Electrochemistry of galvanic couple in an acidic solution

[kamilio]

Homework Statement

From the perspective of corrosion...Gold and Copper II are joined together and immersed in a 1Molar HCl solution. What possible half-cell reactions and full cell reactions can occur? Which one is most likely to proceed?

Homework Equations

Half-cell potentials for
$$Au^{3+} + 3e^- \rightarrow Au(s) = 1.5V$$
$$Cl_2(g) + 2e^- \rightarrow 2Cl^- = 1.36V$$
$$Cu^{2+} + 2e^- \rightarrow Cu(s)=0.34V$$
$$2H^+ + 2e^- \rightarrow H_2(g)=0V$$

The Attempt at a Solution

The metals are exposed to hydrogen and chloride ions. Gold is the most noble and will not oxidize. The hydrogen will not oxidize copper because copper is cathodic wrt hydrogen. For the same reason, hydrogen cannot oxidize gold. Gold will oxidize copper but I don't know the hydrogen ions will steal the incoming electrons from copper and reduce to hydrogen gas.

 

[kamilio]

Can someone please assist me here. It's a rather simple problem but I just can't sort it through.

Hydrogen cannot oxidize gold or copper because it is lower on the emf series. Also, copper and gold cannot react because there are no Au ions in the solution (only H+ and Cl-) right? So then there aren't any half-cell/full-cell reactions?

 

[Blueshift5]

I would like to add some additional information and clarification to the given content. Firstly, the electrochemical reaction that occurs between gold and copper in an acidic solution is known as a galvanic couple. This is a type of electrochemical cell where two different metals are connected through an external circuit and immersed in an electrolyte solution.

In this case, the electrolyte solution is 1M HCl, which is an acidic solution. When gold and copper are joined together and immersed in this solution, the following half-cell reactions can occur:

1) Gold half-cell reaction:
Au^{3+} + 3e^- \rightarrow Au(s)

2) Copper half-cell reaction:
Cu^{2+} + 2e^- \rightarrow Cu(s)

3) Hydrogen half-cell reaction:
2H^+ + 2e^- \rightarrow H_2(g)

4) Chlorine half-cell reaction:
Cl_2(g) + 2e^- \rightarrow 2Cl^-

From these half-cell reactions, we can form two possible full cell reactions:

1) Gold-copper full cell reaction:
Au^{3+} + 3Cu(s) \rightarrow Au(s) + 3Cu^{2+}

2) Hydrogen-chlorine full cell reaction:
2H^+ + 2Cl^- + 3Cu(s) \rightarrow 3Cu^{2+} + H_2(g) + 2Cl^-

Now, to determine which reaction is most likely to proceed, we need to consider the standard electrode potentials for each half-cell reaction. The half-cell potential for gold is 1.5V, while for copper it is 0.34V. This means that the gold half-cell reaction has a higher tendency to occur compared to the copper half-cell reaction.

Moreover, the half-cell potential for hydrogen is 0V, indicating that it has a lower tendency to occur compared to both gold and copper. Similarly, the half-cell potential for chlorine is 1.36V, which is lower than gold but higher than copper.

Based on these values, it is most likely that the gold-copper full cell reaction will proceed, as gold has the highest tendency to oxidize and copper has the highest tendency to reduce. However, it is important to note that the actual reaction that takes place may also depend on other factors such as the concentration of the electrolyte solution and the surface area of the