Oxidation and reduction
Francine Taylor-Campbell, Contributor

We will start this topic by first focusing on the basics that need to be known on oxidation and reduction. Use this lesson to review the important facts so that you can answer the questions.

REVIEW OF FACTS

• A reaction in which oxidation and reduction takes place is called a REDOX reaction.
• Oxidation and reduction can be defined in terms of loss and gain of electrons, oxygen and hydrogen and a change in oxidation number.

Oxidation was originally defined as the gain of oxygen by a substance.

Example

When magnesium burns in oxygen,

2Mg (s) + O₂ (g) = 2MgO ................1
magnesium is oxidised as it gains oxygen to form magnesium oxide

OR

when carbon monoxide burns
2CO (g) + O₂ (g) = 2CO₂ (g) ......................2
it is oxidised to carbon dioxide.

1. In the reaction 2H₂SO₃ (aq) + O₂ (g) = 2H₂SO₄ (aq) give the formulae of the substance which is oxidised and the product of the oxidation.

2. It also occurs when hydrogen is lost by a substance.

E.g. When ammonia burns in pure oxygen
4NH₃ (g) + 3O₂ (g) = 2N₂ (g) + 6H₂O (g) ..................3
Ammonia loses hydrogen and is oxidised to nitrogen.

3. When electronic structures became fully understood, it was realised that when a substance was oxidised it lost electrons and oxidation was defined as a process in which electrons were lost.

Example

In reaction 1, magnesium atoms lose electrons to form magnesium ions

2Mg (s) = 2Mg²⁺(s) + 4e ...................A
These electrons are gained by oxygen molecules
O₂ (g) + 4e = 2O2- (s) ....................B

On adding A and B we get 2Mg (s) + O2 (g) = 2Mg²⁺O^2- or 2MgO (s)
This concept is not so easily applied to covalent compounds e.g. in reactions 2 and 3.

4. The concept of oxidation numbers overcomes this problem.
Oxidation number is numerically the same as valency but has a +ve or a -ve sign.

Thus, in the reaction

2CO (g) + O2 (g) = 2CO2 (g) .................4
The valency of carbon in carbon monoxide is 2 and in carbon dioxide it is 4.

The oxidation numbers are +2 and +4 i.e., the oxidation number of carbon increases from 2 to 4. Thus, oxidation is a process involving an increase in oxidation number.

In the reaction

4NH₃ (g) + 3O₂ (g) = 2N₂ (g) + 3H₂O (g)

The oxidation number of nitrogen in ammonia is -3 and in nitrogen is zero. It increases from -3 to 0.

We now need to explain more fully how oxidation numbers are determined. The following rules apply.

1. The oxidation number of uncombined elements is zero. E.g. H in H2 and Al the oxidation number is 0.

2. The oxidation number of hydrogen in compounds is +1 (except in metallic hydrides e.g. CaH2 when it is -1).

3. The oxidation number of oxygen in its compounds is -2.

4. Some of the other elements may have more than one oxidation number.

5. The oxidation number of elements in simple ions is the same as the charge on the ion, e.g. in FeCl3 Fe₃+ ox # = +3 Cl- ox # is -1.

6. The oxidation numbers of elements in covalent compounds are numerically the same as the valency. E.g. CH₄ C = —4 H = +1

7. The sum of the oxidation numbers of the atoms in a compound is zero. E.g. in NH₄Cl N = —3 H = +1 Cl = —1 —3 + (4 x +1) -1 = 0

8. The sum of the oxidation numbers of the atoms in an ion is equal to the charge on the ion.
E.g. NO₃- O = —2 N = +5 +5 + (3 x -2) = —1

CALCULATION OF OXIDATION NUMBERS FROM FORMULAE

If all but one of the oxidation numbers in a formula unit are known, the unknown oxidation number can be calculated.

Examples

i. MnO₂ 2 x Ox # of O = —4 Ox # of Mn = +4

ii. CH₄ Ox # of C = —4

iii. CO₂ Ox # of C = +4

iv. SO₄^2- Ox # of O = —2, thus 4 x -2 = —8 S = +6

since -8 + 6 = —2

Q2. Calculate the oxidation number of the marked atoms.

i. *Fe₂O₃

ii. *C₂H₆

iii. K*ClO₃

iv.*CrO₄-

v. *Cr₂O₇^2-

vi. Ca(H*SO₃)₂

If an element has more than one oxidation number, this may be included in the name of the compound.

Example

Q3. Write the names of

i. CuSO₄

ii. PbO₂

iii. Cr₂O₇^2-

iv. Fe₂O₃

v. K₂MnO₄

vi. HNO₃

vii. HNO₂

viii. FeSO₄

Reducation is the converse of oxidation. It is a process resulting in the:

i. Loss of oxygen e.g. CuO(s) + H₂(g) = Cu(s) + H₂O(l)
Copper oxide is reduced to copper by the loss of oxygen.

ii. Addition of hydrogen e.g. H₂(g) + Cl₂(g) = 2HCl(g) Chlorine is reduced to hydrogen chloride

iii. Gain of electrons e.g. Cu²⁺ + 2e = Cu ..................5

iv. Decrease in oxidation number e.g. in 5 the oxidation number of copper decreases from +2 in CuO to 0 in Cu.

Francine Taylor-Campbell teaches at Jamaica College. Send questions and comments to kerry-ann.hepburn@gleanerjm.com