Understandings
Oxidation and reduction can be considered in terms of oxygen gain/hydrogen loss, electron transfer or change in oxidation number.
Variable oxidation numbers exist for transition metals and for most main-group non-metals.
Applications
Deduction of the oxidation states of an atom in an ion or a compound.
Deduction of the name of a transition metal compound from a given formula, applying oxidation numbers represented by Roman numerals.
1a ____ Na + ____ O2 → ____ Na2O
1b ____ NaOH + ____ HCl → ____ NaCl + ____ H2O
1c ____ NaOH + ____ H2SO4 → ____ Na2SO4 + ____ H2O
1d ____ Ca(OH)2 + ___ CO2 → ____CaCO3 + ____ H2O
1e ____ N2 + ____ H2 → ____ NH3
1f ____ Al + ____ KMnO4 → ____ Al2O3 + ____ MnO2 + ___ K2O
Redox has a number of different definitions based on its’ use in Chemistry - in all chemical reactions in which this occurs elements are said to either be Oxidised or Reduced. As these processes always happen simultaneously we call these reactions REDOX REACTIONS:
Oxidation is either:
The LOSS of Electrons
The GAINING of OXYGEN OR LOSS of HYDROGEN
The INCREASE in OXIDATION NUMBER
Reduction is either:
The GAIN of Electrons
The GAINING of HYDROGEN or LOSS of OXYGEN
The DECREASE in OXIDATION NUMBER
These definitions of Oxidation and Reduction have to go TOGETHER. I.e.
GAIN OF OXYGEN
2Mg + O2 ——> 2MgO
magnesium has been oxidised as it has gained oxygen
REMOVAL (LOSS) OF HYDROGEN
C2H5OH ——> CH3CHO + H2
ethanol has been oxidised as it has ‘lost’ hydrogen
GAIN OF HYDROGEN
C2H4 + H2 ——> C2H6
ethene has been reduced as it has gained hydrogen
REMOVAL (LOSS) OF OXYGEN
CuO + H2 ——> Cu + H2O
copper(II) oxide has been reduced as it has ‘lost’ oxygen
Used to... tell if oxidation or reduction has taken place
work out what has been oxidised and/or reduced
construct half equations and balance redox equations
OXIDATION NUMBER OR STATES ARE INTERCHANGEABLE IN IBDP CHEMISTRY - BOTH CAN BE USED BUT MEAN THE SAME THING!
ATOMS AND SIMPLE IONS
The number of electrons which must be added or removed to become neutral
atoms Na in Na = 0 neutral already ... no need to add any electrons
cations Na in Na+ = +1 need to add 1 electron to make Na+ neutral
anions Cl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutral
a) C
b) Fe3+
c) Fe2+
d) O2-
e) He
f) Al3+
a. 0
b. +3
c. +2
d. -2
e. 0
f. +3
ELEMENTS H in H2 = 0 both are the same and must add up to Zero
COMPOUNDS C in CO2 = +4
O in CO2 = -2 1 x +4 and 2 x -2 = Zero
Explanation
because CO2 is a neutral molecule, the sum of the oxidation states must be zero
for this, one element must have a positive OS and the other must be negative
HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE?
the more electronegative species will have the negative value
electronegativity increases across a period and decreases down a group
O is further to the right than C in the periodic table so it has the negative value
HOW DO YOU DETERMINE THE VALUE OF AN ELEMENT’S OXIDATION STATE?
from its position in the periodic table and/or
the other element(s) present in the formula
The SUM of the oxidation states adds up to THE CHARGE
e.g. NO3- sum of the oxidation states = - 1
SO42- sum of the oxidation states = - 2
NH4+ sum of the oxidation states = +1
Examples
in SO42- the oxidation state of S = +6 there is ONE S
O = -2 there are FOUR O’s
+6 + 4(-2) = -2 so the ion has a 2- charge
What is the oxidation state (OS) of Mn in MnO4¯ ?
the oxidation state of oxygen in most compounds is - 2
there are 4 O’s so the sum of its oxidation states - 8
overall charge on the ion is - 1
therefore the sum of all the oxidation states must add up to - 1
the oxidation states of Mn four O’s must therefore equal - 1
therefore the oxidation state of Mn in MnO4¯is +7
+7 + 4(-2) = - 1
Many elements can exist in more than one oxidation state
In compounds, certain elements are used as benchmarks to work out other values
HYDROGEN +1 except 0 atom (H) and molecule (H2)
-1 hydride ion, H¯ in sodium hydride NaH
OXYGEN -2 except 0 atom (O) and molecule (O2)
-1 in hydrogen peroxide, H2O2
+2 in F2O
FLUORINE -1 except 0 atom (F) and molecule (F2)
THE TRANSITION ELEMENTS AND MOST MAIN GROUP METALS HAVE VARIABLE OXIDATION STATES
Give the oxidation state of the element other than O, H or F in...
SO2
NH3
NO2
NH4+
IF7
Cl2O7
NO3¯
NO2¯
SO32-
S2O32-
S4O62-
MnO42-
SO2 O = -2 2 x -2 = - 4 overall neutral S = +4
NH3 H = +1 3 x +1 = +3 overall neutral N = - 3
NO2 O = -2 2 x -2 = - 4 overall neutral N = +4
NH4+ H = +1 4 x +1 = +4 overall +1 N = - 3
IF7 F = -1 7 x -1 = - 7 overall neutral I = +7
Cl2O7 O = -2 7 x -2 = -14 overall neutral Cl = +7 (14/2)
NO3¯ O = -2 3 x -2 = - 6 overall -1 N = +5
NO2¯ O = -2 2 x -2 = - 4 overall -1 N = +3
SO32- O = -2 3 x -2 = - 6 overall -2 S = +4
S2O32- O = -2 3 x -2 = - 6 overall -2 S = +2 (4/2)
S4O62- O = -2 6 x -2 = -12 overall -2 S = +2½ ! (10/4)
MnO42- O = -2 4 x -2 = - 8 overall -2 Mn = +6
What is the oxidation state of each element in the following compounds/ions?
CH4
PCl3
NCl3
CS2
ICl5
BrF3
PCl4+
H3PO4
NH4Cl
H2SO4
MgCO3
SOCl2
CH4 C = - 4 H = +1
PCl3 P = +3 Cl = -1
NCl3 N = +3 Cl = -1
CS2 C = +4 S = -2
ICl5 I = +5 Cl = -1
BrF3 Br = +3 F = -1
PCl4+ P = +4 Cl = -1
H3PO4 P = +5 H = +1 O = -2
NH4Cl N = -3 H = +1 Cl = -1
H2SO4 S = +6 H = +1 O = -2
MgCO3 Mg = +2 C = +4 O = -2
SOCl2 S = +4 Cl = -1 O = -2
What is the Oxidation State of O in H2O2?
What is the Oxidation State of H in NaH?
What is the Oxidation State of Cl in HClO?
To avoid ambiguity, the oxidation state is often included in the name of a species
manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO2
sulphur(VI) oxide for SO3 S is in the +6 oxidation state
dichromate(VI) for Cr2O72- Cr is in the +6 oxidation state
phosphorus(V) chloride for PCl5 P is in the +5 oxidation state
phosphorus(III) chloride for PCl3 P is in the +3 oxidation state
Name the following
PbO2
SnCl2
SbCl3
TiCl4
BrF5
PbO2 lead(IV) oxide
SnCl2 tin(II) chloride
SbCl3 antimony(III) chloride
TiCl4 titanium(IV) chloride
BrF5 bromine(V) fluoride