Oxidation Reduction Reactions
Oxidation-Reduction reactions, often called redox reactions involve the transfer of __________.
These reactions must happen together, you can’t have oxidation without _________________.
Each atom is assigned an oxidation state, sometimes called oxidation numbers. The following is a
list of rules for assigning oxidation states to atoms:
Ion or element
Oxidation states
Atoms in elements. Including Dr. H BrONClIF
0
Fluorine (F)
-1
Oxygen (O)
-2
-2
Peroxides (O2 )
-1
Hydrogen (H) in covalent compounds
+1
 Other monatomic ions can have an oxidation state that is the same as their assigned
periodic table charge. Example Na+1 would be +1
 The sum of all oxidation states must equal the overall charge of the compound
Sum oxidation states:
Oxidation states:
+2
-2 = 0
H2O
+1 -2
+5
-8 = -3
PO4
+5
-3
-2
How to tell if an atom has been oxidized or reduced.
__________________________ is an increase in oxidation states (loss of electrons)
__________________________is a decrease in oxidation states (gain of electrons)
2 pneumonic devices to tell if an atom has been reduced or oxidized:
LEO says GER
Loss Electrons Oxidation------Gain Electrons Reduction
OIL RIG
Oxidation Is Loss--------Reduction Is Gain
Oxidizing agents (electron acceptor) is the compound, element, or ion that contains the atom that
has been reduced.
Reducing agents (electron donator) is the compound, element, or ion that contains the atom that
has been oxidized.
2Al + 3I2  2AlI3
Oxidation states:
0
0  +3 -1
Al is oxidized because its oxidation state went up (lost electrons) 0+3, Al is the reducing agent
I is reduced because its oxidation state went down (gain electrons) 0-1, I2 is the oxidizing agent.
Oxidation states:
Balancing Redox Reactions in ACIDIC media
H+ + Cr2O72- + C2H5OH  Cr3+ + CO2 + H2O
+1 +6 -2 -4 +1 -2 +1 +3 +4 -2 +1 -2
Steps in balancing redox reactions in acidic media:
1) Break equation into 2 half reactions: Reduction half reaction and oxidation half reaction.
Ignore any H+ or H2O that are in the original reaction.
Reduction: Cr2O72-  Cr3+
Oxidation: C2H5OH  CO2
2) Balance all elements except hydrogen (H) and oxygen (O)
Reduction: Cr2O72-  2Cr3+
Oxidation: C2H5OH  2CO2
3) Balance oxygens using H2O
Reduction: Cr2O72-  2Cr3+ + 7H2O
Oxidation: C2H5OH + 3H2O  2CO2
Balance Hydrogens using H+
Reduction: 14H+ + Cr2O72-  2Cr3+ + 7H2O
Oxidation: C2H5OH + 3H2O  2CO2 + 12H+
4) Balance charges using electrons (e-)
Reduction: 14H+ + Cr2O72- + 6e-  2Cr3+ + 7H2O
Oxidation: C2H5OH + 3H2O  2CO2 + 12H+ + 12e5) Multiply through one or both equations to make sure electrons will cancel out
Reduction: x2) 28H+ + 2Cr2O72- + 12e-  4Cr3+ + 14H2O
Oxidation: C2H5OH + 3H2O  2CO2 + 12H+ + 12e6) Finally, combine like terms on same side of  and cancel like terms on opposite side of 
Reduction: 16H+ + 2Cr2O72- + 12e-  4Cr3+ + 11H2O
Oxidation: C2H5OH + 3H2O  2CO2 + 12H+ + 12e The electrons completely cancel each other out
 3 H2O on oxd half rxn cancel to zero and reduce the waters on the red from 14 to 11
 The 12 H+ on the oxd half rxn cancel to zero and reduce the H+ from 28 to 16 on red half
7) Final Balanced redox equation in acidic media:
16H+ + 2Cr2O72- + C2H5OH  4Cr3+ + 11H2O + 2CO2
 Note how the total charge on both sides are equal (12+  12+)