Redox Reactions

 

Electrochemistry is one of the units that is neglected in most first year high school chemistry courses. This is a shame because electrochemistry is all around us. Anytime you carry around an electronic device that does not need to be plugged in to work, you are using electrochemistry. 

In addition, most reactions are electrochemical in nature. For instance, the single displacement reaction here

\(Mg + Cu(NO_3)_2 \rightarrow Mg(NO_3)_2 + Cu\) 

can be thought of as a simple trading of places between the copper and the magnesium. However a closer look shows us that there is more going on. 

Any pure metal (actually any pure element) is neutral. That means that the atoms of magnesium on the left side of the reaction have no charge. But, on the right side the magnesium has a charge of +2. (We know it must be a positive ion because it is hanging out with nitrate – a negative ion.) That means that in the course of this reaction, each Mg atom lost 2 electrons. 

If we start to wonder where those electrons went, we don't need to look any further than the copper. On the left side of the reaction, the copper is composed of +2 ions and on the right it is pure, uncharged copper. So, this reaction is not simply a “trading of places”. This reaction involves copper ions taking electrons away from magnesium atoms.

Be sure to avoid the idea that the magnesium “gave” the electrons to the copper. NO positive nucleus has EVER given away negative electrons. This reaction is, pure and simple, electron theft.

Let’s look at another single displacement reaction, this time between copper and silver.

\(AgNO_3 + Cu \rightarrow Cu(NO_3)_2 + Ag\)

Again we can see the electron theft occurring. On the left side of the reaction, the copper has no charge and on the right it has a charge of +2. Something has stolen 2 electrons. The culprit, of course is silver. On the left the silver has a charge of +1 and on the right it is neutral. 

There is a problem however. The silver has stolen 1 electron, but the copper lost 2. The solution of course is that TWO silver ions “gang up” on the copper, each taking one electron, and we could have seen that if the reaction had been properly balanced. Here is it, written correctly.

\(2~ AgNO_3 + Cu \rightarrow Cu(NO_3)_2 + 2~ Ag\)

Let’s look at one more reaction – the combustion of methane.

\(CH_4 + 2~ O_2 \rightarrow CO_2 + 2 ~H_2O\)

Given what we know about elements in general and oxygen in particular, we can find the electron theft occurring here. On the left side, oxygen is a pure element, and therefore has no charge. If we look at the water on the right side we can guess that oxygen has a -2 charge. That guess makes sense for 2 reasons. This first reason is that -2 is the preferred charge on oxygen (since 2 extra electrons fill the energy level). The second reason is that we know that hydrogen is generally +1 (think acid chemistry).

There is a problem with that logic however – water is not ionic, and therefore, doesn’t actually have any charges. That means that electron theft is not as clear here as it was in the other reactions. Rather than outright theft, what has happened here is a transfer of control. The bonds in water are polar, due to the high electronegativity of oxygen. So, in a sense, the oxygen has “taken control” of those electrons, making it “feel” -2ish. This “ish” type of charge is called oxidation number and being able to determine it for separate atoms within compounds is an important skill discussed here.