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Oxidation Number Rules  The oxidation number of any free, uncombined element is zero. n The oxidation number of an element in a simple (monatomic) ion.

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Presentation on theme: "Oxidation Number Rules  The oxidation number of any free, uncombined element is zero. n The oxidation number of an element in a simple (monatomic) ion."— Presentation transcript:

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2 Oxidation Number Rules  The oxidation number of any free, uncombined element is zero. n The oxidation number of an element in a simple (monatomic) ion is the charge on the ion. n In the formula for any compound, the sum of the oxidation numbers of all elements in the compound is zero. In a polyatomic ion, the sum of the oxidation numbers of the constituent elements is equal to the charge on the ion.

3 Oxidation Number Rules n Hydrogen, H, in combined form, has the oxidation number +1, usually. n Oxygen, O, in combined form, has the oxidation number -2, usually.

4 Oxidation Numbers  assign oxidation numbers to each element in these compounds NaNO 3 K 2 Sn(OH) 6 H 3 PO 4

5 Oxidation Numbers  assign oxidation numbers to each element in these compounds NaNO 3 K 2 Sn(OH) 6 H 3 PO 4 +1 +5 -2

6 Oxidation Numbers  assign oxidation numbers to each element in these compounds NaNO 3 K 2 Sn(OH) 6 H 3 PO 4 +1 +5 -2+1 +4 -2 +1

7 Oxidation Numbers  assign oxidation numbers to each element in these compounds NaNO 3 K 2 Sn(OH) 6 H 3 PO 4 +1 +5 -2+1 +4 -2 +1 +1 +5 -2

8 Oxidation Numbers  assign oxidation numbers to each element in these polyatomic ions SO 3 2- HCO 3 - Cr 2 O 7 2-

9 Oxidation Numbers  assign oxidation numbers to each element in these polyatomic ions SO 3 2- HCO 3 - Cr 2 O 7 2- +4 -2

10 Oxidation Numbers  assign oxidation numbers to each element in these polyatomic ions SO 3 2- HCO 3 - Cr 2 O 7 2- +4 -2+1+4 -2

11 Oxidation Numbers  assign oxidation numbers to each element in these polyatomic ions SO 3 2- HCO 3 - Cr 2 O 7 2- +4 -2+1+4 -2 +6 -2

12 Redox Reactions  oxidation - loss of electrons  oxidation number increases  reduction - gain of electrons  oxidation number decreases or reduces!  oxidizing agents - chemical species that gain electrons and oxidize some other species  electron thieves  they are reduced

13 Redox Reactions  reducing agents - chemical species that lose electrons and reduce some other species  victims of electron thieves  they are oxidized  Predicting reactions?

14 Activity Series  Allows us to predict if a certain metal will be oxidized by an acid or a particular salt.  Activity series is arranged in order of decreasing ease of oxidation  Metals at the top lose electrons easily  Metals at the bottom are very stable  Any metal can be oxidized by the ions of the substances below it.

15 Balancing Redox Reactions  Balancing redox reactions.  We will use the Half-Reaction Method

16 Half Reaction Method  Balancing redox reactions by half reaction method rules:  Write the unbalanced reaction  Break into 2 half reactions -  1 for oxidation  1 for reduction  Mass balance each half reaction by adding appropriate stoichiometric coefficents  in acidic solutions we can add H + or H 2 O  in basic solutions we can add OH - or H 2 O

17 Half Reaction Method  Charge balance the half reactions by adding appropriate numbers of electrons  Multiply each half reaction by a number to make the number of electrons added equal to the number of electrons consumed  Add the two half reactions  Eliminate any common terms

18 Half Reaction Method  Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation.

19 Half Reaction Method  Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation.

20 Half Reaction Method  Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation.

21 Half Reaction Method  Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation.

22 Half Reaction Method  Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction.

23 Half Reaction Method  Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction.

24 Half Reaction Method  Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction.

25 Half Reaction Method  Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction.

26 Half Reaction Method  Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction.

27 Half Reaction Method  Example: In basic solution hydrogen peroxide oxidizes chromite ions, Cr(OH) 4 -, to chromate ions, CrO 4 2-. Write and balance the net ionic equation for this reaction.

28 Half Reaction Method

29  Example: When chlorine is bubbled into basic solution, it forms chlorate ions and chloride ions. Write and balance the net ionic equation.

30 Half Reaction Method


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