1. Balancing Redox Equations OBJECTIVE: Describe how oxidation numbers are used to identify, and balance reactions using Oxidation Numbers.

2. Identifying Redox Equations In general, all chemical reactions can be assigned to one of two classes: 1)REDOX – reactions in which electrons are transferred: 2)NON-REDOX - no electron transfer all other reaction types:

3. Identifying Redox Equations In an electrical storm, nitrogen and oxygen react to form nitrogen monoxide: N 2(g) + O 2(g) → 2 NO (g) Is this a redox reaction? How would we know? If the oxidation numbers change for an element, then either oxidation or reduction has taken place.

4. Identifying Redox Equations N 2(g) + O 2(g) → 2 NO (g) What are the oxidation numbers in the above reaction? Do they change? N 2(g)  N 2+ O 2(g)  O 2- 0  +2 0  -2 Nitrogen is oxidizedOxygen is reduced There is a change in the oxidation numbers so a redox reaction has occurred.

5. Lets say we want to balance a reaction! __Cr + + __ Sn 4+  __ Cr 3+ + __ Sn 2+ This one can be done by writing the equation as two half reactions. This one can be done by writing the equation as two half reactions. First assign oxidation numbers. First assign oxidation numbers. Cr +  Cr 3+ = oxidized, 2 electrons are lost. Sn 4+  Sn 2+ = reduced, 2 electrons are gained Write the half reactions from this equation. Be sure to include the electrons!!! Practice Problem #2

6. Cr + + Sn 4+  Cr 3+ + Sn 2+ Add electrons to the correct side of the equation. Oxidation: Cr +  Cr 3+ + 2 e - Reduction: Sn 4+ + 2 e -  Sn 2+ When the half reactions are correct; 1) the number of electrons should be equal 2) and should cancel out leaving you with correct coefficients for a balanced chemical equation. Practice Problem #2

7. __ Au 3+ + __ Ag  __ Au + __ Ag + Write the two half reactions: Oxidation: Ag  Ag + + e - Reduction: Au 3+ + 3 e -  Au The reactions are not balanced in charge. They do not have the same number of electrons. Multiply the Oxidation equation by 3! Balanced Equation: 1 Au 3+ + 3 Ag  1 Au + 3 Ag + Oxidation : 3 Ag  3 Ag + + 3 e -

8. __Cr 3+ + __ Pb  __ Cr + __ Pb 2+ Pb  Pb 2+ + 2 e - (Oxidation) Cr 3+ + 3 e -  Cr (Reduction) The two equations must transfer the same number of electrons. So what must happen? Find a common multiple! Multiply the Oxidation by 3 Multiply the Reduction by 2

9. 3 Pb  3 Pb 2+ + 6 e - (Oxidation) 2 Cr 3+ + 6 e -  2 Cr (Reduction) Combine the two equations to get correct coefficients! 2 Cr 3+ + 3 Pb  2 Cr + 3 Pb 2+ The equation should be balanced both in number of atoms and charges. (Note: The charges don’t always have to be zero on both sides, but need to be the same on both sides!

10. Practice Problem #4 (Special!) __ Mg + __HCl  __MgCl 2 + __H 2 Is this REDOX? Who is being oxidized and who is being reduced? Is this REDOX? Who is being oxidized and who is being reduced? Mg 0  Mg 2+ - lost electrons – oxidized H 1+  H 2 0 - gained electrons – reduced Cl 1-  Cl 1- - No change

11. __ Mg + __HCl  __MgCl 2 + __H 2 Mg  MgCl 2 + 2 e - (Oxidation) 1 HCl + 1 e -  H 2 (Reduction) * Need the Hydrogen atoms to balance first The two equations must transfer the same number of electrons. So what must happen? Mg + 2 HCl + 2 e -  1 MgCl 2 + 2 e - + H 2 Mg + 2 HCl +  MgCl 2 + H 2 2 HCl + 2 e -  H 2 (Reduction)