Why does zinc plate copper in silver penny lab

In class, we performed a lab that involved heating a solution of zinc sulfate with zinc strips at the bottom of the beaker. A penny was placed on top of the zinc strips and soon after the penny became covered in a layer of zinc. What process causes the zinc layer to develop?

The explanation was that the zinc metal transferred electrons to copper which in turn reduced the zinc ions in solution, but this does not seem to make any sense as the zinc is being oxidized to just be reduced again. Why does this happen. Is a thin zinc layer more stable?

Here are the instructions for the experiment

• Was any electricity involved in this plating? – March Ho Sep 5 '15 at 13:51
• No the metals simply contacted each other – popgalop Jun 7 '16 at 23:45
• I made a video on the process of electroplating copper (a penny): youtube.com/watch?v=JOeJpoVqBMQ Using Ni here, not Zn. Same principles though – khaverim Jun 8 '16 at 16:10
• None of the answers to this question account for the electrochemical potential induced by the strain defects present from milling present in the zinc powder. – A.K. Dec 17 '16 at 19:57

The system of $Zn$ metal in a solution of $\ce{Zn^{2+}(aq)}$ is an equilibrium. Which means that the rates of the reactions $\ce{Zn -> Zn^{2+} + 2e-}$ and $\ce{Zn^{2+} + 2e- -> Zn}$ are happening in exactly the same rate and this rate is not zero. It is critical no notice that this rate is non-zero.

By heating up, you are accelerating both reactions, probably at different rates so there will be a new equilibrium point.

The precipitation is happening on any surface that it can nucleate on and this surface might be the surface of the penny or some other zinc. This is immaterial.

The counter reaction doesn't really matter here since it just happens on whichever metal the other reaction is happening and it is just the reduction or oxidation of water.

Electricity is involved: the copper (well, copper plated zinc in the US) penny is in direct contact with the zinc, forming a short-circuited connection. Forcing the plating action "backwards" (copper being more electronegative than zinc) requires current.

Try the same experiment but with a separator, such as a piece of filter paper, between zinc and copper; there should be no plating.

• But i still dont understand why electrons would go from the zinc metal to the copper metal just to go back into the zinc – popgalop Sep 6 '15 at 2:43
• That would make reaction have no net change as you start and end with zinc solution, copper metal, and zinc metal. – popgalop Jun 8 '16 at 15:02

The explanation was that the zinc metal transferred electrons to copper which in turn reduced the zinc ions in solution, but this does not seem to make any sense as the zinc is being oxidized to just be reduced again. Why does this happen. Is a thin zinc layer more stable?

Your intuition is correct and both of the given answers are wrong. The zinc plates the penny because there is a driving force. but the question then is "what is the driving force?" it is an electrical reaction but it is not as simple as one rate just happens to out pace another or is out accelerated by heating. First lets look at the half reaction of the powder: $$\ce{Zn (s) -> Zn^{2+} (aq) + 2e-}\tag{E^\circ = - 0.76}$$

and that of the penny:

$$\ce{Zn^{2+} (aq) + 2e- -> Zn (s)}\tag{E^\circ = + 0.76}$$

as you have already noted the standard electrode potential are equivalent, so what now? Well the operative word here is standard. standard electrode potentials assume that the materials is at 298K, 1 bar of pressure, 1 M electrolyte concentration and also that the surface is flat and well annealed. Zinc powder produced by milling is neither flat nor well annealed, it is quite sharp with a lot of residual stress. These conditions increase the energy of the zinc material which lowers the standard electrode potential for the powder such that it is now lower in magnitude than the electrode half reaction on the penny. i.e.: $$\ce{Zn (s) -> Zn^{2+} (aq) + 2e-}\tag{\color\red {\mathrm E <} - 0.76}$$ This reduced half potential creates a non net zero electrical potential which in turn produces a plating of zinc on the penny which is flatter with less internal stress. Thus your net reaction is: $$\ce{Zn (s, powder) -> Zn (s, plate)} \tag{\color\red {\mathrm E < 0.0}}$$

As for why being sharp and internally stressed cause higher energy, That is beyond the scope of the question, I have have given a brief and hand-wavy explanation below but I strongly encourage you to do your own research for a better explaination.

First for flatness, remember that chemical reactions are kinetic reactions relying on one rate happening faster than another. Knowing this we can infer that for a dissolution reaction, the more directions an atom has to dissociate, the more likely it will be to dissociate (See:Kelvin Equation) since energy directed towards the solid will not result in dissociation. As for the internal stress, stress is caused by atoms not being in a regular low energy state. Some perturbation of the microstructure causes them to be in a non-equilibrium state. Since equilibrium states are the lowest energy states, any configuration not at equilibrium must have higher energy and thus when given the chance will try to return to a lower energy state.

• I'm a little confused by your answer because the question mentions zinc strips not zinc powder. Otherwise your argument makes good sense. – Buck Thorn Oct 24 '20 at 19:18