Why an Oxidising Agent Is the One That Gets Reduced

Burning cooking gas, rusting iron, charging a phone battery and digesting food all move electrons from one substance to another. These redox reactions pair an oxidation with a reduction every time, and once you can track electrons with oxidation numbers, you can balance even complicated ones — such as the reaction that turns purple permanganate colourless in a titration flask.

This lesson covers oxidation and reduction, balancing redox equations, and identifying oxidising and reducing agents.

Oxidation is the loss of electrons, or an increase in oxidation number, and reduction is the gain of electrons, or a decrease in oxidation number — and the two always happen together.

Electron transfer. When zinc is dipped in copper sulphate solution:

$$\mathrm{Zn}(\mathrm{s})+\mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})\to \mathrm{Z}{\mathrm{n}}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$$

- Zinc loses two electrons: oxidation
- Copper ions gain two electrons: reduction

Rules for oxidation numbers:

- An element in its free state has oxidation number 0, as in ${\mathrm{O}}_2$, Na and ${\mathrm{P}}_4$
- A simple ion has an oxidation number equal to its charge
- Oxygen is usually -2, but -1 in peroxides such as ${\mathrm{H}}_2{\mathrm{O}}_2$ and +2 in $\mathrm{O}{\mathrm{F}}_2$
- Hydrogen is +1 with non-metals and -1 in metal hydrides such as NaH
- Fluorine is always -1
- Oxidation numbers add up to zero in a neutral compound, and to the charge in an ion

Worked examples:

- Manganese in $\mathrm{KMn}{\mathrm{O}}_4$: $+1+x+4(-2)=0$, so $x=+7$
- Chromium in ${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$: $2(+1)+2x+7(-2)=0$, so $x=+6$
- Sulphur in ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$: $2(+1)+x+4(-2)=0$, so $x=+6$
- Iron in $\mathrm{F}{\mathrm{e}}_3{\mathrm{O}}_4$: $3x+4(-2)=0$, so $x=+\frac{8}{3}$

An everyday example. An iron gate rusting in the monsoon is being oxidised: iron goes from 0 to +3 as it combines with oxygen and water.

The substance. A fractional oxidation number is an average — in $\mathrm{F}{\mathrm{e}}_3{\mathrm{O}}_4$, two iron atoms are +3 and one is +2, which averages to $+\frac{8}{3}$.

In the oxidation-number method you make the total increase and decrease in oxidation number equal, and in the half-reaction method you balance the electrons lost and gained in separate half-equations before adding them.

Oxidation-number method — permanganate with iron(II) in acid:

- Skeleton: $\mathrm{Mn}{\mathrm{O}}_4^{-}+\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{M}{\mathrm{n}}^{2+}+\mathrm{F}{\mathrm{e}}^{3+}$
- Manganese falls from +7 to +2, a decrease of 5; iron rises from +2 to +3, an increase of 1
- Multiply iron by 5 so the changes are equal
- Balance oxygen with water and hydrogen with ${\mathrm{H}}^{+}$:

$$\mathrm{Mn}{\mathrm{O}}_4^{-}+5F{\mathrm{e}}^{2+}+8{\mathrm{H}}^{+}\to \mathrm{M}{\mathrm{n}}^{2+}+5F{\mathrm{e}}^{3+}+4{\mathrm{H}}_2\mathrm{O}$$

Check. Charge on the left: $-1+10+8=+17$; on the right: $+2+15=+17$.

Half-reaction method — dichromate with iron(II) in acid:

- Reduction half: $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$
- Oxidation half: $\mathrm{F}{\mathrm{e}}^{2+}\to \mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}$, multiplied by 6
- Add the halves, cancelling the electrons:

$$\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+6F{\mathrm{e}}^{2+}+14{\mathrm{H}}^{+}\to 2C{\mathrm{r}}^{3+}+6F{\mathrm{e}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$$

Check. Charge on the left: $-2+12+14=+24$; on the right: $+6+18=+24$.

In basic solution. Balance as in acid, then add as many $\mathrm{O}{\mathrm{H}}^{-}$ ions to both sides as there are ${\mathrm{H}}^{+}$ ions, and combine each ${\mathrm{H}}^{+}$ and $\mathrm{O}{\mathrm{H}}^{-}$ into water.

An everyday example. Water-testing laboratories measure the iron in groundwater by titrating it against potassium permanganate, using exactly this balanced equation to turn a titre into a concentration.

The substance. A redox equation must balance charge as well as atoms — an equation with equal atoms on both sides but unequal charges is still wrong.

The oxidising agent is the substance that is reduced, gaining electrons and falling in oxidation number, while the reducing agent is the substance that is oxidised, losing electrons and rising in oxidation number.

Step by step:

- Assign oxidation numbers to every atom on both sides
- The species containing the atom whose oxidation number falls is the oxidising agent
- The species containing the atom whose oxidation number rises is the reducing agent

Worked example. In a blast furnace, $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3+3CO\to 2Fe+3C{\mathrm{O}}_2$:

- Iron falls from +3 to 0, so $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3$ is the oxidising agent
- Carbon rises from +2 to +4, so CO is the reducing agent

Common oxidising agents. $\mathrm{KMn}{\mathrm{O}}_4$, ${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$, ${\mathrm{O}}_2$, $\mathrm{C}{\mathrm{l}}_2$ and concentrated $\mathrm{HN}{\mathrm{O}}_3$ — each contains an atom in a high oxidation state or with a strong pull on electrons.

Common reducing agents. ${\mathrm{H}}_2$, C, CO, Zn, Na and ${\mathrm{H}}_2\mathrm{S}$ — each gives up electrons readily.

Both roles. Species with an element in an intermediate oxidation state, such as ${\mathrm{H}}_2{\mathrm{O}}_2$ and $\mathrm{S}{\mathrm{O}}_2$, can act as either. In disproportionation, one element is oxidised and reduced at the same time:

$$2{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$$

Oxygen goes from -1 to -2 in water and from -1 to 0 in oxygen gas.

An everyday example. Chlorine added to municipal water tanks acts as an oxidising agent, killing germs by oxidising parts of their cells.

The substance. An oxidising agent is itself reduced — the name describes what it does to its partner, not what happens to it.