Download
calculating and using oxidation numbers n.
Skip this Video
Loading SlideShow in 5 Seconds..
Calculating and using oxidation numbers PowerPoint Presentation
Download Presentation
Calculating and using oxidation numbers

Calculating and using oxidation numbers

94 Vues Download Presentation
Télécharger la présentation

Calculating and using oxidation numbers

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  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 H2, O3 and S8. 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: Cl2(g) + 2Br-(aq) → 2Cl-(aq) + Br2(aq) 0 -1 -1 0 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. 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: -2 -2 2MnO4- + 3H2O2 + 2H+→ 2MnO2 + 3O2 + 4H2O Hydrogen peroxide 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. MnO4-: 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. MnO2: Mn + 2(-2) = 0 Mn - 4 = 0 Mn = +4 0 +1 +1 +1 -2 -2 -2 -1 +7 +4

  5. Oxidation number goes down from +7 to +4: MnO4- has been reduced. +7 -2 +1 -1 +1 +4 -2 0 +1 -2 2MnO4- + 3H2O2 + 2H+→ 2MnO2 + 3O2 + 4H2O Oxidation number goes up from -1 to 0: H2O2 has been oxidised.