Oxidation and reduction
Francine Taylor-Campbell, Contributor

Students of Central High School tour the Gleaner's North Street plant and office, in Kingston, on Tuesday, February 7. Senior research officer Nicolia McDonald demontrates the use of The Gleaner Archives to the group. - Winston Sill/Freelance Photographer

We will continue to review this topic by looking at some questions.

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 A1 the oxidation # is 0.

2. The oxidation # of hydrogen in compounds is +1 ( Except in metallic hydrides e.g. CaH 2 when it is 1 ).

3. The oxidation # of oxygen in its compounds is 2.

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

5. The oxidation # of elements in simple ions is the same as the charge on the ion, for example, in FeCl 3 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. CH4C = --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

Question 1

Calculate the oxidation number of the marked atoms.

i. *Fe 2 O 3
ii. *C 2 H 6
iii. K*ClO 3
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.

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

Question 2

Write the names of

i. CuSO4
ii. PbO2
iii. Cr2O7²-
iv. Fe2O3
v. K2MnO4
vi. HNO3
vii. HNO²
viii. FeSO4

REDUCTION is the opposite of oxidation. It is a process resulting in the

i. Loss of oxygen, e.g. CuO(s) + H2(g) = Cu(s) + H2O(1)

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.