Half Equations at Electrodes

An electron is written as $e^-$ and has a charge of $-1$. Unlike a complete chemical equation, a half-equation includes electrons because it shows just the gain or loss occurring at one electrode.

Two things must balance:

• Atoms: the number of atoms of each element must be the same on both sides.
• Charge: the total electrical charge must be the same on both sides. It does not have to be zero.

To write a half-equation, start with the correct formulae and charges of the reacting ion and its products. Balance the atoms using numbers in front of formulae, then add electrons to balance the charge.

Reduction: copper ions gain electrons

When copper(II) ions form copper at the cathode, start with:

$$\mathrm{Cu^{2+} \rightarrow Cu}$$

There is already one copper atom on each side. However, the left side has a charge of $+2$, while the copper metal on the right is neutral. Adding two electrons to the left balances the charge:

$$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$$

The total charge on the left is $+2 + 2(-1) = 0$, matching the right. Each copper(II) ion gains two electrons, so this is reduction.

Oxidation: bromide ions lose electrons

During electrolysis of molten lead(II) bromide, bromide ions form bromine at the anode. Bromine consists of molecules with the formula $\mathrm{Br_2}$, so two bromide ions are needed:

$$\mathrm{2Br^- \rightarrow Br_2}$$

The atoms balance, but the left side has a total charge of $-2$ and the right side is neutral. Each bromide ion loses one electron, giving:

$$\mathrm{2Br^- \rightarrow Br_2 + 2e^-}$$

Both sides now have a total charge of $-2$. The electrons appear as products because they are lost by the ions: this is oxidation.

Hydrogen formation at the cathode

When hydrogen ions are discharged, they gain electrons and form hydrogen gas. Hydrogen has the molecular formula $\mathrm{H_2}$, so two hydrogen ions are required. Each gains one electron:

$$\mathrm{2H^+ + 2e^- \rightarrow H_2}$$

There are two hydrogen atoms on each side. The left-hand charge is $+2 - 2 = 0$, matching the neutral hydrogen molecule. This is another reduction reaction.

Oxygen formation at the anode

When hydroxide ions are discharged to form oxygen, water is also produced. The balanced half-equation is:

$$\mathrm{4OH^- \rightarrow O_2 + 2H_2O + 4e^-}$$

The four hydroxide ions contain four oxygen atoms and four hydrogen atoms. On the right, $\mathrm{O_2}$ contains two oxygen atoms, while $\mathrm{2H_2O}$ contains the other two oxygen atoms and all four hydrogen atoms.

The total charge on the left is $-4$. Oxygen and water are neutral, but the four electrons on the right give a total charge of $-4$, so charge also balances. Electrons are lost, making this oxidation.

Electron loss can alternatively be written by subtracting electrons on the left:

$$\mathrm{4OH^- - 4e^- \rightarrow O_2 + 2H_2O}$$

This represents the same change: four electrons are removed from the hydroxide ions. Writing the electrons as products usually makes their loss easier to see.