Balancing half equations in acidic solution

Why the simple method is not enough

• Half equations like $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ are easy because only one element changes.

• Many important oxidants contain oxygen — MnO₄⁻, Cr₂O₇²⁻, IO₃⁻, NO₃⁻, H₂O₂. When these are reduced, the oxygen has to go somewhere.

• In acidic solution, that oxygen leaves as water, and the hydrogen for the water comes from the H⁺ ions in the acid.

• This is why these reactions are always done in acid. Without H⁺ there is nothing to mop up the oxygen.

The five-step method

Use this order every time. It never fails and it never needs guessing.

• 1. Balance the element being oxidised or reduced.
• 2. Balance the oxygen by adding H₂O to the side short of oxygen.
• 3. Balance the hydrogen by adding H⁺ to the side short of hydrogen.
• 4. Balance the charge by adding e⁻ to the more positive side.
• 5. Check the atoms and the charge.

• The order matters. If you add electrons before the H⁺, you will get the wrong number every time.

Worked through: the permanganate half equation

This is the one you will use most, so learn the process on it.

Step 1 — balance manganese. One Mn on each side already:

$$\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$$

Step 2 — balance oxygen. Four O on the left, none on the right, so add 4H₂O to the right:

$$\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

Step 3 — balance hydrogen. Eight H on the right now, none on the left, so add 8H⁺ to the left:

$$\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

Step 4 — balance charge. Left is $(-1) + (+8) = +7$; right is $+2$. The left is more positive by 5, so add 5e⁻ to the left:

$$\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$$

Step 5 — check. Mn: 1 = 1. O: 4 = 4. H: 8 = 8. Charge: $+7 - 5 = +2$ on the left, $+2$ on the right. Correct.

• The 5 electrons agree with the oxidation-number change: Mn goes from +7 to +2, a fall of 5.

The half equations worth knowing

You can derive all of these with the five steps, but knowing them saves time:

Species	Half equation	Electrons
Permanganate in acid	$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$	5
Dichromate in acid	$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$	6
Peroxide as oxidant	$\text{H}_2\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow 2\text{H}_2\text{O}$	2
Peroxide as reductant	$\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2e^-$	2
Iodate in acid	$2\text{IO}_3^- + 12\text{H}^+ + 10e^- \rightarrow \text{I}_2 + 6\text{H}_2\text{O}$	10
Hypochlorite in acid	$\text{OCl}^- + 2\text{H}^+ + 2e^- \rightarrow \text{Cl}^- + \text{H}_2\text{O}$	2
Concentrated nitric acid	$\text{NO}_3^- + 2\text{H}^+ + e^- \rightarrow \text{NO}_2 + \text{H}_2\text{O}$	1
Sulfite oxidised	$\text{SO}_3^{2-} + \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 2\text{H}^+ + 2e^-$	2
Sulfur dioxide oxidised	$\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ + 2e^-$	2
Hydrogen sulfide oxidised	$\text{H}_2\text{S} \rightarrow \text{S} + 2\text{H}^+ + 2e^-$	2

• Notice that the oxidation half equations put H⁺ on the right — the same five steps, run on a species that is gaining oxygen from water rather than losing it.