Oxidation-Reduction (Redox) Reactions

Oxidation-Reduction Reactions  Electron transfer between ionic compounds, change in oxidation numbers  One compound wants to GIVE electrons, other compound wants to TAKE electrons.  Reduction—  GAIN of electrons in element/compound  Compound becomes more NEGATIVE  DECREASE in oxidation number  Oxidation  LOSS of electrons in element/compound  Compound becomes more POSITIVE  INCREASE in oxidation number

 Oxidation CANNOT happen without Reduction.  Reduction CANNOT happen without Oxidation  Both have to happen in a redox reaction ! ! !

Example 1:  Mg (s) + Cu +2 (aq) Mg +2 (aq) + Cu (s)

Ex. 2: 2Mg + O 2  2MgO  Where does reduction happen?  Which does oxidation happen?

Example 3:  2 Na + Cl 2  2NaCl  Assign oxidation numbers  Identify where oxidation occurs, where reduction occurs.

Determine whether each of these reactions is a oxidation-reduction reaction. If so, identify what is oxidized and what is reduced. 1)Mg + 2HCl  MgCl 2 + H 2 2)I 2 + 3Cl 2  2 ICl 3 3)NaOH + HCl  NaCl + H 2 O 4)2Na + 2H 2 O  2NaOH + H 2

Half-Reaction  Describes the oxidation or reduction process going on in a redox reaction  2 half-reactions make ONE redox reaction  Contains “half” of a redox reaction

Half-Reactions  A method of separating the OXIDATION part of a chemical reaction from the REDUCTION part.  Oxidation process---one equation  Reduction process---one equation  Then we put it all together.

Half-Reaction Example  Cu +2 (aq) + 2e -1 Cu (s) Reduction  Mg (s) Mg +2 (aq) + 2e -1 Oxidation Total: Mg (s) + Cu +2 (aq) Mg +2 (aq) + Cu (s)

Example 4: Oxidation: Cu (s) Cu +2 (aq) + 2e -1 Reduction: Ag +1 (aq) + 1e -1 Ag (s) *When electrons do not balance, multiply one or both half- reactions when needed to balance the amount of electrons gained and lost in the reaction. **

How do we write half-reactions?  Original Equation: 2 Na + Cl 2  2NaCl  Oxidation Half-Reaction:  2Na  2Na + + 2e -  Show electrons “lost” through process, 2 because you have 2 Na  Reduction Half-Reaction:  Cl 2 + 2e -  2Cl -  Show electrons “gained” through process, 2 because you have 2 Cl **Electrons MUST equal in number for BOTH half- reactions!!!

Putting Redox Equation back together……  Original Equation:  2 Na + Cl 2  2NaCl  Assign Oxidation Numbers  Identify where oxidation/reduction happening  Half-Reactions:  Oxidation: 2Na  2Na + + 2e -  Reduction: Cl 2 + 2e -  2Cl -  Cancel elements and electrons existing on both sides, Add half-reactions together.

RESULT ! !  2 Na + Cl 2  2NaCl

Example 5:  2Mg + O 2  2MgO

Example 6:  Cu (s) + NO 3  Cu +2 + NO

Example 7:  Zn + 2HCl  ZnCl 2 + H 2

Practice!  Write half-reactions and balance if the reaction is redox ! 1)Mg + 2HCl  MgCl 2 + H 2 2)I 2 + 3Cl 2  2 ICl 3 3)NaOH + HCl  NaCl + H 2 O 4)2Na + 2H 2 O  2NaOH + H 2