1. Chapter 20: Oxidation – Reduction Reactions

2. Oxidation and Reduction
Oxidation involves the loss of electrons reduction involves the gain of electrons
O xidation L osing
I s E lectrons
L osing O xidation
R eduction G aining
G aining R eduction

3. Oxidation and Reduction
Oxidation states/number: the distribution of electrons or the charge of an element
Any chemical process in which elements undergo changes in oxidation number is an oxidation-reduction, or redox reaction

4. Oxidation and Reduction
The part of the reaction involving oxidation or reduction alone can be written as a half-reaction
Cu → Cu+2 + 2e –
oxidation half-reaction
2NO3 – + 2e– + 4H+ → NO2 + 2H2O
reduction half-reaction
Cu + 2NO3 – + 4H+ → Cu+2 + 2NO2 +2H2O
redox reaction

5. Oxidation Number Rules
The oxidation number of any uncombined element is 0
The oxidation number of a monatomic ion equals the charge on the ion
The more electronegative element in a binary compound is assigned the number equal to the charge it would have if it were an ion
The oxidation number of fluorine in a compound is always -1

6. Oxidation Number Rules
Oxygen has an oxidation number of -2 unless it is combined with F, in which it is +1 or +2, or it is in a peroxide, in which it is -1
Hydrogen’s oxidation state in most of its compounds is +1 unless it is combined with a metal, in which case it is -1
In compounds, Group 1 and 2 elements and aluminum have oxidation numbers of +1, +2, and +3, respectively

7. Oxidation Number Rules
The sum of the oxidation numbers of all atoms in a neutral compound is 0
The sum of the oxidation numbers of all atoms in a polyatomic ion equals the charge of the ion

8. Examples: state the oxidation number of every element
 Cl2 _________________ SO3 __________________
Fe3+ _________________ H2O2 __________________
NO2 _________________ MgH2 ____________________
OF _________________

9. Examples: state the oxidation number of every element
 Cl2 ______Cl = 0_____ SO3 _S = +6 _ O = – 2
Fe3+ ______Fe = +3___ H2O2 _H = +1 _ O = – 1
NO2 _N = +4 _O = – 2 MgH2 _Mg = +2 _ H = – 1
OF _O = +1 _ F = – 1

10. Examples: state the oxidation number of every element
Li4Pb ____________________ MnBr62– ____________________
AlBr3 ____________________ H3AsO3 ___________________________
PO43– ___________________ Be(BrO3)2 ______________________________
H5P3O10 ___________________________

11. Examples: state the oxidation number of every element
Li4Pb _Li = +1 _Pb = – 4 MnBr62– _Mn = +4 _Br = – 1
AlBr3 _Al = +3 _Br = – 1 H3AsO3 H = +1 _As = + 3 _O = – 2
PO43– _P = +5 _O = – Be(BrO3)2 _Be = +2 _Br = + 5 _O = – 2
H5P3O10 _H = +1 _P = + 5 _O = – 2

12. Example Problem: Balance the following redox equation.
HI + HNO2 → NO + I2 + H2O

13. Example Problem: Balance the following redox equation.
+1 – – – –2
HI + HNO2 → NO + I2 + H2O
Step 1 – Assign oxidation numbers

14. Example Problem: Balance the following redox equation.
oxidation
+1 – – – –2
HI + HNO2 → NO + I2 + H2O
reduction
Step 2 – Identify which elements are changing oxidation state and write the half reactions
ox: HI → I2
red: HNO → NO

15. Example Problem: Balance the following redox equation.
oxidation
+1 – – – –2
HI + HNO2 → NO + I2 + H2O
reduction
Step 3 – Write the half reaction for oxidation.
a.) balance the atoms: use H2O to balance O’s, then H+ to balance the H’s
b.) balance the charge: add electrons to the right side of the equation
ox: HI → I2 + 2H+ + 2e–
red: HNO → NO

16. Example Problem: Balance the following redox equation.
oxidation
+1 – – – –2
HI + HNO2 → NO + I2 + H2O
reduction
Step 4 – Write the half reaction for reduction.
a.) balance the atoms: use H2O to balance O’s, then H+ to balance the H’s
b.) balance the charge: add electrons to the right side of the equation
ox: HI → I2 + 2H+ + 2e–
red: HNO2 + H+ + e– → NO + H2O

17. Example Problem: Balance the following redox equation.
oxidation
+1 – – – –2
HI + HNO2 → NO + I2 + H2O
reduction
Step 5 – Multiply the coefficients in the half reactions so that the number of electrons gained = the numbers of electrons lost
ox: (2HI → I2 + 2H+ + 2e–) x1
red: (HNO2 + H+ + e– → NO + H2O) x 2

18. Example Problem: Balance the following redox equation.
oxidation
+1 – – – –2
HI + HNO2 → NO + I2 + H2O
reduction
Step 6 – Combine the half reactions and cancel out anything common to both sided of the equation.
ox: HI → I2 + 2H+ + 2e–
red: 2HNO2 + 2H+ + 2e– → NO H2O
overall equation: 2HI + 2HNO2 → 2NO + I2 + 2H2O

19. Oxidizing and Reducing Agents
Reducing agent: is a substance that has the potential to cause another substance to be reduced; they lose electrons (are oxidized)
Oxidizing agent: is a substance that has the potential to cause another substance to be oxidized; they gain electrons (are reduced)

20. Oxidizing and Reducing Agents
Relative strengths of oxidizing and reducing agents on page 643
Disproportionation: the process in which a substance acts as both an oxidizing agent and a reducing agent; the substance is self-oxidizing and self-reducing