclear, english is not my native language. – JR 19 Mar 18 at 20:51
• @JR19 did you try adding $\ce{NaCl}$ to cathode side you can see the color change from blue too green the copper complex formation copper aqua complex is blue where as copper chloro complex is yellow mix of these colors show green but $\ce{Zn^{2+}}$ d orbitals are filled so no color effect too bad you don't see this in anode side. – Avon97 Mar 19 at 14:54

## 3 Answers

This is due to the complex formation of $$\ce{Zn^2+}$$ with $$\ce{Cl^-}$$ which will result a different half cell equilibrium in this case the anode,

Original half cell reaction,

$$\ce{Zn^2+ (aq) + 2e^- <=> Zn(s) E^o = 0.762 V (reduction)}$$

Modified half cell reaction,

$$\ce{[ZnCl_4]^2- (aq) + 2e^- <=> Zn(s) + 4Cl^- (aq)}$$

(Since you are using a voltmeter it has infinite resistance (hypotheticaly) so the cells are in equilibrium)

The precipitation must be due to $$\ce{Na2SO4}$$ not $$\ce{ZnCl2}$$ because $$\ce{ZnCl2}$$ is more soluble than $$\ce{Na2SO4}$$ $$\ce{(281 g/L vs 357 g/L (25 °C))}$$ $$\ce{Na2SO4}$$ $$\ce{ZnCl2}$$. It should dissolve after few minutes and form the complex , try stirring the flask.

You should increase the voltage up and down with decreasing or increasing the concentration of $$\ce{Zn^2+}$$ ions with same solvent $$\ce{ZnSO4 (aq)}$$ to use the Nerst equation. Addition of different chemicals will change the cell equilibrium thus don't use the Nerst equation for those cases like this scenario. Have to refer a book for the new cell equilibrium half cell value to verify the result with the new $$\ce{E^{o}_{cell}}$$ value.

• I don't agree with $\ce{NaCl}$ precipitation. It the precipitation really happens, that should be as $\ce{Na2SO4}$ because it has the lowest solubility among all other salts in display. – Mathew Mahindaratne Mar 19 at 6:17
• @MathewMahindaratne Yeah correct I have edited that – Avon97 Mar 19 at 14:43
• @Avon97. No ! If you consider that some complex with $\ce{Zn}$ and $\ce{Cl-}$ is formed, this effect will decrease the concentration of the non complexed $\ce{Zn^{2+}}$ ion. And if this concentration decreases, the potential changes from $\pu{-0.76 V}$ to a more negative value. And this is the contrary of what JR 10 has found. – Maurice Mar 19 at 18:11

@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 "activity". At high concentration, concentration must be replaced by "activity". 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 "activity". 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 "concentration" given by an electrode is the number of moles divided NOT by the volume of the solution, but divided by the volume of the "free water", namely the volume of water not engaged in adsorption around the ions. This sort of "new concentration" 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.

• @Buck Thorn. OK. I shall edit the text, and modify this sentence – Maurice Mar 19 at 11:03

Although we have two good answers for this question, I'd like to point out some other information to explain what's going on when you add $$\ce{NaCl}$$ to the anode. Understandably, increasing $$\ce{NaCl}$$ would make $$\ce{Zn^2+}$$ ions to act as they are from $$\ce{ZnCl2}$$. According to Ref.1, aqueous $$\ce{ZnCl2}$$ solutions contain at least four $$\ce{Zn}$$ species indicated by Ramon IR spectroscopy:

Raman studies of aqueous solutions of zinc chloride indicate the presence of $$\ce{Zn(H2O)6^++},$$ $$\ce{ZnCl_{(aq)}+},$$ linear $$\ce{ZnCl2_{(aq)}},$$ and $$\ce{ZnCl4(H2O)2^=}$$ in aqueous solutions less concentrated than ten molar. Ion concentrations are reported as functions of zinc chloride molarity. The tetrachloro complex dominates for stoichiometric zinc chloride concentrations between ∼0.5 to ∼10M with significant concentrations of $$\ce{ZnCl+}$$ below ∼4M zinc chloride and $$\ce{ZnCl2}$$ above ∼4M. For stoichiometric solutions greater than 10M, the evidence suggests a polymeric species or aggregate with structural characteristics similar to those found in the crystal.

The presence of these four species were confirmed later by X-ray scattering studies (Ref.2). Thus, it is safe to say that the standard potential of zinc ion would be adjusted significantly due to the presence of these four species (Ref.3). In addition, Ref.4 reports the increment of the conductivity of aqueous $$\ce{ZnCl2}$$ solutions according to their concentrations.

References:

1. D. E. Irish, Bruce McCarroll, T. F. Young, "Raman Study of Zinc Chloride Solutions," J. Chem. Phys. 1963, 39(12), 3436-3444 (DOI: https://doi.org/10.1063/1.1734212).
2. M. Maeda, T. Ito, M. Hori, G. Johansson, "The Structure of Zinc Chloride Complexes in Aqueous Solution," Verlag der Zeitschrift für Naturforschung A 1996, 51(1-2), 63-70 (DOI: https://doi.org/10.1515/zna-1996-1-210) (PDF).
3. Lutfullah, Helen S. Dunsmore, Russell Paterson, "Re-determination of the standard electrode potential of zinc and mean molal activity coefficients for aqueous zinc chloride at 298.15 K," Journal of the Chemical Society, Faraday Transactions 1: Physical Chemistry in Condensed Phases 1976, 72, 495-503 (DOI: https://doi.org/10.1039/F19767200495).
4. B. K. Thomas, D. J. Fray, "The conductivity of aqueous zinc chloride solutions," Journal of Applied Electrochemistry 1982, 12(1), 1-5 (DOI: https://doi.org/10.1007/BF01112058).