Chemistry 30 AP Redox Reactions: Oxidation Numbers
Name: _______________________
Electrochemical reactions involve the transfer of electrons. Mass and charge are conserved when balancing these
reactions, but you need to know which atoms are oxidized and which atoms are reduced during the reaction.
The oxidation state of an atom in is defined as the apparent charge that an atom would have if electron pairs in covalent
bonds belonged entirely to the more electronegative atom.
The oxidation number is a positive or negative number corresponding to the oxidation state assigned to an atom in a
covalently bonded entity. Oxidation numbers are used to track the number of electrons lost or gained by each atom. These
oxidation numbers are assigned using the following rules:
1. The oxidation number of a free element is always 0. E.g. the atoms in He and N2 have oxidation numbers of 0. 2.
The oxidation number of a monatomic ion equals the charge of the ion. E.g. the oxidation number of Na+is +1 and
the oxidation number of N3-is -3.
3. The oxidation number of hydrogen is usually +1. E.g. in NH3 the oxidation number of hydrogen is +1 (N has an
oxidation number of -3)
Exception: The oxidation number of hydrogen is -1 in metal hydrides (compounds containing elements
that are less electronegative than hydrogen, as in CaH2).
4. The oxidation number of oxygen in compounds is usually -2. E.g. the oxidation number of oxygen in CO2 is -2 (C
has an oxidation number of +4)
Exceptions: The oxidation number of oxygen is -1 in peroxides like H2O2 and the oxidation number of
oxygen is +2 in OF2 (Fluorine has a high electronegativity so it is assigned its charge of -1, making O
+2)
5. The oxidation number of a Group 1 element in a compound is +1.
6. The oxidation number of a Group 2 element in a compound is +2.
7. The oxidation number of a Group 17 element (halogens) in a compound is -1 (except when that element is
combined with one having a higher electronegativity). E.g. the oxidation number of Cl is -1 in HCl, but the
oxidation number of Cl is +1 in HOCl.
8. The sum of the oxidation numbers of all of the atoms in a neutral compound is 0. E.g. the oxidation numbers
in CO2 are: C (+4) and O (-2), making the sum = 0
9. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion. E.g. the sum of the
oxidation numbers for SO42-is -2. (S = +6 and O = -2)
Example: Photosynthesis
Determine the oxidizing and reducing agents using oxidation numbers for photosynthesis:
CO2(g) + H2O(l) → C6H12O6(aq) + O2(g)
Step 1: Assign individual oxidation numbers to each atom within the reaction
+4 -2 +1 -2 0 +1 -2 0
CO2(g) + H2O(l) → C6H12O6(aq) + O2(g)
Step 2: Determine the change in oxidation numbers for each element
∙ An increase in number is oxidation (therefore the reducing agent).
o Oxygen’s oxidation number changes from -2 to 0 (increase = oxidation)
∙ A decrease in oxidation number is reduction (therefore the oxidizing agent)
o Carbon’s oxidation number changes from +4 to 0 (decrease = reduction)
Step 3: Complete the half reactions by adding electrons to the appropriate side of the equation CO2(g) + 4 e- →
C6H12O6(aq) **Carbon went from 4+ to 0 (4 e- were gained - carbon dioxide is the OA) H2O(l) → O2(g) +
2 e- **Oxygen went from 2- to 0 (2 e- were lost – water is the RA)
Step 4: Balance the half reactions by balancing the electrons and complete the net redox reaction (add water and hydrogen
ions when necessary in acidic solutions)
Reduction Half Reaction (OA) CO2(g) + 4 e- → C6H12O6(aq)
Oxidation Half Reaction (RA) 2 H2O(l) → 2 O2(g) + 4 e
Overall Redox Reaction Equation: CO2(g) + 2 H2O(l) → 2 O2(g) + C6H12O6(aq)
1. Determine the oxidation number of S in the following compounds:
c. SO42a. SO2
d. SO32b. SO3
2. Determine the oxidation number of N in the following compounds:
c. N2 d. NO
a. N2O b. NH3
e. S2O32f. S2O82-
e. N2H4 f. NH4Cl
3. Use oxidation numbers to determine if the following reactions are redox reactions. Is it redox?
a. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
b. H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l)
c. Pb(NO3)2(aq) + 2 KI(aq) → PbI2(aq) + 2 KNO3(aq)
d. 5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)
e. 2 Cr2O72-(aq) + 6 I-(aq) + 14 H+(aq) → 2 Cr3+(aq) + 3 I2(aq) + 7 H2O(l)