Download Comparing Oxidation and Reduction

p.5 Assigning Oxidation Numbers (Charge: +2, -1...)
Rules for Oxidation Numbers
Rule Statement
1.
Atoms in elemental form = 0
An element has no charge
2.
Monoatomic ions = the ion’s charge
3.
4.
5.
6.
7.
8.
9.
Oxygen almost always = -2
Except in peroxides (O22-) = -1
and OF2 = +2
Hydrogen almost always +1
Except in metal hydrides H = -1
Oxidation states in compounds must
sum to zero
Oxidation states in polyatomic ions must
sum to the ion charge
Always assign the more electronegative
element a negative oxidation number
In a compound or ion containing more
than two elements, the element written
the farthest to the right takes its most
common oxidation state.
Alkali always +1
Alkaline-Earth always +2
Examples
Fe: 26 p and 26 e
Na, O2, P4, As, Zn
Fe3+: 26p and 23 e (lost 3 e)
K+. Ca2+, Cl : ions!!!
Na2O2 and OF2
H2SO4, CaH2, LiH
(H2O) 0, [Ca(NO3)2] 0
ClO4 1-, ClO3 1- contain chlorine
= +7 and +5
PF5 contains F = -1 and thus
P = +5
SCN- contains N = -3(most
common) S = -2 (negative
value), thus C = +4
NaCl, CaCl2
Comparing Oxidation and Reduction
Processes
Oxidation
Reduction
1. Number of
electrons
Loss (Leo)
e- : products
Gain (Ger)
e- : reactants
2. Where electrons go
3. Status of oxygen
Away from the atom
Zn0
 Zn2+ + 2eAtom
ion
product
Gain
Towards the ion
Cu2+ +
2e-  Cu0
Ion
reactant atom
Loss
4. Status of hydrogen
Loss
Gain
5. Oxidation number
Increases
(charge)
Zn (0)  Zn (+2)
Example:
CH3CH2OH  CH3COOH
Decreases
Cu (+2)  Cu(0)
ethanol
acetic acid (vinegar) (la piquette)
Oxygen status: 1 Oxygen  2 Oxygen
oxidation
Hydrogen status: 6 Hydrogen  4 Hydrogen oxidation
For the equations below:
.balance them
.write the oxidation half reaction
.write the reduction half reaction
.identify the oxidizing and reducing agents
1) C2H4
+
O2

CO2 +
H2O
4+
2+
C2H4
+
O2
 C + 2O + 2H + O24+
+
C2H4  2C + 4H + 12e(4e- + O2 2O2- ) X 3
oxidation hr: C2H4  2C4+ + 4H+ + 12ereduction hr: 12e- + 3O2 6O2overall balanced: C2H4 + 3O2  2CO2 + 2H2O
OA (being reduced): O2
RA (being oxidized): C2H4
2) KClO3

KCl +
O2
K+ + Cl5+ + 3O2-  K+ + Cl +
O20
+
Spectator: K
Cl5+ + 3O2- 
K+ + Cl- +
O20
(6e- + Cl5+  Cl-) X 2
(2O2- 
O20 + 4 e--) X 3
oxidation hr: 6O2- 
3O20 + 12ereduction hr: 12e- + 2Cl5+  2Cladding 2 K+ to each reactant and product sides:
overall balanced: 2KClO3 
2KCl +
3O2
OA (being reduced): Cl5+
RA (being oxidized): O23) H2 +
O2

H2O
H2
+
O2

2H+ + O2(H2

2H+ + 2e-) X 2
4e- + O2

2O2oxidation hr: 2H2

4H+ + 4ereduction hr: 4e- + O2

2O2overall balanced: 2H2
+ O2 
2 H2O
OA (being reduced): O2
RA (being oxidized): H2