1. Redox Reactions
Chapter 20

2. Electron Transfer & Redox Rxn
One of the defining characteristics of single-replacement and combustion reactions is that they always involve the transfer of electrons from one atom to another
Synthesis too!
Complete Chemical Equation:
2 Na(s) + Cl2(g)  2 NaCl(s)
Net Ionic Equation:
2 Na(s) + Cl2(g)  2 Na+ + 2 Cl-(ions in crystal)
In this reaction, an electron is transferred from each of the two sodium atoms to the chlorine atoms to form two Cl- ions

3. Combustion(Synthesis) Example
Complete Chemical Equation:
2 Mg(s) + O2(g)  2 MgO(s)
Net Ionic Equation:
2 Mg(s) + O2(g)  2 Mg O2-
Again, electrons are transferred from the magnesium solid to the oxygen atoms.
How many?

4. Redox Reactions
A reaction in which electrons are transferred from one atom to another is called an oxidation-reduction reaction
We use Redox for simplicity
How do oxidation and reduction differ?
Oxidation is referred to as the loss of electrons from atoms of a substance
Na  Na+ + e-
Reduction is defined as the gain of electrons by atoms of a substance
Cl2 + 2e-  2Cl-

5. Redox Reactions An easy way to remember the difference is the saying:
LEO the lion says GER or LEO GER
Loss of Electrons is Oxidation
Gain of Electrons is Reduction
Remember - The oxidation number of an atom is the charge made by ions forming
Group 1 is +1, Group 2 is +2
Nobel Gases are 0, Group 17 is -1, etc.

6. Oxidizing and Reducing Agents
In Redox Reactions, one substance will act as the oxidizing agent.
Thus the substance that oxidizes another substance by accepting its electrons is called an oxidizing agent
The reducing agent is the substance that reduces another substance by losing electrons
A disproportionation is a substance that acts as both a reducing and oxidizing agent.
2 K(s) + Br2(g)  2 KBr(s)
oxidized
Potassium was oxidized so it is
the reducing agent
Bromine was reduced so it is the
oxidizing agent
reduced

7. Redox and Electronegativity
Recall:
Electronegativity is an atom in a chemical compound that wants to have electrons
The trend on the PT increases to the upper right, not Nobel Gases
These will create partial charges

8. Ionic vs. Covalent Oxidations
Ionic bonds will form ions and thus have oxidation charges
Covalent do not have formal oxidations because they share electrons
Thus, their oxidation numbers in redox reactions are based purely on electronegativity values in relation to the bond

9. Determining an Oxidation Numbers
Rules for Determining Oxidation Numbers
The oxidation number of an uncombined atom is zero. This is also true for polyatomic molecules such as O2, Cl2, S8, etc.
The oxidation number of a monatomic ion is equal to the charge on the ion Na1+ is +1, Cl1- is -1
The oxidation number of the more electronegative atom in a molecule or a complex ion is the same as the charge it would have if it were an ion
The most electronegative element, fluorine, always has an oxidation number of -1 when it is bonded to another element
The oxidation number of oxygen in compounds is always -2, except in peroxides (H2O2) and when bonded to the only element more electronegative than itself, F (it will be +2 in this case)
The oxidation number of hydrogen in most cases is +1, unless its bonded to a less electronegative atom such as metals in group 1, 2 and 13
The metals of groups 1 and 2 and Aluminum in 13 are still always +1, +2 and +3, respectively
The sum of the oxidation numbers in a neutral compound must be zero
The sum of oxidation numbers of the atoms in a polyatomic ion is equal to the charge on the ion

10. Balancing Redox Reactions
Simple redox reactions require the inspection technique
The skill we have been using this year, which is done by just counting up the atoms and balancing them on both sides
More complex redox reactions will require a more systematic (mathematical) approach
Oxidation Number Method
Equation-Balance Method

11. Oxidation Number Method
Assign oxidation numbers to all atoms in the equation
Identify the atoms that are oxidized and the atoms that are reduced
Determine the change in oxidation number for the atoms that are oxidized and for the atoms that are reduced
Make the change in oxidation numbers equal in magnitude by adjusting the coefficients in the equation
If necessary, use the conventional method to balance the remainder of the equation
Ex. Cu + HNO3 → Cu(NO3)2 + NO + H2O

12. Cu + HNO3 → Cu(NO3)2 + NO + H2O
Assign oxidation numbers to all atoms in the equation
Oxidation:
Cu + H N O3 → Cu(N O3)2 + N O + H2O
Identify the atoms that are oxidized and the atoms that are reduced
Oxidized: Cu0  Cu+2 + 2e-
Reduced: N+5 + 3e-  N+2
Determine the change in oxidation number for the atoms that are oxidized and for the atoms that are reduced
Copper: Net Change = +2 Nitrogen: Net Change = -3
Make the change in oxidation numbers equal in magnitude by adjusting the coefficients in the equation with the elements whose oxidation is changing
3Cu + HNO3 → Cu(NO3) NO + H2O
If necessary, use the conventional method to balance the remainder of the equation
3Cu HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

13. Practice Problems Al(s) + MnO2(s) --> Al2O3(s) + Mn(s)
SO2(g)  +  HNO2(aq)  -->  H2SO4(aq)  +  NO(g)
HNO3(aq)  +  H2S(aq)  -->  NO(g)  + S(s)  +  H2O(l)
Al(s)  +  H2SO4(aq)  -->  Al2(SO4)3(aq)  +  H2(g)

14. Equation-Balance Method
This method can be broken down into two subsets:
Acidic Condition
HNO3(aq)  +  H2S(aq)  --> 
An acid as one of the reactants
Basic Condition
Cr(OH)3(s) + ClO3-(aq)  --> 
A base as one of the reactants

15. Acidic Method
Step 1: Write the skeletons of the oxidation and reduction half-reactions. (They contain the formulas of the compounds oxidized and reduced, but the atoms and electrons have not yet been balanced.) Step 2: Balance all elements other than H and O. Step 3: Balance the oxygen atoms by adding H2O molecules where needed. Step 4: Balance the hydrogen atoms by adding H+ ions where needed. Step 5: Balance the charge by adding electrons, e-. Step 6: If the number of electrons lost in the oxidation half-reaction is not equal to the number of electrons gained in the reduction half-reaction, multiply one or both of the half- reactions by a number that will make the number of electrons gained equal to the number of electrons lost. Step 7: Add the 2 half-reactions as if they were mathematical equations. The electrons will always cancel. If the same formulas are found on opposite sides of the half-reactions, you can cancel them. If the same formulas are found on the same side of both half-reactions, combine them. Step 8: Check to make sure that the atoms and the charges balance.

16. Practice Problems MnO4-(aq) + Br-(aq) --> MnO2(s) + BrO3-(aq)
I2(s) + OCl-(aq)  -->  IO3-(aq) + Cl-(aq)
Cr2O72-(aq) + C2O42-(aq)  -->  Cr3+(aq) + CO2(g)
Mn(s) + HNO3(aq)  -->  Mn2+(aq) + NO2(g)

17. Basic Method
Steps 1-7: Begin by balancing the equation as if it were in acid solution. If you have H+ ions in your equation at the end of these steps, proceed to Step #8. Otherwise, skip to Step #11.
Step 8: Add enough OH- ions to each side to cancel the H+ ions. (Be sure to add the OH- ions to both sides to keep the charge and atoms balanced.)
Step 9: Combine the H+ ions and OH- ions that are on the same side of the equation to form water.
Step 10: Cancel or combine the H2O molecules.
Step 11: Check to make sure that the atoms and the charge balance. If they do balance, you are done. If they do not balance, re-check your work in Steps 1-10.

18. Practice Problems CrO42-(aq) + S2-(aq) --> Cr(OH)3(s) + S(s)
MnO4-(aq) + I-(aq)  -->  MnO2(s) + IO3-(aq)
H2O2(aq) + ClO4-(aq)  -->  O2(g) + ClO2-(aq)
S2-(aq) + I2(s)  -->  SO42-(aq) + I-(aq)