Oxidation and Reduction
C020
A broad class of reactions in chemistry involves the exchange of electrons
and are known as reduction-oxidation (redox) reactions
An atom is oxidized when it looses electrons, and reduced when it gains electrons
An oxidation process is always accompanied by a reduction process
Divide these reactions into two halves that represent the corresponding
loss and gain of electrons
Examples of oxidation half reactions:
Na → Na+ + e- Fe2+ → Fe3+ = e-
Examples of reduction half reactions:
Cl2 + 2e- → 2Cl- Cu2+ + 2e- → Cu
Redox Reactions 2
1
Two mnemonics to remember this:
OIL RIG = oxidation is loss; reduction is gain
LEO GER = lose electrons oxidize; gain electrons reduce
The species supplying the electron is said to be the reducing agent
(which itself is oxidized)
The species accepting electrons is said to be the oxidizing agent
(which is itself reduced)
Redox Reactions 3
Oxidation numbers
An oxidation number (ON) represents “the number of electrons theoretically
lost or gain by each atom in a molecule during a reaction
ON can (and usually are) integers but they also can be fractional!
How to assign oxidation numbers
1.) All neutral elements and their allotropes have an ON = 0
For example: O20, O30, Cl20, Fe0, S80
2.) For monatomic ions, ON = ion charge
For example: O2- (-2), Al3+ (+3), Na+ (+1)
Redox Reactions 4
2
3.) In a polyatomic species, individual ONs are usually as follows
in order of importance:
i) F = -1
ii) Group I metals = +1 (Li, Na, K, Rb, Cs)
iii) Group II metals = +2 (Be, Mg, Ca, Sr, Ba)
iv) H = +1
v) O = -2
vi) Group VII halogens = -1 (Cl, Br, I)
Redox Reactions 5
4.) In a polyatomic species the sum of all individual ONs =
overall charge on the species
Examples:
+1 -2
H2O Neutral = 0 = 2x(+1)-2
+1 -2?
H2O2 Neutral = 0= 2x(+1)+2x(-2) =-2
No problem. H has priority over O so solve for ON for O
Neutral = 0= 2x(+1)+2x(ON)
∴ ON = -1
The O in peroxide forms an exception to the expectations of point 3).
Redox Reactions 6
3
+1 -2
H2PO4- Net charge = -1 = 2 x (+1) + ON(P) + 4 x (-2)
∴ ON(P) = -1 -2 + 8 = +5
In general groups I and II and other metals act as reducing agents; that is,
they donate electrons, and groups VI and VII act as oxidizing agents; that is,
they accept electrons
Redox Reactions 7
More examples
+2 -2
1) MgO Using ON(O) = -2
-3 3x(+1)
2) PH3 Using ON(H) = +1
+1 -2
3) HS- Using ON(H) = +1 and ON(overall) = -1
(+1) + ON(S) = -1 ∴ ON(S) = -2
+5 -2
4) ClO3- Using ON(O) = -2 which has higher priority over Cl
ON(Cl) + 3 x (-2) = -1 ∴ ON(Cl) = +5
+2 -2
Using ON(O) = -2
5) S2O32-
2 x ON(S) + 3 x (-2) = -2 ∴ 2 x ON(S) = +4
ON(S) = +2
+1 -1
6) NaCl Using ON(group I) = +1 and/or ON(group VII) = -1
Redox Reactions 8
4
Balancing Redox Equations
There are a number of ways to balance redox reactions;
we will use the half-reaction method since it does not require judging
which species is reduced or oxidized
A redox equation is fully balanced only when
1.) there is a complete mass balance
2.) the # electrons liberated by the reducing agent
= the # electrons accepted by the oxidizing agent
Redox Reactions 9
Half-
Half-Reaction Method Steps
1.) Recognize the reaction as an oxidation-reduction process and separate
the reaction, preferably written in ionic form, into two half reactions:
one an oxidation, the other a reduction
This can usually be done by inspection, but oxidation numbers can guide you
2.) balance each half-reaction separately as follows:
a) Balance coefficients for all atoms except H and O
b) Add sufficient H2O to side deficient in O to balance O
c) Calculate the number of H atoms by which side is deficient
i) in acidic solution add H+ to side deficient in H
ii) in basic solution add the # of H2O to side deficient in
H, and the same # of OH- to other side
d) Balance charges by adding e- to side deficient in negative charge
Redox Reactions 10
5
3.) Multiply half reactions by suitable coefficients to balance the e-
and add the two half reactions together
4.) Cancel out species common to both sides of the equation
5.) Check both the atom and charge balance
Redox Reactions 11
Oxidation
Example:
2x+1 +3 4x-2 +2 +4 2X-2
+7 4x-2
MnO4- (aq) +H2C2O4(aq) → Mn2+(aq) + CO2(g)
Reduction
∴ Half reactions are:
a) MnO4- → Mn2+
b) H2C2O4(aq) → CO2 (g)
a)
5e- + 8H+ + MnO4- → Mn2+ + 4H2O
ChargeLHS = 8 – 1 = +7 ChargeRHS = +2 ∴ Add 5e- to reactant side
(reduction)
Redox Reactions 12
6
b) H2C2O4 (aq) → 2 CO2 (g) + 2H+ +2e-
∴ Products gain 2 e ;
-
ChargeLHS = 0 ChargeRHS = +2 reactants oxidized
c) Multiply reduction reaction by x2 and the oxidation reaction by x 5 so that both
reactions involve 10 e-
10 e- +16H+ + 2MnO4- → 2Mn2+ + 8H2O
+
5H2C2O4 → 10CO2 + 10H+ + 10e-
6 16H+ + 5H2C2O4 + 2 MnO4- →2Mn2+ + 10CO2 + 10H+ + 8H2O
6H+ + 5H2C2O4 + 2 MnO4- → 2Mn2+ + 10CO2 + 8H2O
Mass check: 2Mn, 28O, 10C and 16H on both sides
ChargeLHS = +6 – 2 = +4 ChargeRHS = +4
Redox Reactions 13
Disproportionation Reactions
For elements with 3 or more oxidation states, an element in
an intermediate oxidation state can be both oxidized and reduced
(a redox reaction know as disproportionation)
Example: 0 2x-1
+1 -1 +2
2CuCl → Cu +CuCl2
Here oxidation numbers help establish if redox reaction is also
a disproportionation reaction
Such reactions can sometimes be difficult to balance
Example: Balance P4 → PH3 +H2PO2- in basic solution
Hint Rewrite the equation with the disproportionating species
written as reacting with itself
Redox Reactions 14
7
P4 + P4 → PH3 +H2PO2-
Half reactions
12e- + 12H2O + P4 → 4 PH3 +12OH x4
-
Reduction
8OH- + P4 → 4 H2PO2- + 4e- Oxidation x12
48H2O + 16P4 +96OH- → 16PH3 +48H2PO2- + 48OH-
48
Divide equation by 16
3H2O + P4 +3OH- → PH3 + 3H2PO2-
Mass check: 9H, 6O and 4 P on both sides of the equation
ChargeLHS = -3 ChargeRHS = -3
Balanced
Redox Reactions 15
