1. Redox Reactions

2. Oxidation numbers identify and indicate which element is oxidized and which is reduced. Here's an example - the reaction between sodium metal and chlorine gas:
It is often useful to write the oxidation number for every element above the element in the equation. Thus for our reaction we have:
balancing coefficients in the equation  do not affect the value of the oxidation numbers.
2 Na +
Cl2 →
2 NaCl
   +1   -1
2 Na +
Cl2 →
2 NaCl

3. example Add in the oxidation numbers
2 Mg + O2 →
2 MgO
Add in the oxidation numbers
which are oxidized and which are reduced?
An increase in oxidation number indicates oxidation
A decrease in oxidation number indicates reduction

4. Reducing agent
the substance that is oxidized. It allows another element to be reduced.
Oxidizing agent
the substance that is reduced. It allows another element to be oxidized.

5. example N2 + 2H2 → 2 NH3 element initial ox no final ox no e-
final ox no e-
oxidized or reduced
Agent

6. Assignments
Practice 6.1.3
Assignment 6.1.3

7. Balancing Redox reactions
Many redox reactions cannot easily be balanced just by counting atoms. Consider the following net ionic equation
Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s)
If you simply count atoms, the equation is balanced but the charges aren’t balanced! Charges represent gain or loss of electrons, and, like atoms, electrons are conserved during chemical reactions.

8. 1. Balancing Equations using Oxidation Numbers
Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s
element
initial ox no
final ox no
change in e-
balance for electrons

9. example MnO41- + Fe2+ + H1+ → Mn2+ + Fe3+ + H2O element initial ox no
final ox no
change in e-
balance for electrons

10. example
NH3 + O2 → NO2 + H2O
Before using a multiplier to get the electrons to match, notice the subscript with oxygen - O2. In our summary chart we base our oxidation number changes on a single atom, but our formula tells us that we must have at least two oxygen.  
element
initial ox no
final ox no
change in e-
balance for electrons

11. element
initial ox no
final ox no
change in e-
No. atoms
No. e-
balance for electrons 28

12. Since we were counting oxygen atoms in the O2 molecule on the reactant side of the equation, that's where we'll use the "7". Since nitrogen's oxidation number also changed we will use nitrogen's balancing coefficient there
4 NH3 + 7 O2 → 4 NO2 + H2O
The last step is to balance for hydrogen atoms (and finishing oxygen), which will mean placing a 6 in front of H2O:
4 NH3 + 7 O2 → 4 NO2 + 6 H2O

13. K2Cr2O7 + NaI + H2SO4 → Cr2(SO4)3 + I2 + H2O + Na2SO4 + K2SO4

14. K2Cr2O7 + NaI + H2SO4 → Cr2(SO4)3 + I2 + H2O + Na2SO4 + K2SO4
Next, check for any subscripts associated with either of these two elements - we see that Cr always has a subscript of "2" (in both K2Cr2O7 and Cr2(SO4)3), and I has a subscript in I2. So we'll add that to our summary chart to get a total number of electrons transferred, and then balance
element
initial ox no
final ox no
change in e-
No. atoms
No. e-
balance for electrons

15. element
initial ox no
final ox no
change in e-
No. atoms
No. e-
balance for electrons

16. One more example One more tricky one. Balance
Zn + HNO3 → Zn(NO3)2 + NO2 + H2O
Determine oxidation numbers and create your summary chart:
The main thing to notice is that N appears in two separate products - Zn(NO3)2 and NO2. Should we consider the subscript for nitrogen from Zn(NO3)2? In this case no, because this compound also contains Zn, the oxidized element. Also, the oxidation number for nitrogen does not change from HNO3 to Zn(NO3)2 .
element
initial ox no
final ox no
change in e-
No. atoms
No. e-
balance for electrons

17. We now get our balancing coefficients from our summary table
We now get our balancing coefficients from our summary table. A "1" will be placed in front of Zn, but which N should we use for the "2"? If you put it in front of both HNO3 and NO2 you'll find you cannot balance for nitrogen atoms. Since the oxidation number for nitrogen changed in becoming NO2, we will try it there first. Some trial-and-error may be required: