Kinds of redox reaction

Five kinds of redox reaction

Once you can find oxidation numbers, you can sort redox reactions into five kinds.

Combination

Two substances join to form one, and at least one of them is a free element.

In the first, carbon goes from 0 to $+4$ and oxygen from 0 to $-2$. In the second, magnesium goes from 0 to $+2$ and nitrogen from 0 to $-3$.

Decomposition

One compound breaks into two or more substances. It is a redox reaction only when some oxidation number changes.

In the second, chlorine goes from $+5$ to $-1$ and oxygen from $-2$ to 0. Compare $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$. Calcium stays at $+2$, carbon at $+4$ and oxygen at $-2$, so this decomposition is not a redox reaction.

Displacement

An element in a compound is replaced by another element. There are three groups.

A metal displaces a metal.

This is the zinc and copper reaction of lesson 1. A more active metal takes the place of a less active one.

A metal displaces hydrogen.

Very active metals such as sodium and calcium react even with cold water. Less active ones, such as magnesium and iron, need steam or an acid. Copper, silver and gold do not displace hydrogen at all.

A halogen displaces a halogen.

The oxidising power of the halogens falls in the order $\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2$. A halogen displaces a lower one from its salt. Fluorine is so strong that it even takes oxygen out of water: $2\text{F}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{HF} + \text{O}_2$.

Disproportionation

In a disproportionation, one element in one oxidation state is both oxidised and reduced in the same reaction.

Oxygen starts at $-1$. It goes down to $-2$ in water and up to 0 in oxygen gas.

Chlorine gives a pair of examples. In cold dilute alkali:

The chlorine goes from 0 to $-1$ and to $+1$. In hot concentrated alkali:

Now the chlorine goes from 0 to $-1$ and to $+5$. Each atom that becomes $+5$ loses 5 electrons. Each atom that becomes $-1$ gains only 1. So 5 atoms must be reduced for each one oxidised.

Set the oxidation number of the oxidised chlorine to 1, then to 5. How many atoms are reduced for each atom oxidised? Do the electrons lost and gained stay equal?

An element can disproportionate only if it has a state above and a state below the one it starts in. Fluorine cannot. It has only $-1$ in compounds and 0 as the element, so it has nothing higher to go to.

Comproportionation

This is the opposite change. Two species hold the same element in different oxidation states, and both turn into one state in between.

Sulfur is $-2$ in hydrogen sulfide and $+4$ in sulfur dioxide. Both end as sulfur at 0. Another example is the decomposition of ammonium nitrate:

Nitrogen is $-3$ in the ammonium ion and $+5$ in the nitrate ion. Both end as $+1$ in $\text{N}_2\text{O}$.

Sorting a reaction

1. Find the oxidation numbers of every element on both sides.
2. If none changes, it is not a redox reaction.
3. If the same element goes both up and down from one starting state, it is a disproportionation.
4. If two states of one element become one, it is a comproportionation.
5. Otherwise, decide between combination, decomposition and displacement from the shape of the equation.

The limits of oxidation numbers

• The number assumes that the bonding electrons belong wholly to the more electronegative atom. That is only true for ionic bonds. In covalent compounds the charge is not real.
• A fractional value is an average. The formula alone does not tell you which atoms differ.
• An oxidation number is not the same as the number of bonds. Oxygen in $\text{H}_2\text{O}_2$ has two bonds, but its oxidation number is $-1$.
• The rules are a tool for bookkeeping. They are not a picture of what the molecule looks like.

Practise with new numbers

Try it in Python

The program decides from the oxidation numbers of one element, before and after, whether it is oxidised, reduced or disproportionated.

Common mistakes

• Calling every decomposition a redox reaction. Check the oxidation numbers first.
• Thinking disproportionation needs two different elements. It needs one element in one starting state.
• Mixing up disproportionation and comproportionation. One goes from one state to two. The other goes from two to one.
• Forgetting that a displacement is also a redox reaction. An element is oxidised and another is reduced.

Remember these

• The five kinds are combination, decomposition, displacement, disproportionation and comproportionation.
• A displacement or a combination with a free element is always a redox reaction.
• In a disproportionation one element goes up and down together.
• Oxidation numbers are a bookkeeping tool, and fractional values are averages.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What kind of reaction is $2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$?

2. Why can fluorine not disproportionate?

3. In $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$, which oxidation numbers does oxygen take?

4. Is $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ a redox reaction?