Transcript Chapter 18.2

Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Chapter 18.2 - Balancing redox equations
Redox reactions: reduction – oxidaton reactions.
Reduction: lowering of positive charge (increase of negative charge)
Oxidation: increase of positive charge (lowering of negative charge)
If an electron is transferred, then
the negative charge must decrease somewhere (where the electron comes
from),
and it must increase somewhere else (where the electron goes),
hence reduction and oxidation must occur together.
There is a balance in such changes,
and reduction – oxidaton reactions provide good examples
for balancing chemical equations.
Redox Reactions, Acid, Base,
Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Chapter 18: Electrochemistry
• Chapter 18: Electrochemistry
• 18.2 Balancing Oxidation-Reduction
Equations [19.1]
• (Note: This section is usually covered with
a very brief review of oxidation numbers
and oxidizing/reducing agents, as covered
in Chemistry 1010, which is found in
section 4.9 of the Tro text).
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
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.
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
In a very simplistic way, the basis of oxidation numbers is the following idea:
Each (single) chemical bond between any two atoms X and Y is formed by
two electrons.
There are three possibilities:
1. X attracts electrons more than Y
2. Y attracts electrons more than X
3. X and Y attract electrons exactly the same way (important: this
happens only if X = Y)
Case 1. Even if the attraction is only slightly stronger by X than by Y, we
pretend that the entire electron pair belongs to X.
Case 2. Similarly to case 1, we pretend that the entire electron pair belongs
to Y.
Case 3. We assign one electron to X, and the other electron to Y.
If we carry out this formal assignment of electrons for each bond of the
molecule or ion, then the “charge” obtained on each atom is the
oxidation number of the given atom.
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
A numerical measure of how strongly atoms attract electrons is
electronegativity
We may phrase the concept of oxidation number in terms of
electronegativity:
In calculating oxidation numbers the two electrons in a bond
are completely assigned to the more electronegative
element,
UNLESS
the bond is between two atoms of the same element,
in this case the electrons are shared equally so that
one electron is assigned to each of the atoms.
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
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.
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
Oxidation number rules (Rule 2)
Monatomic ions (like Mg2+, Li+, F-, S2-, etc.) have an oxidation number
equal to the charge
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)
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
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
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, or if it is bonded to another H atom, where it will
have an oxidation number of 0 (Rule 1).
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.
Oxidation number rules (Rule 7)
The sum of the oxidation 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).
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
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.
If in any doubt, please remember that the main idea is to artificially
assign the entire bonding electron pair to the more electronegative atom
in each bond. If all bonding electron pairs are distributed this way, then
the charge obtained on each atom becomes the oxidation number.
(Of course, the charge on each atom includes the protons in the nucleus
as well as all the electrons of the atom).
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Problems:
Assign oxidation numbers to every atom
in the following compounds and ions:
S8
TiO2
H 2 O2
Cr2O72-
LiH
H 2O
HSO4CaCO3
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Using oxidation numbers, the balancing of redox chemical equations
becomes simpler.
The main steps in balancing redox equations:
1. Using oxidation numbers, identify
what is oxidized (what loses electrons) and
what is reduced (what gains electrons).
2. What are the products after the oxidation and reduction take place?
3. Is the redox reaction done under acidic or basic conditions?
The information obtained from these 3 steps results only in an unbalanced
skeleton equation where we know generally what reactants and products
are specifically involved in the electron transfer (redox) process.
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Examples
Examples of skeleton equations with
oxidation numbers shown:
Cr 2 O
6
2
7
aq   C 2 O aq  
2
4
3
-2
Al s   NO
0
 5 -2
3
Cr
3
-2
g  
Al OH
 3 - 2 1
3
aq   CO 2 g 
 4 -2
4 s   NH 3 aq 
-
 3 1
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
How to proceed further?
Brake up the problem to two formal half-reactions :
We break the skeleton reaction into two unbalanced halfreactions 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 becomes more negative.
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Example: Half-reactions from skeleton reaction
2
Cr 2 O 7
6
aq   C 2 O 24  aq  
3
-2
-2
Cr
3
aq   CO 2 g 
3
 4 -2
Oxidation half-reaction:
C 2O
3
2
4
aq  
CO 2 g 
each C atom should lose 1e
-
 4 -2
-2
Reduction half-reaction:
2
Cr 2 O 7
6
-2
aq  
Cr
3
aq 
each Cr atom should gain 3 e

3
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Steps for balancing half-reactions
in ACIDIC solution:
1.
Balance all atoms EXCEPT H and O in
each half reaction:
C 2 O 4 aq   2 CO 2 g  oxidation
2
3
Cr 2 O 7 aq   2 Cr aq  reduction
2
2.
Balance O atoms by adding water to the
side missing O atoms:
2
C 2O 4
aq   2 CO 2 g 
2
Cr 2 O 7
aq  
2 Cr
3
already balanced
for O!
aq   7 H 2 O (l)
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Steps for ACIDIC solution
3.
Balance H atoms by adding H+ to the
side missing H atoms:
Oxidation half-reaction
C 2O
2
4
aq   2 CO 2 g 
already balanced
for H!
Reduction half-reaction
2

3
Cr 2 O 7 aq   14 H aq   2 Cr aq   7 H 2 O (l)
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Steps for ACIDIC solution
4.
Balance charge by adding electrons to
the side with more total positive charge:
Oxidation half-reaction
C 2 O aq   2 CO 2 g   2 e
 
 

2
4
total charge is - 2
-
total charge is zero
Reduction half-reaction
Cr 2 O 7 aq   14 H aq   6 e  2 Cr aq   7 H 2 O (l)
      
       

2
total charge is  12

-
3
total charge is  6
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Steps for 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 C 2O
2
4
aq   6 CO 2 g   6 e
-
Reduction half-reaction
2
Cr 2 O 7
aq   14
H

aq   6 e -
 2 Cr
3
aq   7 H 2 O (l)
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Steps for ACIDIC solution
6.
Add the half reactions together and then
simplify by cancelling out species that
show up on both sides:
Added together
3 C 2O
2
4
aq   Cr 2 O aq   14 H aq   6 e
3
 6 CO 2 g   2 Cr aq   7 H 2 O (l)
2
7

-
6e
-
Simplified (should have NO extra electrons!)
3 C 2O
2
4
aq   Cr 2 O aq   14 H aq 
3
 6 CO 2 g   2 Cr aq   7 H 2 O (l)
2
7

Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Steps for 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:
2
3 C 2O 4
aq   Cr 2 O 72  aq   14 H  aq 
3
 6 CO 2 g   2 Cr aq   7 H 2 O (l)
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Steps for balancing half-reactions in BASIC solution:
First follow steps one to seven as seen in the
case of acidic solution.
8.
Add the same number of OH- groups as
there are H+ present to BOTH sides of
the equation:
2
3 C 2O 4
aq   Cr 2 O 72  aq   14
H


aq   14
OH aq 
  
added
 6 CO
3


aq   7 H 2 O (l)
g

2
Cr
2
 14 OH aq 
  

added
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Steps for BASIC solution
One side of the reaction has BOTH OHand H+ present in equal amounts.
Combine these together to make an equal
amount of water:
9.
2
3 C 2O 4


aq   Cr 2 O 72  aq   14
H aq   14 OH aq 
       

becomes 14 H 2 O (l)
 6 CO
3


aq   7 H 2 O (l)
g

2
Cr
2
 14 OH

aq 
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Steps for BASIC solution
10.
3 C 2O
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   Cr 2 O aq   7 H 2 O (l)
3





 6 CO 2 g  2 Cr
aq  14 OH aq 
2
7
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
Problem
Balance the following
unbalanced redox skeleton
equation in BASIC solution
Al s   NO
0
 5 -2
3
g  
Al OH
 3 - 2 1
4 s   NH 3 aq 
-
 3 1
Redox Reactions, Acid, Base , Paul G. Mezey
Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]
End of section
Redox Reactions, Acid, Base , Paul G. Mezey