Oxidation is the loss of electrons, reduction is the gain of electrons.

There are also some other definition that can be used to describe reduction and oxidation. They are summarised nicely in the table below:

An example of an oxidation half equation… Mg Mg^{2+} + 2e^{–}.

Reduction half equation… O + 2e^{–} O^{2-}.

### Calculation of oxidation number

There are a few rules to remember

- The oxidation number of an element is zero.
- The oxidation number of an ion is equal to the charge of the ion.
- Hydrogen has an oxidation number of +1 (this doesn’t apply to hydrides when hydrogen is -1).
- Oxygen has an oxidation number of -2 (except in peroxides when it is -1).

It is then simply a matter of adding together all of the oxidation numbers of the elements in a compound and making sure that the total is = 0 if the compound is neutral, otherwise they need to add to the charge of the ion.

### Balancing redox half equations

## Example:

Iron(II) ions are oxidised to iron(III) ions, by reaction with acidified potassium permanganate.

This can be worked out in a number of steps. In this example:

MnO_{4}^{–} → Mn^{2+}

**Step 1** Balance all the elements, except hydrogen and oxygen. In this case, there are the same number of ‘Mn’ on each side of the equation (one):

MnO_{4}^{–} → Mn^{2+}

**Step 2** If oxygen is unbalanced, add H_{2}O to the side short of oxygen; in this case, the right-hand side (RHS). Multiply until the number of oxygens on each side is the same:

MnO_{4}^{–} → Mn^{2+} + 4H_{2O}

**Step 3** If hydrogen is unbalanced, add H^{+} to the side short of hydrogen; in this case, the left-hand side (LHS). Multiply until the number of hydrogens on each side is the same:

MnO_{4}^{–} + 8H^{+} → Mn^{2+} + 4H_{2}O

**Step 4** The total charge must be the same on both sides of the equation. Only electrons can be added to one side or the other. Balance the charge, by adding the appropriate number of electrons to the side which is short of negative charge; in this case the LHS:

MnO_{4}^{–} + 8H^{+} + 5e^{–}→ Mn^{2+} + 4H_{2}O

The charge on the left-hand side = (one negative) + (eight positive) + (five negative) = two positive.

The charge on the right-hand side = (two positive) + (no charge) = two positive.

*And for the iron ions…*

Fe^{2+} → Fe^{3+}

The iron is balanced on both sides, so there is no need to add any other species. So we just need to balance the charges by adding electrons to the appropriate side.

Fe^{2+} → Fe^{3+} + e^{–}

We would know that something had gone wrong with our half equations if the electrons had both ended up being on the same side (all good here).

**Adding the half equations together to create a balanced overall redox equation**

**Step 5** The number of electrons must be the same for both half equations, so you need to multiple the half equations to obtain the same number of electrons. In this case we only need to multiple the iron half equation by 5.

5Fe^{2+} → 5Fe^{3+} + 5e^{–}

**Step 6** Add the half equations together and cancel the electrons (and any other species if required).

5Fe^{2+} → 5Fe^{3+} + 5e^{–}

MnO_{4}^{–} + 8H^{+} + 5e^{–}→ Mn^{2+} + 4H_{2}O

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5Fe^{2+} + MnO_{4}^{–} + 8H^{+} → 5Fe^{3+} + Mn^{2+} + 4H_{2}O

You can check if you have done it correctly by making sure that the atoms and charges are balanced on both sides…. in this case they are!

All done. A balanced overall redox equation. Not so hard after all…?