Oxidation and reduction II
Francine Taylor-Campbell, Contributor

Reduction is the opposite of oxidation. It is a process resulting in the:

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

Copper oxide is reduced to copper by the loss of oxygen.

ii. Addition of hydrogen, eg H2(g) + Cl2(g) = 2HCl(g)

Chlorine is reduced to hydrogen chloride

iii. Gain of electrons eg Cu2+ 2e = Cu - Eqn 5

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

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

1. The oxidation number of uncombined elements is 0. For example H in H2 and Al the oxidation number is 0.

2. The oxidation number of hydrogen in compounds is +1 (except in metallic hydrides, example 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, example in FeCl3 Fe3+ ox number = +3 Cl- ox number is -1.

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

7. The sum of the oxidation numbers of the atoms in a compound is 0. Example 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. Example, 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 number of O = —4 Ox number of Mn = +4

ii. CH4 Ox number of C = —4

iii. CO2 Ox number of C = +4

iv. SO42- Ox number of O = —2, thus 4 x -2 = —8 S = +6

since -8 + 6 = —2

QUESTION 2

Calculate the oxidation number 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.

Example 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 3

Write the names of

i. CuSO4

ii. PbO2

iii. Cr2O72-

iv. Fe2O3

v. K2MnO4

vi. HNO3

vii. HNO2

viii. FeSO4

Jodi Spence, process development officer, Scientific Research Council, explains the process of tissue culture technology to interested clients (not pictured) at the science and technology conference and exposition earlier this year. - FILE

Francine Taylor-Campbell is an independent contributor.