1. Calculating and using oxidation numbers

2. 1 The oxidation number of any free, uncombined element is zero. This includes polyatomic molecules of elements such as H 2, O 3 and S 8. 2 The charge on a simple (monatomic) ion is the oxidation number of the element in that ion. In a polyatomic ion, the sum of the oxidation numbers of the atoms in that ion is equal to the charge on the ion. 3 In compounds (whether ionic or covalent), the sum of the oxidation numbers of all atoms in the compound is zero. 4 The oxidation number of oxygen is –2, except in the case of peroxides where it is –1. 5 The oxidation number of hydrogen is +1, except in the case of metallic hydrides where it is –1.

3. State the oxidation numbers for each element in this equation: Cl 2 (g) + 2Br - (aq) → 2Cl - (aq) + Br 2 (aq) 1 The oxidation number of any free, uncombined element is zero. 2 The charge on a simple (monatomic) ion is the oxidation number of the element in that ion. 00 Each Br - ion has an oxidation number of -1: Br - doesn’t have an ON of -2 because there are two of them.

4. State the oxidation numbers for each element in this equation: 2MnO 4 - + 3H 2 O 2 + 2H + → 2MnO 2 + 3O 2 + 4H 2 O 1 The oxidation number of any free, uncombined element is 0. 5 The oxidation number of hydrogen is +1, 4 The oxidation number of oxygen is –2 except… in the case of peroxides where it is –1. 2 In a polyatomic ion, the sum of the oxidation numbers of the atoms in that ion is equal to the charge on the ion. MnO 4 - : Mn + 4(-2) = -1 Mn - 8 = -1 Mn = +7 3 In compounds (whether ionic or covalent), the sum of the oxidation numbers of all atoms in the compound is zero. MnO 2 : Mn + 2(-2) = 0 Mn - 4 = 0 Mn = +4 Hydrogen peroxide +4 +7 +1 0 -2

5. +7 -2 +1 -1 +1 +4 -2 0 +1 -2 2MnO 4 - + 3H 2 O 2 + 2H + → 2MnO 2 + 3O 2 + 4H 2 O Oxidation number goes down from +7 to +4: MnO 4 - has been reduced. Oxidation number goes up from -1 to 0: H 2 O 2 has been oxidised.