Balancing Redox Reactions by the Ion-Electron (Half-Reaction) Method

The ion-electron method (also called the half-reaction method) is the more systematic and more widely applicable of the two balancing techniques, and it is the one that generalizes cleanly to both acidic and basic media. It works by treating oxidation and reduction as two entirely separate, independently balanced equations, joined only at the final step.

The procedure for acidic medium:

1. Split the skeletal ionic equation into two half-reactions: one containing the species that is oxidized, one containing the species that is reduced.
2. In each half-reaction, balance every atom except O and H first.
3. Balance oxygen by adding $\text{H}_2\text{O}$ to whichever side is short of oxygen.
4. Balance hydrogen by adding $\text{H}^{+}$ to whichever side is short of hydrogen.
5. Balance the charge in each half-reaction by adding electrons ($e^{-}$) to the more positive side.
6. Multiply each half-reaction by the smallest whole number needed so that both half-reactions involve the same number of electrons, then add them together, cancelling the electrons and any species that appear identically on both sides.

A complete, six-step worked example of this exact procedure, applied to $\text{MnO}_4^{-} + \text{C}_2\text{O}_4^{2-} \to \text{Mn}^{2+} + \text{CO}_2$ (the reaction underlying permanganometry against oxalic acid, covered later in this chapter), is given in the reference note alongside this section.

Basic (alkaline) medium is handled with one extra step appended to the same procedure: balance the skeletal equation exactly as if the medium were acidic (steps 1–6 above), then for every $\text{H}^{+}$ that appears in the final combined equation, add an equal number of $\text{OH}^{-}$ to both sides — the $\text{H}^{+}$ and the newly added $\text{OH}^{-}$ on the same side combine to $\text{H}_2\text{O}$, which is then simplified against any $\text{H}_2\text{O}$ already present on the opposite side. This trick works because adding equal amounts of $\text{OH}^{-}$ to both sides of a correctly balanced equation cannot change its balance, while it converts every stray $\text{H}^{+}$ (which should not appear in a basic solution) into water.

For example, balancing $\text{MnO}_4^{-} + \text{I}^{-} \to \text{MnO}_2 + \text{I}_2$ (basic medium) using the direct basic-medium versions of the half-reactions gives:

$2\text{I}^{-} \to \text{I}_2 + 2e^{-} \quad \text{(oxidation)}$ …