I did an experiment where I dissolved 30 grams of zinc sulfate in water and boiled it with strips of solid zinc metal and strips of copper metal and a layer of zinc was deposited on the copper.

I know that if you place solid copper in zinc sulfate aqueous without the solid zinc nothing would happen because zinc is more reactive than copper. Why does the addition of zinc sulfate cause the zinc to coat the copper? What is happening?


4 Answers 4


Zinc, being in contact with copper, is easily oxidized, because copper drives electrons out of it.

Metallic copper becomes electron-rich and is therefore capable to yield electrons to $\ce{Zn^2+}$ (stemming from $\ce{ZnSO4}$), which reduces to $\ce{Zn^0}$. The latter is finally observed to adsorb at the copper strips.

In other words, copper metal works as an auxiliary electrode whose function is to transport electrons, without being $-$ by no means $-$ transformed during the experiment.

In conclusion, the overall reaction uniquely involves zinc and its ions.

This is similar to what is observed when copper, in contact with zinc, is dipped into an acidic solutions. Hydrogen bubbles are seen even at the copper surface, without the latter being oxidized.

  • $\begingroup$ Not standard conditions so you can't be so sure. Tin can are coated inside and outside with tin (simpler to manufacture) Tin only physical protects the outside of tin cans (a bit of damage and moisture and the steel corrodes quickly- tin is anodic to steel) and but with acid food and complexation agents it electrochemically protects the steel (under those condition it is cathodic). $\endgroup$ Dec 16, 2013 at 9:45
  • $\begingroup$ Of course these are not standard conditions. But the fact is that copper, in contact with zinc (that's the essential point), drives electrons out of metallic zinc....This excess of electrons is available for $Zn^{2+}$ to reduce. The situation is very well explained in this article $\endgroup$
    – mannaia
    Dec 16, 2013 at 10:08
  • $\begingroup$ Sorry for commenting on something this old but... If the zinc strips are dissolving and the zinc ions are depositing, does this proceed infinitely? After all, the newly deposited zinc is just a purer version of the previous strip, right? $\endgroup$
    – peruca3d
    Jul 6, 2020 at 1:40

Here is my qualitative answer. As already stated by others, looking at the Electromotive Series (similar to activity series), under standard conditions of $\pu{77^\circ F}$ and concentrations of $\pu{1 M}$, Zn has an electrical potential of $\pu{+0.76 V}$. Cu has an electrical potential of $\pu{-0.34 V}$. The potential difference of approximately $\pu{1.1 V}$, of these two samples in a cell will be a primary driving force for ionic/charge movement.

Qualitatively speaking, Zn is more active, and dissolves to a greater concentration than would Cu under similar conditions, i.e., Cu is more noble than Zn. Therefore, very few $\ce{Cu^2+}$ ions will dissolve into the $\ce{ZnSO4}$ solution. However, the $\ce{Zn^2+}$ ions in solution (from the dissolved $\ce{ZnSO4}$) will be attracted to the "negatively" charged Cu sample, plating out on its surface. The pure Zn sample will then provide new $\ce{Zn^2+}$ ions, to replace those that have plated out on the Cu sample. The driving force for the dissolution of new $\ce{Zn^2+}$ ions into solution, are the negatively charged sulfate ions present. If you weighed the pure Zn and pure Cu samples before and after the experiment, you would likely see the Cu weight increase equal the Zn weight decrease.


In looking at the reaction, it seems our reactants are zinc sulfate, zinc metal, and copper metal. Qualitatively, this reaction yielded what looks like a zinc layer deposited on to the copper.

One familiar reaction is that of zinc and copper sulfate.

$$\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s) }$$

Zinc metal is above copper in the activity series and therefore, zinc will replace copper in solution.

The zinc metal is being oxidized as it loses electrons, and the copper(II) ions are reduced to copper metal as they acquire electrons.

What is interesting is that you are correct, since zinc metal is above copper in terms of the reactivity series, if you just placed copper metal in zinc sulfate solution, you would not expect the reaction to go in reverse.

Using Wikipedia's definition, a galvanic cell, or voltaic cell, named after Luigi Galvani, or Alessandro Volta respectively, is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell. It generally consists of two different metals connected by a salt bridge, or individual half-cells separated by a porous membrane.

The zinc metal in the zinc sulfate solution along with the copper metal, zinc metal loses its electrons more easily, and its electrons are received by copper. We can write these reactions by:

$$\begin{align}\ce{Zn(s) &-> Zn^{2+}(aq) + 2e^-}\\ \ce{Cu^{2+}(aq) + 2e^- &-> Cu(s)}\end{align}$$

Here we can see that the zinc metal that we placed in our solution is oxidized and our copper metal is reduced. This reaction is able to proceed as the energy released from the oxidation of zinc is more than that of the reduction of the copper metal.

As zinc loses its electrons, it forms ions, and I assume that this is the deposited zinc metal that you see on the copper metal that gains electrons.

  1. I am not sure an electrochemical reaction is occurring; it could be a simple dissolution and deposition. It happens all the time but it is so slow at room temperature that it is not noticeable. Measure the weight of the zinc and actual copper before and after the experiment.

  2. If it is an electrochemical reaction, then the answer is in 3 parts

    1. Standard electrochemical half cell potential (usually expressed as reduction such as the first reaction but I'll reverse it to give the oxidation reaction that may occur)

    $$\begin{align}\ce{Zn^2+(aq) + 2e- &-> Zn (s)}&& E^\circ=\pu{-0.7628V}\\ \ce{Cu(s)&->Cu^2+ (aq) + 2e-} && E^\circ=\pu{-0.3402 V}\\ \ce{Zn^2+ (aq) + Cu(s) &-> Cu^2+ (aq) + Zn(s)} && E^\circ=\pu{-1.103V}\end{align}$$ so this would not occur spontaneously; electrical energy is needed for it to occur.

    1. This occurs at standard condition which are usually $\pu{1mol/L}, \pu{25 ^\circ C}, \pu{1 atm}$ (other ions also have an effect)

    2. To understand it properly you will need to calculate the actual concentration in mol/L (http://drfus.com/chapter-20/20-6-cell-potentials-under-nonstandard-conditions) or you can just measure the actual potential difference between the pieces of metal to find the potential difference.


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