1. Balancing redox reactions

2. Balance oxidation-reduction reactions using redox methods
Include: oxidation number method, and half-reaction method
Additional KEY Terms

3. K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3
Oxidation number method
Step 1: Assign oxidation numbers and determine oxidized and reduced +1 +6 -2 +1 -2 +4 -2 +1 -2 +1 +3 -2
K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3
Step 2: Calculate number of electrons lost/gained
lose 4e-
K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 -2 +4 +1 +6 +3
gains 3e- x 2 atoms = 6e-

4. K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3
Step 3: Add coefficients to balance the number of electrons lost/gained
Use common multiple to make them equal
lose 4e-
x 3 atoms = 12e-
K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 +4 +6 +3 2 3 3 2
gains 3e- x 2 atoms = 6e-
x 2 atoms = 12e-
So with the coefficients and subscripts you actually have a total of 4 atoms of Cr each gaining 3e- for a total of 12
Step 4: Balance other atoms by conventional method 2
K2Cr2O7 + H2O + S → SO2 + KOH + Cr2O3 2 3 3 4 2

5. P + HNO3 + H2O → NO + H3PO4 3 5 2 5 3 +1 +5 -2 +1 -2 +2 -2 +1 +5 -2
Balance the following using the oxidation number method:
lose 5e-
x 3 atoms = 15e- +1 +5 -2 +1 -2 +2 -2 +1 +5 -2
P + HNO3 + H2O → NO + H3PO4 3 5 2 5 3
gains 3e-
x 5 atoms = 15e-

6. H2SeO3 + HClO3 → H2SeO4 + Cl + H2O
Always look at the reactant and product to see if the numbers match– you might need to massage the coefficient to make the atoms equal
lose 2e-
x 5 atoms = 10e- +1 +4 -2 +1 +5 -2 +1 +6 -2 +1 -2 5
H2SeO HClO3 → H2SeO Cl + H2O 2 5 1 2
gains 5e-
x 2 atoms = 10e-
Since there is already 2 Cl on the product side (subscript) I don’t need to add the coefficient to it.

7. CAN YOU / HAVE YOU?
Balance oxidation-reduction reactions using redox methods
Include: oxidation number method, and half-reaction method
Additional KEY Terms