Redox and Electrons

• oxidation is the loss of electrons;
• reduction is the gain of electrons.

The memory aid OIL RIG summarises this: Oxidation Is Loss; Reduction Is Gain.

Oxidation and reduction must occur together. An electron lost by one species has to be gained by another, so the complete change is called a redox reaction. The species losing electrons is oxidised, while the species gaining them is reduced.

For example, when a magnesium atom forms a magnesium ion, it loses two electrons:

$\mathrm{Mg(s) \rightarrow Mg^{2+}(aq) + 2e^-}$

Magnesium is therefore oxidised. A copper(II) ion can gain those two electrons to become a copper atom:

$\mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)}$

The copper(II) ions are therefore reduced. Notice that the charge is balanced in each half equation as well as the number of atoms.

Why displacement reactions involve electron transfer

A more reactive metal has a greater tendency to form positive ions by losing electrons. If magnesium is placed in copper(II) sulfate solution, magnesium is more reactive than copper, so it displaces copper from the compound:

$\mathrm{Mg(s) + CuSO_4(aq) \rightarrow MgSO_4(aq) + Cu(s)}$

At particle level, magnesium atoms lose electrons and enter the solution as $\mathrm{Mg^{2+}}$ ions. Copper(II) ions gain those electrons and form copper atoms. The sulfate ions remain unchanged.

The two electron changes occur simultaneously. Adding the half equations together cancels the two electrons because the number lost equals the number gained:

$\mathrm{Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)}$

This is the ionic equation for the displacement reaction. It shows only the species that undergo the chemical change.

From a symbol equation to an ionic equation

An ionic equation removes ions that are present but do not change. These unchanged ions are called spectator ions.

Begin with the balanced symbol equation:

$\mathrm{Mg(s) + CuSO_4(aq) \rightarrow MgSO_4(aq) + Cu(s)}$

Separate the dissolved ionic compounds into their ions. Do not split solid metals into ions:

$\mathrm{Mg(s) + Cu^{2+}(aq) + SO_4^{2-}(aq) \rightarrow Mg^{2+}(aq) + SO_4^{2-}(aq) + Cu(s)}$

The $\mathrm{SO_4^{2-}}$ ions appear unchanged on both sides, so cancel them. This leaves:

$\mathrm{Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)}$

A correct ionic equation must conserve both matter and charge. Here there is one magnesium atom and one copper atom on each side, while the total charge is $2+$ on each side.

Identifying what is oxidised and reduced

Use the movement of electrons rather than relying only on where a substance appears in the overall equation.

In

$\mathrm{Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)},$

magnesium changes from a neutral atom to a $2+$ ion. It must lose two electrons, so magnesium is oxidised. The copper(II) ion changes from a $2+$ ion to a neutral atom. It must gain two electrons, so copper(II) ions are reduced.

The same reasoning works when a question supplies half equations directly:

$\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-}$

Electrons are products, so zinc loses them and is oxidised.

$\mathrm{Ag^+(aq) + e^- \rightarrow Ag(s)}$

Electrons are reactants, so silver ions gain them and are reduced.

If the numbers of electrons in two half equations differ, multiply one or both complete half equations until equal numbers are lost and gained. The electrons can then cancel when the equations are combined. Never alter the formula or charge of a species merely to make the electron numbers match.