What happens when a piece of copper is placed in 1M HCl?
Can I just clarify, without going into the detail of the specific problem, if you have Cu(s) in solution (like you do here) can it ONLY be oxidised? So essentially you're looking for whether there's anything in solution that can oxidise Cu?


Let us take a longer approach to address your query. For electrochemical problems like these, first you have to make a list of starting materials and possible products. Do a thought experiment even before even looking up the tables. It is long way but it will help you in doing future problems.

Your question is: What happens when a piece of copper is placed in 1M HCl?

Do a thought experiment +  use some basic intuitive chemistry.

Making a list of starting materials, a) Cu metal, b) H$^+$, c) Cl$^-$.

Now there are only two possibilities for each substance.

a) Copper Cu can be oxidized to Cu$^+$ or Cu$^{2+}$ or reduced to Cu$^-$. Basic chemistry tells us that metals like to form cations so Cu$^-$ can be eliminated. Similarly, there is possibility for Cu$^+$ or Cu$^{2+}$. However, say, you are just interested in Cu$^{2+}$.

Now you have the problem well defined: Will Cu -> Cu$^{2+}$?

b) Follow the same reasoning: H$^+$, it can be oxidized to  H$^{2+}$ or reduced to H$_2$. Basic chemistry will tell you that H$^{2+}$ is not possible. Now your question is more well defined: Will H$^+$ -> H$_2$

c) In the same way, ask the same question for chloride ion. It can be oxidized to Cl$_2$ or reduced to  Cl$^{2-}$. Basic chemistry would tell you that Cl$^{2-}$ is not feasible. So your only concern is: Will Cl$^-$ -> Cl$_2$

Now you can ask only two questions:

Can copper metal reduce H$^+$ to H$_2$?

or

Can copper metal oxidize Cl$^-$ to Cl$_2$? This can be easily eliminated because in order for copper to oxidize it must reduce itself further, which is not possible.

At this stage you can utilize the electrode potentials. Recall cathode refers to reduction and anode refers to oxidation
Ecell= E$_{cathode}$-E$_{anode}$

Ecell= E(for hydrogen half cell bc it is being reduced) - E (copper half cell)

Ecell= 0.00- (+0.34) = - 0.34 V

The negative sign shows this not possible under these conditions.

Do the same for chloride ion.

Ecell= E(Reduction of Cu to Cu$^-$) - E (chlorine half cell)
Ecell= Undefined- (+1.36) = Undefined because Cu$^-$ does not exist in solution.

In short nothing will happen to copper in HCl.

P.S. Practically, HCl slowly dissolves copper in the presence of oxygen. This not relevant here.