Classical Idea of Redox Reactions

Every time you light a candle, watch iron slowly turn to rust, or charge a battery, a particular type of chemical reaction is at work. These are redox reactions, and they form one of the most important families of chemical processes in all of chemistry. Before we can understand them deeply, we need to trace how chemists first recognised and defined them.

Why Redox Reactions Matter

Redox reactions are not just a classroom topic. They show up across nearly every area of science and daily life:

• Energy from fuels — Burning coal, petrol, natural gas, or wood for domestic, transport, and commercial energy all involve redox processes
• Metal extraction — Highly reactive metals like sodium and aluminium are obtained through electrochemical processes, which are redox reactions
• Industrial manufacturing — Chemical compounds such as caustic soda ($NaOH$) are produced via redox-based methods
• Batteries — Both dry cells and wet batteries operate through controlled redox reactions
• Corrosion — The rusting of iron and tarnishing of silver are redox processes that cost industries billions every year
• Agriculture — Soil nutrient cycles, nitrogen fixation, and the action of many fertilisers and pesticides involve redox chemistry
• Pharmaceutical and biological systems — Countless biochemical reactions in living organisms, from respiration to photosynthesis, are redox in nature
• Environmental issues — The concept of a Hydrogen Economy (using liquid hydrogen as a clean fuel) and the development of the Ozone Hole both fall under redox chemistry

With that context, let us see how chemists built up the definitions of oxidation and reduction step by step.

Oxidation: How the Definition Grew

Stage 1: Addition of Oxygen

The story begins with oxygen. The very word “oxidation” comes from “oxide.” In the earliest understanding, oxidation simply meant the addition of oxygen to an element or compound.

This made sense because Earth’s atmosphere contains about 20% dioxygen ($O_2$), and many elements readily combine with it. That is the main reason so many elements are found in nature as their oxides rather than in pure form.

Here are classic examples of oxidation under this original definition:

$2Mg(s) + O_2(g) \rightarrow 2MgO(s) \qquad \text{(7.1)}$

Magnesium combines with oxygen to form magnesium oxide. Oxygen has been added to magnesium, so magnesium is oxidised.

$S(s) + O_2(g) \rightarrow SO_2(g) \qquad \text{(7.2)}$

Sulphur combines with oxygen to form sulphur dioxide. Again, oxygen is added to sulphur, so sulphur is oxidised.

$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \qquad \text{(7.3)}$

When methane burns, oxygen adds to both carbon and hydrogen. Methane is oxidised.

Stage 2: Removal of Hydrogen

Chemists noticed something interesting in reaction (7.3). Look at what happens to the hydrogen originally bonded to carbon in methane: it gets replaced by oxygen and ends up in water. The hydrogen has effectively been stripped away from the carbon. This observation pushed chemists to broaden the definition: oxidation also includes the removal of hydrogen from a substance.

A clearer illustration of this idea:

$2H_2S(g) + O_2(g) \rightarrow 2S(s) + 2H_2O(l) \qquad \text{(7.4)}$

Here, hydrogen sulfide ($H_2S$) loses its hydrogen (which goes into water), leaving behind solid sulphur. The $H_2S$ is oxidised through the removal of hydrogen.

Stage 3: Addition of Any Electronegative Element

As chemical knowledge expanded, chemists realised that reactions very similar to the ones above can happen with elements other than oxygen. Magnesium, for example, reacts with fluorine, chlorine, and sulphur in much the same way it reacts with oxygen:

$Mg(s) + F_2(g) \rightarrow MgF_2(s) \qquad \text{(7.5)}$

$Mg(s) + Cl_2(g) \rightarrow MgCl_2(s) \qquad \text{(7.6)}$

$Mg(s) + S(s) \rightarrow MgS(s) \qquad \text{(7.7)}$

In each case, an electronegative element (one that strongly attracts electrons) is added to magnesium. Since the chemistry is essentially the same as when oxygen is added, chemists extended the definition: oxidation includes the addition of any electronegative element, not just oxygen.

Stage 4: Removal of Any Electropositive Element

The broadening did not stop there. Consider this reaction:

$2K_4[Fe(CN)_6](aq) + H_2O_2(aq) \rightarrow 2K_3[Fe(CN)_6](aq) + 2KOH(aq)$

Potassium ferrocyanide ($K_4[Fe(CN)_6]$) is converted to potassium ferricyanide ($K_3[Fe(CN)_6]$). Notice that potassium, an electropositive element (one that readily gives up electrons), has been removed from the compound. Chemists recognised this as oxidation too: removal of an electropositive element from a substance is oxidation.

The Complete Classical Definition of Oxidation

Putting all four stages together:

Oxidation is the addition of oxygen or any electronegative element to a substance, or the removal of hydrogen or any electropositive element from a substance.

Reduction: The Mirror Image

Every broadening of the oxidation definition was matched by a corresponding broadening of the reduction definition. Originally, reduction meant only the removal of oxygen from a compound. Over time, it grew to encompass much more.

Here are the key examples that illustrate the full classical definition of reduction:

Removal of Oxygen

$2HgO(s) \xrightarrow{\Delta} 2Hg(l) + O_2(g) \qquad \text{(7.8)}$

When mercuric oxide is heated, it breaks down into liquid mercury and oxygen gas. Oxygen is removed from the compound, so $HgO$ is reduced.

Removal of an Electronegative Element

$2FeCl_3(aq) + H_2(g) \rightarrow 2FeCl_2(aq) + 2HCl(aq) \qquad \text{(7.9)}$

Ferric chloride ($FeCl_3$) loses one of its three chlorine atoms per formula unit, becoming ferrous chloride ($FeCl_2$). Chlorine is an electronegative element, and its removal from the compound makes this a reduction of $FeCl_3$.

Addition of Hydrogen

$CH_2{=}CH_2(g) + H_2(g) \rightarrow CH_3{-}CH_3(g) \qquad \text{(7.10)}$

Ethene gains hydrogen to become ethane. Adding hydrogen to a substance is reduction.

Addition of an Electropositive Element

$2HgCl_2(aq) + SnCl_2(aq) \rightarrow Hg_2Cl_2(s) + SnCl_4(aq) \qquad \text{(7.11)}$

In this reaction, mercury (an electropositive element) is effectively added to mercuric chloride to form mercurous chloride ($Hg_2Cl_2$). This counts as reduction of $HgCl_2$.

The Complete Classical Definition of Reduction

Reduction is the removal of oxygen or any electronegative element from a substance, or the addition of hydrogen or any electropositive element to a substance.

Oxidation and Reduction Always Come in Pairs

Look back at reaction (7.11). While $HgCl_2$ is being reduced (gaining mercury), $SnCl_2$ is simultaneously being oxidised (gaining two extra chlorine atoms to become $SnCl_4$). If you go back and examine every single reaction listed above, you will find the same pattern: whenever one substance is oxidised, another is reduced in the same reaction.

This is not a coincidence. It is a fundamental rule. Oxidation cannot happen in isolation, and neither can reduction. They are two sides of the same coin. Because of this inseparable pairing, chemists coined the term “redox” (from reduction + oxidation) to describe this entire class of chemical reactions.

Worked Example: Problem 7.1

Identify the species undergoing oxidation and reduction in each reaction below.

(i) $H_2S(g) + Cl_2(g) \rightarrow 2HCl(g) + S(s)$

Solution: Look at $H_2S$. It starts with hydrogen bonded to sulphur. In the products, that hydrogen has moved to chlorine (forming $HCl$), and sulphur is left behind as a free element. Hydrogen, a more electropositive element, has been removed from sulphur. Alternatively, you can view it as chlorine (a more electronegative element) being added to hydrogen. Either way, $H_2S$ is oxidised.

Now look at $Cl_2$. It ends up combined with hydrogen in $HCl$. Hydrogen (an electropositive element) has been added to chlorine. $Cl_2$ is reduced.

(ii) $3Fe_3O_4(s) + 8Al(s) \rightarrow 9Fe(s) + 4Al_2O_3(s)$

Solution: Aluminium starts as a free element and ends up combined with oxygen in $Al_2O_3$. Oxygen has been added to aluminium, so $Al$ is oxidised.

Ferrous ferric oxide ($Fe_3O_4$) loses all of its oxygen, leaving behind free iron metal. Oxygen has been removed from it, so $Fe_3O_4$ is reduced.

(iii) $2Na(s) + H_2(g) \rightarrow 2NaH(s)$

Solution: This reaction requires more careful thinking. No oxygen or halogen is involved. The key is electronegativity: hydrogen is more electronegative than sodium. So when hydrogen combines with sodium:

• From sodium’s perspective: an electronegative element (hydrogen) is added to it. Sodium is oxidised.
• From hydrogen’s perspective: an electropositive element (sodium) is added to it. Hydrogen is reduced.

This example highlights that even when common electronegative elements like oxygen or chlorine are absent, the classical framework still works as long as you correctly apply the concept of relative electronegativity. It also hints at why chemists later developed an even more powerful definition of redox based on electron transfer, which we will explore in the next topic.