Download Oxidation Numbers - Spanish Point Chemistry

Oxidation Numbers
&
Balancing equation
Oxidation Number
• Oxidation number is defined as
 The
charge an atom has
 Or appears to have
 When electrons are distributed
 according to certain rules
Oxidation Number Rules
• The oxidation number of
• an Element is
0
• group One elements is +1
• group Two elements is +2
in
compounds
The oxidation number of
an
ion is equal to the
charge on the ion
• halogens is
-1
(except ……????)
(in binary compounds)
• The oxidation number of
H in a compound
is +1
– except
in metal hydrides when it is -1
• The oxidation number of
O in a compound is
-2
– except (x2)
in peroxides when it is -1 (H2O2)
in OF2 when it is +2 (why?)
• Oxidation numbers
• add up to zero in a compound
• add up to the charge of a complex ion
• What is the oxidation number of each element
in :- (write down before you go on)
H20
MnO4¯
I2
KBrO3
Na2S2O3
H2O2
NaOCl
The oxidation number of each element is :H20
+1
-2
MnO4¯
+7 -2
I2
0
KBrO3
+1 +5 -2
Na2S2O3
+1 +2 -2
H2O2
+1
-1
NaOCl
+1 -2
+1
Learning Check
Can I give the oxidation number RULE for
a) Oxygen
b) Hydrogen
c) free element
d) Neutral atom (sum)
e) Ion (simple and complex)
f) Group 1 element
g) Group 2 element
h) HALOGEN
STILL NOT
The End - click to go on
Balancing Equations with oxidation numbers
STEPS
1. Assign oxidation numbers
2. Identify what is oxidised and reduced
3. Write half equation SIDE by SIDE for each
(showing number of electrons on the move for one atom of each)
4.
5.
6.
7.
8.
Rewrite for the number of atoms given e.g. Cr2
Balance the electrons
REWRITE the original equation using these “prefixes”
Balance remainder by inspection
CHECK – do the charges on each side cancel out??
Example
Assign & Identify
Cr2O72- + Fe2+ + H+ Cr3+ + Fe3+ + H20
+6 -2
+2
+1
+3
Oxidised x1
Reduced x3
+3
+1 -2
2 Identify oxidised or reduced &
number of electrons lost or gained per atom & as given
Reduced
Cr + 3e- Cr
Cr2 + 6e- 2Cr
Oxidised
Fe – e-  Fe
Fe – e-  Fe
ATOM
GIVEN
• Balance Electrons
Cr2 + 6e- 2Cr
6
Fe –
6 e-
6
 Fe
• Rewrite and sub back
Cr2O72- + 6Fe2+ + H+ 2Cr3+ + 6Fe3+ + H20
• Balance remainder by inspection
14 2Cr3+ + 6Fe3++ H 0
Cr2O72- + 6Fe2++ H+
2
7
Check
• Charges on each side should balance
• Cr2O72- + 6Fe2++ 14 H+ 2Cr3+ + 6Fe3++ 7 H20
• LEFT
212+
14+
24+
RIGHT
6+
18+
___
24+