# Reaction of solid metals with a solution of metal ions

Explain the following experiments and their results. What is their chemical equations as well?

1. When immersing a Magnesium (Mg) chunk inside aqueous solution of Copper ions (Cu(II)), a red-brown sediment of Copper appears. Why?

2. When immersing a Silver (Ag) stripe inside aqueous solution of Calcium ions (Ca(II)), no chemical reaction takes place. Why?

• Homework questions should show the work or thinking the poster has already done in an initial attempt to answer the question. This way a responder can better see where help is needed and provide a more meaningful response. For help asking a good homework question, see: How do I ask a homework question on Chemistry Stack Exchange – ron Dec 31 '14 at 20:02
• I'm sorry for the confusion, this is not homework. I found an old chemistry textbook which I find very interesting, would be glad to have help with these questions.. – Josh Dec 31 '14 at 20:25
• @Josh not your problem, this is the usual thinking of some people here, will hellp you! – RE60K Jan 1 '15 at 10:09

You can explain these experiments with the Standard Electrode Potential. With these experimental values you know which element/compound releases more energy upon receiving a comparable amount of electrons. As a rule of thumb, alkalimetals and alkaline earth metals release less energy than, for example, elements from the halogen-group.

With this table at hand, you can predict that $\ce{Mg}$ will be oxidized and the $\ce{Cu^2+}$ reduced, because of the potential for the two following reactions: $$\ce{Mg^2+ + 2e^- -> Mg(s)} \quad E^\circ=−2.372\,\text{V}$$ $$\ce{Cu^2+ + 2e^- -> Cu(s)} \quad E^\circ=0.337\,\text{V}$$

Reducing copper-ions will release more energy, so the following chemical reaction will apply: $$\ce{Mg +Cu^2+-> Cu + Mg^2+}$$

In an analogous fashion, for the second question: $$\ce{Ag^2+ + 2e^- -> Ag(s)} \quad E^\circ=+1.98\,\text{V}$$ $$\ce{Ca^2+ + 2e^- -> Ca(s)} \quad E^\circ=−2.868\,\text{V}$$

Reducing $\ce{Ag^2+}$ will release more energy - since silver is already in its solid state, the following reaction will not take place:

$$\ce{Ag +Ca^2+-> Ag^2+ + Ca}$$

Note: To see how those potentials or potential difference relate to energy see, for example, the Nernst Equation / Gibbs free energy

This is based on chemical reactivity series based upon electrode potential (electrochemistry topic): $${\bf More\;Reactive}:\ce{K\;Ca\;Mg\;Al\;Zn\;Fe\;Ni\;Sn\;Pb\;H\;Cu\;Hg\;Ag\;Au\;Pt}:{\bf Less\;Reactive}$$

So Magnesium[Mg] displaces Copper[Cu] and all other less reactive than it, but since Calcium[Ca] is more reactive than Silve[Ag], silver can't displace Calcium: $$\ce{Mg +Cu^2+-> \underbrace{Cu}_{red}+Mg^2+}\\\ce{Ca +Ag+\not\!-> Ag+Ca^2+}$$