Download Electrochemistry Oxidation – Reduction and Oxidation Numbers

Electrochemistry
Oxidation – Reduction and Oxidation Numbers
Many important chemical reactions involve the
transfer of electrons.
e.g.
2Na + Cl2 Æ 2 NaCl
Many do not.
e.g. Pb(NO3)2(aq) + 2KI(aq) Æ PbI2(s) + 2KNO3(aq)
Chemists have devised a useful bookkeeping method
to determine if electron transfer is involved in a
chemical reaction. It is referred to as the oxidation
state method.
If an atom loses electrons, it is oxidized.
If an atom gains electrons, it is reduced.
The oxidation state or equivalently, the oxidation
number, of an atom is the charge assigned to an atom
using the oxidation state method.
Rules for assigning oxidation numbers:
1. Elements in their most abundant naturally occurring
form are assigned an oxidation number of zero.
e.g. Na, Fe, Cl2, O2
2. The sum of the oxidation numbers for a compound or
formula unit is zero.
3. For a polyatomic ion, the oxidation numbers of the
constituent atoms sums to the ion charge.
4. Monatomic ions are assigned oxidation numbers equal
to the ion charge.
e.g., Na+ = +1, Al3+ = +3, Cl- = -1, N3- = -3
5. Oxygen in a compound or ion usually has an oxidation
state of –2. (Peroxides are the exception, in which case the
oxidation number is –1.)
6. Hydrogen in a compound or ion usually has an
oxidation state of +1. (Hydrides are the exception, in
which case the oxidation number is –1.)
7. For covalently bonded substances, shared electrons are
assigned to the more electronegative element.
8. When two atoms of the same element share electrons,
they are divided equally between the atoms (each atom gets
half.)
Determining oxidation numbers from a formula is usually
an exercise in solving for the unknown oxidation number of
one if the elements in the formula.
Procedure for assigning an oxidation number:
1.
List the known oxidation numbers of all atoms in
the compound or ion.
2.
Multiply each oxidation number by the relevant
number of atoms.
3.
Determine the difference between the sum of the
oxidation numbers of the known atoms and the total
charge of the molecule or ion.
4.
Divide the difference by the number of atoms of the
unknown oxidation state.
e.g.
K2Cr2O7
O = -2; K = +1; solve for Cr
0 = 7(-2) + 2(+1) + 2(oxidation number of Cr)
+12 = 2(oxidation number of Cr)
+6 = oxidation number of Cr
e.g
NaNO3
O = -2; Na = +1; solve for N
0 = 3(-2) + 1(+1) + (oxidation number of N)
+5 = oxidation number of N
Acid-Base reactions: H+ transfer
AcidA + BaseB
BaseA + AcidB
AcidA,BaseA are protonated and deprotonated forms
of A. They constitute a conjugate acid-base pair.
(The same holds for BaseB, AcidB.)
Oxidation-Reduction reactions: e- transfer
AOX + BRED
ARED + BOX
AOX, ARED are oxidized and reduced forms of A.
BOX, BRED are oxidized and reduced forms of B.
AOX and BOX are oxidizing agents
ARED and BRED are reducing agents
AOX oxidizes BRED and BRED reduces AOX
In acid-base chemistry we speak in terms of strong
acids and bases and weak acids and bases. We also
speak of strong and oxidizing agents and strong and
weak reducing agents….
Strong oxidizing agents have high electron affinities.
Strong reducing agents readily donate electrons.
Balancing oxidation/reduction reactions:
as always, the conservation laws must be obeyed
the half reaction method
(two examples: acid solution, base solution)
In acid solution:
consider the unbalanced net reaction:
MnO4- + C2O42-
Mn2+ + CO2
Step 1: write the half reactions (for each redox pair)
(i.e. AOX
ARED; BRED
BOX)
MnO4-
C2O42-
Mn2+
CO2
Step 2: balance each half reaction using H2O, H+,
and e-, as needed
MnO4- + 8H+ + 5eC2O42-
Mn2+ + 4H2O
2CO2 + 2e-
Step 3: multiply each half reaction as needed to
balance the number of electrons on each side
and add the half reactions
2(MnO4- + 8H+ + 5e-
Mn2+ + 4H2O)
5(C2O422CO2 + 2e-)
___________________________________________
2MnO4- + 16H+ + 5C2O42-
2Mn2+ + 8H2O + 10CO2
In base solution:
MnO4- + C2O42-
Mn2+ + CO2
Step 1: write the half reactions (for each redox pair)
(i.e. AOX
ARED; BRED
BOX)
MnO4-
C2O42-
Mn2+
CO2
Step 2: balance each half reaction using H2O, OH-,
and e-, as needed
MnO4- + 4H2O + 5eC2O42-
Mn2+ + 8OH-
2CO2 + 2e-
Step 3: multiply each half reaction as needed to
balance the number of electrons on each side
and add the half reactions
2(MnO4- + 4H2O + 5e-
Mn2+ + 8OH-)
5(C2O422CO2 + 2e-)
___________________________________________
2MnO4- + 8H2O + 5C2O42-
2Mn2+ + 16OH- + 10CO2
n = #electrons transferred = 10
Ex. 2
H3AsO3 + I2
H3AsO4 + I-
In acid:
H3AsO3 + H2O
H3AsO4 +2H+ + 2e-
I2 + 2e2I___________________________________________
H3AsO3 + I2 + H2O
H3AsO4 + 2I- + 2H+
In base:
H3AsO3 + 2OH-
H3AsO4 + H2O + 2e-
I2 + 2e2I___________________________________________
H3AsO3 + I2 + 2OHH3AsO4 + 2I- + H2O
n = #electrons transferred = 2