Half Equations

A half equation represents the change at just one electrode. It names the reacting ion or atom, the product formed and the electrons transferred. A complete electrolysis process therefore has two half equations: one for the reaction at the cathode and one for the reaction at the anode.

Every half equation must conserve both atoms and electric charge. Electrons may be added to one side to balance charge, but they do not balance atoms.

Writing cathode half equations

The cathode is connected to the negative terminal of the direct-current supply. Positively charged ions arrive there and gain electrons. The electrons therefore appear on the left-hand side of a cathode half equation.

For a monatomic metal ion, the size of its positive charge tells you how many electrons it needs to become a neutral atom:

$Na^+ + e^- \rightarrow Na$

$Pb^{2+} + 2e^- \rightarrow Pb$

$Al^{3+} + 3e^- \rightarrow Al$

Consider the lead equation. The total charge on the left is $+2 + 2(-1)=0$, matching the charge of the neutral lead atom on the right. There is also one lead atom on each side.

Some elements form molecules containing more than one atom. Hydrogen is diatomic, so two hydrogen ions are needed to make one $H_2$ molecule:

$2H^+ + 2e^- \rightarrow H_2$

There are two hydrogen atoms on each side, while the total charge on the left is $+2-2=0$, matching the neutral product.

Writing anode half equations

The anode is connected to the positive terminal. A species reacting there loses electrons, so electrons appear on the right-hand side of its half equation.

Halogens form diatomic molecules. Two chloride ions are therefore required to make one chlorine molecule:

$2Cl^- \rightarrow Cl_2 + 2e^-$

The atoms balance: there are two chlorine atoms on each side. The left-hand charge is $-2$ and the right-hand charge from two electrons is also $-2$.

The same pattern applies to bromide ions during the electrolysis of molten lead bromide:

$2Br^- \rightarrow Br_2 + 2e^-$

Oxide ions can form oxygen molecules:

$2O^{2-} \rightarrow O_2 + 4e^-$

Two oxide ions have a total charge of $-4$, so four electrons must appear on the right. The two oxygen atoms form one $O_2$ molecule.

In an aqueous electrolyte, hydroxide ions may form oxygen and water at an inert anode:

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

Count the atoms first: both sides contain four oxygen atoms and four hydrogen atoms. Then check charge: the left is $-4$ and the four electrons give the right a charge of $-4$.

An anode reaction does not always begin with an anion. When a copper anode takes part in electrolysis, copper atoms leave the electrode as copper ions:

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

The neutral copper atom has charge zero. On the right, $+2$ from the copper ion and $-2$ from the electrons also total zero.

A reliable method

Use this sequence whenever you write a half equation:

1. Write the formula of the reacting particle and the product.
2. Balance the number of atoms using coefficients in front of the formulae.
3. Calculate the total charge on each side.
4. Add electrons to the more positive side until the charges are equal.
5. Check that both atoms and charge balance.

For example, begin an iodide equation as $I^- \rightarrow I_2$. Balancing iodine gives $2I^- \rightarrow I_2$. The left now has charge $-2$, so place two electrons on the right:

$2I^- \rightarrow I_2 + 2e^-$

Never alter the formula $I_2$ or change the charge on $I^-$ merely to make the equation balance.

Linking the two electrodes

The half equations show that electrons released by the reaction at the anode pass into the external circuit, while the cathode reaction receives electrons from the external circuit. Ions—not electrons—carry charge through the molten electrolyte.

For molten lead bromide, the electrode equations are:

$Pb^{2+} + 2e^- \rightarrow Pb$

$2Br^- \rightarrow Br_2 + 2e^-$

The same number of electrons appears in both equations. If the half equations were combined, those electrons would cancel because they are transferred rather than created or destroyed.