Chapter 19.1
Redox Reactions in Acidic or Basic Solutions
Dr. Peter Warburton
peterw@mun.ca
http://www.chem.mun.ca/zcourses/1011.php
Oxidation numbers
Redox reactions are easier to
balance if we understand where
the electrons are coming from and
where they are ending up.
Oxidation numbers help us figure
this out.
© Peter Warburton 2008
All media copyright of their respective
owners
2
Oxidation numbers
The two electrons in a bond are
completely assigned to the more
electronegative element.
UNLESS
there are bonds between two atoms of
the same element. The electrons are
shared equally so that one electron
is assigned to each of the atoms.
© Peter Warburton 2008
All media copyright of their respective
owners
3
Oxidation numbers
Based on this simple idea, we can
follow a set of rules to assign
oxidation numbers.
We go through the set of rules until
we find the FIRST rule that
applies to our specific atom in the
compound or ion of interest.
© Peter Warburton 2008
All media copyright of their respective
owners
4
Oxidation number rules (Rule 1)
Atoms of pure elemental
compounds (e.g. metals,
solid carbon, O2 gas, Br2
liquid, I2 solid, etc.) have an
oxidation number of ZERO
© Peter Warburton 2008
All media copyright of their respective
owners
5
Oxidation number rules (Rule 2)
Monatomic ions (like
2+
+
2Mg , Li , F , S , etc.) have
an
oxidation number equal
to the charge
© Peter Warburton 2008
All media copyright of their respective
owners
6
Oxidation number rules (Rule 3)
Fluorine, as the most
electronegative element, will
ALWAYS have an
oxidation number of -1
EXCEPT in F2
where it has an oxidation
number of ZERO (Rule 1)
© Peter Warburton 2008
All media copyright of their respective
owners
7
Oxidation number rules (Rule 4)
Oxygen, as the second most
electronegative element, will
usually have an
oxidation number of -2
UNLESS it is bonded to
another oxygen or fluorine
© Peter Warburton 2008
All media copyright of their respective
owners
8
Oxidation number rules (Rule 5)
Hydrogen, will have an
oxidation number of +1
unless it is bonded to a metal
atom, where it will have an
oxidation number of -1
© Peter Warburton 2008
All media copyright of their respective
owners
9
Oxidation number rules (Rule 6)
Halogens (Cl, Br, I, and At),
generally have an
oxidation number of -1
EXCEPT when bonded to F,
O, or halogens of the same
type or above it on the
periodic table.
© Peter Warburton 2008
All media copyright of their respective
owners
10
Oxidation number rules (Rule 7)
The sum of the oxidations
numbers for ALL the atoms
in a compound or ion MUST
ADD UP to match the total
charge on the compound
(zero) or ion (ion charge).
© Peter Warburton 2008
All media copyright of their respective
owners
11
Applying the rules
Apply the rules in the order
given. Any atoms not
specifically covered in the
rules can usually be assigned
oxidation numbers by
applying Rule 7 and some
logic.
© Peter Warburton 2008
All media copyright of their respective
owners
12
Problem
Assign oxidation numbers to
every atom in the following
compounds and ions:
S8
LiH
TiO2
H2O
H2O2
HSO4Cr2O72CaCO3
© Peter Warburton 2008
All media copyright of their respective
owners
13
Balancing redox equations
1.
2.
3.
Using oxidation numbers, identify
what is oxidized (loses electrons)
and reduced (gains electrons).
What are the products after the
oxidation and reduction take place?
Is the redox reaction done under
acidic or basic conditions?
© Peter Warburton 2008
All media copyright of their respective
owners
14
Balancing redox equations
The information from the previous
slide results in an unbalanced
skeleton equation where we
know generally what reactants
and products are specifically
involved in the electron transfer
(redox) process.
© Peter Warburton 2008
All media copyright of their respective
owners
15
Balancing redox equations
Examples of skeleton equations with
oxidation numbers shown:
Cr2O
6
2
7
aq   C2O aq   Cr aq   CO2 g 
-2
2
4
3 -2
3
3
 4 -2
Al s   NO g   AlOH s   NH3 aq 
4
3
0
© Peter Warburton 2008
5 -2
 3 - 2 1
All media copyright of their respective
owners
3 1
16
Half-reactions
We then break the skeleton reaction into
two unbalanced half-reactions where the
oxidation half-reaction has an atom
where the oxidation number becomes
more positive
and the
reduction half reaction has an atom
where the oxidation number become
more negative.
© Peter Warburton 2008
All media copyright of their respective
owners
17
Half-reactions from skeleton rxn
Cr2O72 aq   C2O 24 aq   Cr 3 aq   CO2 g 
6
3 -2
-2
3
 4 -2
Oxidation half-reaction
C2O
3
2
4
aq   CO2 g 
-2
each C atom should lose 1e
 4 -2
Reduction half-reaction
Cr2 O 72 aq   Cr 3 aq  each Cr atom should
6
-2
© Peter Warburton 2008
-
gain 3 e 
3
All media copyright of their respective
owners
18
Balancing half-reactions in ACIDIC solution
1.
Balance all atoms EXCEPT H and O in
each half reaction:
C2O24 aq   2 CO2 g  oxidation
Cr2 O 72 aq   2 Cr 3 aq  reduction
2.
Balance O atoms by adding water to the
side missing O atoms:
C2O24 aq   2 CO2 g  already balanced for O!
Cr2 O 72 aq   2 Cr 3 aq   7 H 2 O (l)
© Peter Warburton 2008
All media copyright of their respective
owners
19
Balancing half-reactions in ACIDIC solution
3.
Balance H atoms by adding H+ to the
side missing H atoms:
Oxidation half-reaction
C2O
2
4
aq   2 CO2 g  already balanced for H!
Reduction half-reaction
Cr2 O 72 aq   14 H  aq   2 Cr 3 aq   7 H 2 O (l)
© Peter Warburton 2008
All media copyright of their respective
owners
20
Balancing half-reactions in ACIDIC solution
4.
Balance charge by adding electrons to
the side with more total positive charge:
Oxidation half-reaction
C 2O aq   2 CO2 g   2 e




2
4
total charge is - 2
-
total charge is zero
Reduction half-reaction
Cr2O72 aq   14 H  aq   6 e-  2 Cr 3 aq   7 H 2O (l)




total charge is  6
total charge is 12
© Peter Warburton 2008
All media copyright of their respective
owners
21
Balancing half-reactions in ACIDIC solution
5.
Make the number of electrons the same
in both half-reactions by multiplication,
while avoiding a fractional number of
electrons:
Oxidation half-reaction
3 C2O
2
4
aq   6 CO2 g   6 e
-
Reduction half-reaction
Cr2 O 72 aq   14 H  aq   6 e -  2 Cr 3 aq   7 H 2O (l)
© Peter Warburton 2008
All media copyright of their respective
owners
22
Balancing half-reactions in ACIDIC solution
6.
Add the half reactions together and then
simplify by cancelling out species that
show up on both sides:
Added together
2
2

3 C2O4 aq   Cr2O7 aq   14 H aq   6 e
 6 CO2 g   2 Cr 3 aq   7 H 2O (l)  6 eSimplified (should have NO electrons!)
3 C2O
2
4
© Peter Warburton 2008
aq   Cr2O aq   14 H aq 
 6 CO2 g   2 Cr 3 aq   7 H 2O (l)
2
7
All media copyright of their respective
owners

23
Balanced reaction in ACIDIC solution
7.
Confirm that the reaction is
balanced in number of atoms and
total charge on both sides of the arrow.
If the reaction stoichiometry can be
simplified by division without giving
fractional coefficients, you can simplify
further:
3 C2O
2
4
aq   Cr2O72 aq   14 H  aq 
3
 6 CO2 g   2 Cr aq   7 H 2O (l)
© Peter Warburton 2008
All media copyright of their respective
owners
24
Balanced reaction in BASIC solution
First follow steps one to seven as seen in
acidic solution.
8.
Add the same number of OH- groups as
there are H+ present to BOTH sides of
the equation:
3 C 2 O 24 aq   Cr2 O 72 aq   14 H  aq   14 OH  aq 

added
 6 CO 2 g   2 Cr 3 aq   7 H 2 O (l)  14 OH  aq 

added
© Peter Warburton 2008
All media copyright of their respective
owners
25
Balanced reaction in BASIC solution
9.
One side of the reaction has BOTH OHand H+ present in equal amounts.
Combine these together to make an equal
amount of water:
3 C 2 O 24 aq   Cr2 O 72 aq   14 H  aq   14 OH  aq 


becomes 14 H 2 O (l)
 6 CO 2 g   2 Cr 3 aq   7 H 2 O (l)  14 OH  aq 
© Peter Warburton 2008
All media copyright of their respective
owners
26
Balanced reaction in BASIC solution
10.
3 C2O
Simplify by cancelling out an equal
number of water from each side until one
side has no water and confirm that the
reaction is balanced in number of atoms
and total charge on both sides of the
arrow:
2
4
aq   Cr2O aq   7 H 2O (l)
3

 6 CO2 g   2 Cr aq   14 OH aq 
© Peter Warburton 2008
2
7
All media copyright of their respective
owners
27
Problem
Balance the following
unbalanced redox skeleton
equation in BASIC solution
Al s   NO g   AlOH s   NH3 aq 
4
3
0
5 -2
© Peter Warburton 2008
 3 - 2 1
All media copyright of their respective
owners
3 1
28