Types of Redox Reactions: Combination and Decomposition

Now that we have oxidation numbers as a tool for tracking electron shifts, let us see how they help us sort redox reactions into recognisable categories. Chemists group redox reactions into four broad types: combination, decomposition, displacement, and disproportionation. This topic focuses on the first two, which are closely related since one is essentially the reverse of the other.

Combination Reactions: Building Something New

A combination reaction is one where two or more substances come together to form a single product:

$A + B \rightarrow C$

Think of it as a construction process: separate pieces join to make one finished item.

Here is the key question: when is a combination reaction also a redox reaction? The answer is straightforward. At least one of the reactants (either A, or B, or both) must be a free element. If a reactant starts as a free element, its oxidation number is zero. When it becomes part of a compound, its oxidation number must change to some non-zero value. That change is exactly what makes the reaction redox.

Combustion: The Most Familiar Combination Redox

Every combustion reaction is a redox combination reaction because it uses elemental $O_2$ as a reactant. Let us track the oxidation numbers in the burning of carbon:

$\overset{0}{C}(s) + \overset{0}{O_2}(g) \xrightarrow{\Delta} \overset{+4 \quad -2}{CO_2}(g) \qquad \text{(7.24)}$

• Carbon starts at $0$ (free element) and ends at $+4$ in $CO_2$. Its oxidation number increases, so carbon is oxidised.
• Oxygen starts at $0$ (free element) and ends at $-2$ in $CO_2$. Its oxidation number decreases, so oxygen is reduced.

Both reactants are in the elemental form here, and both undergo oxidation-number changes. Oxygen acts as the oxidising agent (it gets reduced), and carbon acts as the reducing agent (it gets oxidised).

Combination Without Oxygen

Redox combination reactions are not limited to combustion. Any reaction where free elements combine counts. Consider magnesium burning in nitrogen gas:

$\overset{0}{3Mg}(s) + \overset{0}{N_2}(g) \xrightarrow{\Delta} \overset{+2 \quad -3}{Mg_3N_2}(s) \qquad \text{(7.25)}$

• Magnesium goes from $0$ to $+2$. Oxidation number increases, so magnesium is oxidised.
• Nitrogen goes from $0$ to $-3$. Oxidation number decreases, so nitrogen is reduced.

No oxygen is involved at all, yet this is still a clear redox reaction because both elements change oxidation number when they combine into a compound.

Combustion of a Compound: Methane Burning in Oxygen

What happens when one of the reactants is already a compound rather than a free element? Consider the combustion of methane:

$\overset{-4 \; +1}{CH_4}(g) + \overset{0}{2O_2}(g) \xrightarrow{\Delta} \overset{+4 \quad -2}{CO_2}(g) + \overset{+1 \quad -2}{2H_2O}(l)$

Let us track every atom:

• Carbon goes from $-4$ (in $CH_4$) to $+4$ (in $CO_2$). That is a large jump of 8 units upward, so carbon is oxidised.
• Oxygen goes from $0$ (in $O_2$) to $-2$ (in both $CO_2$ and $H_2O$). Oxidation number decreases, so oxygen is reduced.
• Hydrogen stays at $+1$ on both sides ($+1$ in $CH_4$ and $+1$ in $H_2O$). Its oxidation number does not change.

This is worth pausing on. Even though hydrogen is part of the reaction, it is neither oxidised nor reduced here. Only carbon and oxygen undergo oxidation-number changes. The reaction still qualifies as redox because at least some elements changed, but not every atom in a redox reaction has to participate in the electron shift. This is an important and subtle point.

Decomposition Reactions: Breaking Things Apart

Decomposition reactions are the mirror image of combination reactions. Instead of pieces joining together, a single compound breaks down into two or more simpler substances:

$C \rightarrow A + B$

The rule for when a decomposition reaction is also a redox reaction mirrors the combination rule: at least one of the products must be a free element. If a product ends up as an uncombined element (oxidation number zero), then its oxidation number must have changed from whatever it was in the starting compound, and that change makes it redox.

Water Splitting Apart

The electrolysis of water is a clean example:

$\overset{+1 \quad -2}{2H_2O}(l) \xrightarrow{\Delta} \overset{0}{2H_2}(g) + \overset{0}{O_2}(g) \qquad \text{(7.26)}$

• Hydrogen goes from $+1$ (in $H_2O$) to $0$ (in $H_2$). Oxidation number decreases, so hydrogen is reduced.
• Oxygen goes from $-2$ (in $H_2O$) to $0$ (in $O_2$). Oxidation number increases, so oxygen is oxidised.

Both products are free elements, and both undergo oxidation-number changes. Notice how the decomposition of water is the exact reverse of the formation of water from its elements, a classic combination reaction.

Sodium Hydride Breaking Down

$\overset{+1 \quad -1}{2NaH}(s) \xrightarrow{\Delta} \overset{0}{2Na}(s) + \overset{0}{H_2}(g) \qquad \text{(7.27)}$

• Sodium goes from $+1$ to $0$. It is reduced.
• Hydrogen goes from $-1$ to $0$. Its oxidation number increases, so it is oxidised.

Remember that $NaH$ is a metal hydride, one of the exceptions where hydrogen carries $-1$ instead of the usual $+1$. When the compound decomposes, hydrogen’s oxidation number rises from $-1$ to $0$, meaning hydrogen is the species being oxidised in this case.

Potassium Chlorate Decomposition

$\overset{+1 \quad +5 \quad -2}{2KClO_3}(s) \xrightarrow{\Delta} \overset{+1 \quad -1}{2KCl}(s) + \overset{0}{3O_2}(g) \qquad \text{(7.28)}$

Let us track all three elements:

• Potassium is $+1$ in $KClO_3$ and $+1$ in $KCl$. No change. Potassium is neither oxidised nor reduced.
• Chlorine goes from $+5$ (in $KClO_3$) to $-1$ (in $KCl$). Its oxidation number drops by 6 units, so chlorine is reduced.
• Oxygen goes from $-2$ (in $KClO_3$) to $0$ (in $O_2$). Its oxidation number rises, so oxygen is oxidised.

Just like the methane combustion example, not every element participates in the redox change. Potassium simply goes along for the ride, keeping its $+1$ oxidation number throughout.

When Decomposition is NOT Redox

This is an important distinction. Not every decomposition reaction involves oxidation-number changes. Consider the thermal decomposition of calcium carbonate:

$\overset{+2 \quad +4 \quad -2}{CaCO_3}(s) \xrightarrow{\Delta} \overset{+2 \quad -2}{CaO}(s) + \overset{+4 \quad -2}{CO_2}(g)$

Check every element:

• Calcium is $+2$ in $CaCO_3$ and $+2$ in $CaO$. No change.
• Carbon is $+4$ in $CaCO_3$ and $+4$ in $CO_2$. No change.
• Oxygen is $-2$ in $CaCO_3$, $-2$ in $CaO$, and $-2$ in $CO_2$. No change.

Not a single oxidation number moves. The compound breaks apart, but no electrons shift. This tells us that this decomposition is not a redox reaction. The atoms rearrange into different compounds, but they keep the same oxidation states throughout.

The critical difference? Neither product is a free element. Both $CaO$ and $CO_2$ are compounds, so there is no atom that needs to reach oxidation number zero, and no oxidation-number change is forced.

Combination vs. Decomposition: A Side-by-Side View

Feature	Combination	Decomposition
General form	$A + B \rightarrow C$	$C \rightarrow A + B$
Direction	Simpler substances join to form one product	One compound splits into simpler substances
Redox condition	At least one reactant must be a free element	At least one product must be a free element
Relationship	Forward process	Reverse process
Example (redox)	$C + O_2 \rightarrow CO_2$	$2H_2O \rightarrow 2H_2 + O_2$
Example (not redox)	Two compounds combining without element change	$CaCO_3 \rightarrow CaO + CO_2$

The pattern is symmetric. For combination, you need an element going in; for decomposition, you need an element coming out. In both cases, the element’s oxidation number changes from or to zero, and that is what makes the reaction redox.