The half-reaction method

A battery on paper

A battery has two ends. One end pushes electrons into the wire, and the other end takes them back. A redox reaction is also two jobs done together. One species loses electrons and another gains them.

The half-reaction method writes the two jobs on separate lines. Each line is a half-reaction. We balance each half on its own and then join them. It is the most reliable method for reactions in acid or in base.

The steps in acid

For one half-reaction in acidic solution:

1. Balance every atom except oxygen and hydrogen.
2. Balance oxygen by adding $\text{H}_2\text{O}$ to the side that needs it.
3. Balance hydrogen by adding $\text{H}^+$ to the side that needs it.
4. Balance charge by adding electrons to the side with the more positive charge.

Then, for the whole reaction:

5. Multiply each half so that the electrons are equal.
6. Add the two halves, and cancel what appears on both sides.
7. Check atoms and charge.

Worked example 1: dichromate and iron(II)

Orange dichromate turns green when it meets iron(II) in acid. The two halves are:

Chromium half. Two chromium atoms are balanced. Add $7\ \text{H}_2\text{O}$ on the right for the oxygen. Add $14\ \text{H}^+$ on the left for the hydrogen. The charge is $-2 + 14 = +12$ on the left and $+6$ on the right, so add $6e^-$ on the left:

Iron half. The charge is $+2$ on the left and $+3$ on the right, so the right side takes one electron:

Join them. The chromium half takes 6 electrons, so multiply the iron half by 6. The electrons cancel:

Check. Charge: $-2 + 14 + 12 = +24$ on the left, and $6 + 18 = +24$ on the right. Oxygen: 7 and 7. Hydrogen: 14 and 14.

Practise with new numbers

Hydrogen peroxide plays both parts

Hydrogen peroxide can be an oxidising agent or a reducing agent. The oxygen in it has an oxidation number of $-1$, between $-2$ in water and $0$ in $\text{O}_2$.

• As an oxidising agent in acid: $\text{H}_2\text{O}_2 + 2\text{H}^+ + 2e^- \rightarrow 2\text{H}_2\text{O}$
• As a reducing agent: $\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2e^-$

With permanganate it is the reducing agent. Multiply the permanganate half by 2 and the peroxide half by 5, so that both carry 10 electrons:

Ten $\text{H}^+$ appear on both sides, so six are left on the left:

Reactions in base

A basic solution has almost no $\text{H}^+$. So we balance as if in acid, and then remove the $\text{H}^+$:

1. Balance the half-reaction as in acid.
2. Add to both sides as many $\text{OH}^-$ as there are $\text{H}^+$.
3. On the side with $\text{H}^+$ and $\text{OH}^-$ together, combine them into $\text{H}_2\text{O}$.
4. Cancel the water that now appears on both sides.

Example: permanganate with iodide, giving brown manganese dioxide. In acid, the permanganate half is $\text{MnO}_4^- + 4\text{H}^+ + 3e^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}$. Add $4\ \text{OH}^-$ to both sides. The four $\text{H}^+$ and four $\text{OH}^-$ on the left make $4\ \text{H}_2\text{O}$. Cancel two water molecules:

The iodide half is $2\text{I}^- \rightarrow \text{I}_2 + 2e^-$. The smallest number of electrons that both 3 and 2 divide into is 6. So multiply the first half by 2 and the second by 3:

Chlorine splits in two directions

When chlorine passes into hot concentrated alkali, some chlorine atoms are reduced to chloride and others are oxidised to chlorate. This is a disproportionation. We write the two halves in base:

Both halves must carry 10 electrons, so multiply the first by 5. Adding gives $6\text{Cl}_2 + 12\text{OH}^- \rightarrow 10\text{Cl}^- + 2\text{ClO}_3^- + 6\text{H}_2\text{O}$. Every coefficient divides by 2:

Of every six chlorine atoms, five are reduced and one is oxidised.

Try it in Python

The program balances an ionic equation. It writes one row for each element and one for charge, and then solves for the coefficients.

Common mistakes

• Forgetting to multiply the electrons. When you multiply a half-reaction, multiply every term, electrons included.
• Leaving electrons in the final equation. In the sum they must cancel completely.
• Using $\text{OH}^-$ in an acid. Acid has almost no hydroxide. Use $\text{H}^+$ and water.
• Adding $\text{OH}^-$ to only one side. In base, add the same number to both sides.
• Putting the electrons on the wrong side. They go on the side with the more positive charge.

Remember these

• Acid: balance atoms, then oxygen with $\text{H}_2\text{O}$, hydrogen with $\text{H}^+$, charge with electrons.
• Make the electrons equal, add the halves, and cancel.
• Base: balance as in acid, add $\text{OH}^-$ to both sides, combine with $\text{H}^+$ into water, and cancel.
• Hydrogen peroxide is an oxidising agent or a reducing agent, depending on its partner.
• In a disproportionation, one element is both oxidised and reduced.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Balance the half-reaction $\text{Sn}^{2+} \rightarrow \text{Sn}^{4+}$ for charge.

2. In acid, the sulfate ion becomes sulfur dioxide: $\text{SO}_4^{2-} \rightarrow \text{SO}_2$. How many $\text{H}^+$ and how many electrons are needed?

3. Why does a half-reaction in acid never show $\text{OH}^-$?

4. A half-reaction balanced in acid has $6\ \text{H}^+$ on the left. What do you do to write it for base?

5. In $3\text{Cl}_2 + 6\text{OH}^- \rightarrow 5\text{Cl}^- + \text{ClO}_3^- + 3\text{H}_2\text{O}$, which chlorine atoms are oxidised, and which are reduced?