Oxidation and reduction
Francine Taylor-Campbell, Contributor

St. Mary High School defeated Jamaica College to emerge victors of the 2005 National Schools' Debate competition held at the LOJ Auditorium, New Kingston. - File

POINTS TO NOTE

* 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.

1. Oxidation can be defined as the gain of oxygen by a substance. E.g., 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.

Q1 In the reaction 2H2SO3 (aq) + O2 (g) = 2H2SO4 (aq), give the formula of the substance which is oxidised and the product of the oxidation.

2. Oxidation can also occur 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. In chemical reactions electrons are usually transferred from one substance to another. When a substance is oxidised it loses electrons, hence, oxidation can be defined as a process in which electrons are lost.

E.g., In reaction one, 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+O²- or 2MgO (s)

This concept is not so easily applied to covalent compounds, e.g., in reactions two and three.

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 is 4.

The oxidation numbers are +2 and +4 ie, the oxidation number of carbon increases from two to four. 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 is 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.

For example, 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. SO4²- Ox # of O = --2, thus 4 x -2 = --8 S = +6

since -8 + 6 = --2

Q2 Calculate the oxidation # of the marked atoms.
i. *Fe2O3
ii. *C2H6
iii. K*ClO3
iv.*CrO4-
v. *Cr2O7²-
vi. Ca(H*SO3)2

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

E.g., CuO Copper II oxide Cu2O Copper I oxide
FeCl3 Iron III chloride FeCl2 Iron II chloride
H2SO4 Sulphuric Acid VI H2SO3 Sulphuric Acid IV
KMnO4 Potassium manganate VII HClO Chloric Acid I

Q3 Write the names of
i. CuSO4
ii. PbO2
iii. Cr2O7²-
iv. Fe2O3
v. K2MnO4
vi. HNO3
vii. HNO2
viii. FeSO4

Reduction 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 is an independent contributor.