1.1.3 Redox Reaction in Terms of Change of Oxidation Number
Oxidation States (Oxidation Numbers)
- Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state.
Oxidation State of Some Elements
1. The oxidation state of an element is zero.
Example
Element | Oxidation State |
Mg | 0 |
H2 | 0 |
Br2 | 0 |
2. For a simple ion with single atom, the oxidation state is equal to the charge.
Example
| Ion | Oxidation State |
| Cu2+ | +2 |
| Br– | -1 |
| O2- | -2 |
| Al3+ | +3 |
3. Some elements almost always have the same oxidation states in their compounds:
Example 1: The oxidation state of oxygen is always -2 except peroxide, which is -1.
| Compound | Oxidation state of oxygen |
| H2O | -2 |
| H2SO4 | -2 |
| ZnO | -2 |
| KClO3 | -2 |
| H2O2 | -1 |
Example 2: The oxidation state of hydrogen is always +1 except hydride, which is -1.
| Compound | Oxidation state of hydrogen |
| NH3 | +1 |
| HCl | +1 |
| NaOH | +1 |
| MgH2 | -1 |
| NaH | -1 |
4. The sum of the oxidation states of all the atoms or molecule in a neutral compound is zero.
Example:
| Ion | Sum of Oxidation State |
| H2O | 0 |
| CO2 | 0 |
| NH3 | 0 |
5. The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
Example:
| Ion | Sum of Oxidation State |
| NO3– | -1 |
| CO32- | -2 |
| PO43- | -3 |
| NH4+ | +1 |
Working Out the Unknown Oxidation State of an Element in a Compound
- The sum of the oxidation state of each element in a compound are equal to the charge of the compound.
- This rule can be used to find the unknown oxidation number of an element is a compound.
Example 1
Find the oxidation state of all the elements in a Chlorate(V), ClO3– ion.
Answer:
Oxidation number of O = -2
Oxidation number of Cl = x
Example 2
Find the oxidation state of all the elements in a Potassium manganate(VII), KMnO4 ion.
Answer:
Oxidation number of K = +1`
Oxidation number of O = -2
Oxidation number of Mn = x
Example 3
Find the oxidation state of all the elements in an Ammonium ion, NH4+ ion.
Answer:
Oxidation number of H = +1
Oxidation number of N = x
Using the Oxidation States in Naming Compounds
- You will have come across names like iron(II) sulphate and iron(III) chloride. The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe2+ and Fe3+ ions.
Example
Formula
Name of the compound FeCl2 Iron(II) chloride FeCl3 Iron (III) chloride MnO2 Manganese(IV) oxide Mn(NO3)2 Manganese (II) nitrate PbCl2 Lead(II) chloride - Transition metals always show difference oxidation state as shown in the table below.
Metal
Oxidation state
Fe +2, +3 Cu +1, +2 Mn +2, +4, +6, +7 - Non-metal elements (except fluorine) usually have more than one oxidation state.
Oxidation and Reduction in Term of Changes of Oxidation State
Oxidation involves an increase in oxidation state
Reduction involves a decrease in oxidation state
- Another way to determine oxidation and reduction is to see the change of the oxidation state after a reaction.
- An atom is said to be oxidised when its oxidation state increases.
- An atom is said to be reduced when its oxidation state decreases.
Example:
- The magnesium’s oxidation state has increased by 2, from 0 to +2. Therefore, it has been oxidised.
- The hydrogen’s oxidation state has decreased by 1, from +1 to 0. Therefore it has been reduced.
- The chlorine is in the same oxidation state on both sides of the equation – it hasn’t been oxidised nor reduced.
Example:
There is no change of oxidation state for all elements. This isn’t a redox reaction.
Example:
- In this example, we can see that the oxidation state of chlorine has increased and also decreased.
- Chlorine is oxidised and reduced, at the same time.
- This is a good example of a disproportionation reaction. A disproportionation reaction is one in which a single substance is both oxidised and reduced.