Transcript Oxidation – Reduction Reactions

Chapter 14
Oxidation
and
Reduction
Oxygen is the most
abundant element on Earth
and is involved in many of
the most important chemical
reactions in our lives.
Earlier Theories
Originally oxidation was defined as the addition of
oxygen to a substance.
e.g. 1 - the burning of coal:
Carbon + Oxygen → Carbon Dioxide
C + O2
CO2
C gains oxygen
The carbon is said to be oxidised – oxidation reaction
e.g. 2 - the burning of magnesium
2Mg + O2 → 2MgO
Mg gains oxygen
The magnesium is said to be oxidised – oxidation reaction
It was also discovered that it was possible to remove
oxygen from some substances. Reduction is the opposite
of oxidation and was defined as the removal of oxygen
from a substance
e.g. when hydrogen gas is passed over heated copper(II)
oxide
Copper(II) Oxide + Hydrogen
CuO
+ H2
CuO loses oxygen
Copper + Water
Cu
+ H2O
Addition of
hydrogen – using
H2 to remove
oxygen
The copper oxide is said to be reduced – reduction rxn
‘OIL RIG’ Theory
(Oxidation & Reduction involving Electron
Transfer)
After the discovery of the electron it was noted that
what was common to both of these reactions was the
transfer of electrons from one substance to another.
From this discovery grew a new understanding and
definition of the terms Oxidation and Reduction!!!
Oxidation
Definition:
Oxidation of an element takes place when it loses
electrons
e.g. the burning of magnesium
2Mg2+O2-
2Mg + O2
Mg loses 2 electrons
Oxidation
Reduction
Definition:
Reduction of an element takes place when it gains
electrons
e.g. when hydrogen gas is passed over heated
copper(II) oxide
Cu2+O2- + H2
Cu2+ gains two electrons
Reduction
Cu + H2O
O
I
L
R
I
G
– oxidation
– is
– loss
– reduction
– is
– gain
Oxidation – Reduction Reactions
(Redox Reactions)
If one substance loses electrons (oxidation) then there
must be another substance to accept these electrons
(reduction). Thus, it makes sense that oxidation and
reduction must always occur together and such
reactions are known as oxidation-reduction reactions
Oxidation-Reduction Reactions
e.g. 1 – formation of sodium chloride
2Na + Cl2
→ 2Na+Cl-
• Each Na atom loses 1e- - oxidation
• Each Cl atom gains 1e- - reduction
*Note: - In the above rxn neither oxygen or hydrogen
are present. Thus allowing for a broader definition of
oxidation and reduction than originally developed
e.g. 2 – the burning of magnesium
2Mg + O2
2Mg2+O2-
• Each Mg atom loses 2e- - oxidation
• Each O atom gains 2e- - reduction
e.g. 3 – visible results of an oxidation-reduction rxn!!!
Reaction between zinc metal and copper ions
Place a piece of zinc metal in a solution of copper
sulfate and observe
Copper has left the solution to become plated
onto the zinc plate
Explanation: The zinc has become plated with copper
metal (reddish deposit) and at the same time the blue
colour of the solution has faded. This fading implies that
the Cu2+ ions are being used up. On analysis of the faded
solution it now contains Zn2+ ions. The above reaction
can be explained by the following oxidation-reduction
reaction:
Zn + Cu2+ → Zn2+ + Cu
• Each Zn atom loses 2e- - oxidation
• Each Cu2+ atom gains 2e- - reduction
Oxidising & Reducing Agents
A substance that allows oxidation to take place by
gaining electrons itself is called the oxidising agent
e.g. Cu2+
Definition
An oxidising agent is a substance that brings about
oxidation in other substances and is itself reduced
e.g. Cu2+
Definition
A reducing agent is a substance that brings about
reduction in other substances and is itself oxidised
e.g Zn
Complete the following redox reaction in terms of
oxidation and reduction:
Zn + Cu2+ → Zn2+ + Cu
Useful Hints:
Oxidation involves reactions such as:
- addition of oxygen or other electronegative
element e.g. Mg → MgO
- removal of hydrogen e.g. HCl → Cl
- increase in valency e.g. FeCl2 → FeCl3
Reduction involves reactions such as:
- removal of oxygen e.g. MgO → Mg (Mg2+ + 2e- → Mg)
- addition of hydrogen e.g. Cl → HCl (Cl + e- → Cl-)
- decrease in valency e.g. FeCl3 → FeCl2 (Fe3+ + e- → Fe2+)
Oxidation & Reduction Reactions in
terms of Electron Transfer
Please leave space (2 pages) for
examples
Common Oxidising and Reducing Agents
Oxidising Agents:
1. Hydrogen Peroxide (H2O2) – used in the bleaching of
hair converts the coloured pigment in hair to a
colourless hair pigment.
The H2O2 is reduced and is itself the oxidising
agent
2. Tincture of iodine – often found in first aid boxes
used to prevent the infection of cuts.
Live Germ + I2
→
2I-
+ Dead Germ
Oxidising
Agent
- The iodine, I2, is the oxidising agent and it is
itself reduced
- The iodine oxidises chemicals in the cells of
germs – kills them.
3. Chlorine is added to swimming pool and drinking
water to oxidise chemicals in the cells of germs. This
kills the germs and disinfects the water.
Live Germ + Cl2 → 2Cl- + Dead Germ
Oxidising
Agent
- Chlorine, Cl2, is an oxidising agent and it is itself
reduced
Learn:
4. Sodium Hypochlorite (NaClO) – is a chlorine
compound found in household bleach e.g. domestos.
The hypochlorite ions ClO- oxidise the coloured
material in fabrics to a colourless material
Coloured Fabric + ClO- → Cl- + Colourless fabric
Oxidising
Agent
- When NaClO is added to water the oxidisng agent
HOCl is formed
- HOCl acts as an oxidising agent and is itself
reduced to the chloride ion – Cl-
Reducing Agents:
Learn:
1. Carbon Monoxide – is used in industry to remove
the oxygen from iron ore in order to convert it to
pure iron
Fe2 O3 + 3CO → 2Fe + 3CO2
Reducing
Agent
- CO is the reducing agent and is itself oxidised
(gives electrons to the 2nd O atom)
- Fe2O3 is the oxidising agent and is itself
reduced
Learn:
2. Sulphur Dioxide (SO2) – is used to bleach straw and
paper. E.g. Used in making straw hats – converts the
yellow colour of the dye in straw to a colourless form
of the dye
Coloured form + SO2
of dye
Reducing
Colourless form + SO3
of dye
Agent
- SO2 is the reducing agent and is itself oxidised –
SO3
Quick Review
Oxidising Agents:
I2 is an oxidising agent and is reduced to ICl2 is an oxidising agent and is reduced to ClClO- (HOCl) is an oxidising agent and is reduced to ClReducing Agents:
CO is a reducing agent and is oxidised to CO2
SO2 is a reducing agent and is oxidised to SO3
Must know one example of bleach as an
oxidising agent (NaClO) and as a reducing agent (SO2 )
Oxidation Numbers
With many developments in chemistry it became clear
that the definitions of oxidation and reduction in terms
of electron transfer was no longer sufficient.
Consider the reaction between carbon and oxygen
C + O2
CO2
According to the earlier theory, here the carbon is
oxidised.
But as CO2 is a covalent molecule and NO IONS here,
there is NO complete transfer of electrons from Carbon
to form Carbon Dioxide.
i.e. CO2 is a covalent molecule where the electrons are
shared.
In order to overcome this problem oxidation number
(state) was introduced:
Definition:
Oxidation Number: is the charge that an atom has or
appears to have when electrons are distributed
according to certain rules.
Oxidation number is also described as the charge
which an atom appears to have in a compound if the
bonding is assumed to be completely ionic.
H2 O
The two electrons in
each of the covalent
bonds are counted
with the more
electronegative atom
i.e. oxygen, therefore,
Oxidation number of
X
oxygen is -2
H
O
Since each hydrogen has
now formally lost an e- oxidation number of each
hydrogen is +1
X
H
Water in terms of oxidation number:
H2 O
+1 -2
Rules for Assigning Oxidation Numbers
Rule 1:
The oxidation number of any uncombined element is
zero:
e.g. O.N. of sodium Na is 0
O.N. of hydrogen (H) in H2 is 0
O.N. of fluorine (F) in F2 is 0
Rule 2:
The sum of the oxidation numbers of all the atoms in a
molecule must add up to zero:
e.g.
O.N.
Sum O.No’s =
NaCl
+1 -1
0
H2O
2(+1)-2
0
Rule 3:
The oxidation number of an ion of an element is the
same as its charge:
e.g. O.N. of Na+ is +1
O.N. of N3- is -3
O.N. of Na in Na+Cl- is +1
O.N. of Cl in Na+Cl- is -1
The alkali metals (Li, Na, K etc.) and the alkaline earth
metals (Be, Mg, Ca etc.) nearly always have the same
O.N. in their compounds:
Alkali Metals O.N. = +1
Alkaline Earth Metals O.N. = +2
Rule 4:
The O.N. of oxygen is always -2 except:
- in peroxides (H2 O2) when it has an O.N. of -1
- in the compound OF2 when it has an O.N. of +2
*Note: Fluorine is the only element that is more
electronegative than oxygen. For this reason in the OF2
molecule the electrons in the two covalent bonds are
assigned to the fluorine atoms. So, in order for the sum of
the oxidation numbers to equal 0 oxygen is given an O.N.
of +2
Rule 5:
The O.N. of hydrogen is always +1 except in the metal
hydrides when it is assigned an O.N. of -1:
e.g. NaH – H atom O.N. of -1 because sodium
hydride is an ionic compound Na+H-. The
O.N. of -1 comes from the one negative
charge on the hydrogen atom
MgH2
Rule 6:
The oxidation number of a haolgen (Group 7) when
bonded to a less electronegative atom is -1.
e.g. - Fluorine is the most electronegative
element and always has an oxidation
number of -1
- Chlorine has an oxidation number of -1 in
compounds where it is not bonded to
oxygen or fluorine (both more
electronegative than chlorine)
Cl2O - O atom has O.N. of -2
- each Cl atom must have O.N. of +1
Cl2O7 - each O atom has O.N. of -2 = total of -14
- each Cl atom must have O.N. of +7
Rule 7:
The sum of the O.N.’s of all the elements in a complex
ion must equal the charge on the ion.
e.g. NO3- - in the nitrate ion
O.N.
Sum of O.N.
NO3-1
(+5)+(3)(-2)
-1
Calculating Oxidation Numbers
Example 1:
What is the oxidation number of the sulphur atom in
the H2SO4 molecule?
*Note: Must be very familiar with the seven rules
Solution:
- Let the O.N. of S
=x
- The O.N. of H
= +1
- The O.N. of O
= -2
So,
2(1) + x + 4(-2) = 0
H2 SO4
2 + x -8 = 0
x -6 = 0
2(+1) +6 4(-2)
x = 6 → O.N. of S atom = 6
Example 2:
What is the O.N. Of carbon in CO2?
Solution:
- Let the O.N. of C
=x
- The O.N. of O
= -2
So,
x + 2(-2) = 0
x -4 = 0
x = 4 → O.N. of C atom = 4
CO2
+4 2(-2)
Example 3:
What is the oxidation number of each of the elements in
CuSO4
Note: Cu is a transition metal and can have variable
oxidation states
Of the three elements in CuSO4 only one is
mentioned in the rules for assigning O.N.
- To work out the O.N.’s on Cu and S remember the ion
SO42- Therefore, in order to balance molecule Cu must have
+2 charge – Cu2+
- SO42- ion:
- O.N. of S
=x
-O.N. of O
= -2
x + 4(-2) = -2
x -8 = -2
x=6
Now:
O.N. Cu = +2
O.N. S = +6
O.N. O = -2
CuSO4
+2 +6 4(-2)
Try the following:
i)
What is the oxidation number of sulphur in
Na2S2O3
ii) What is the oxidation number of sulphur in
Na2S4O6
Transition Metals and their Oxidation
Numbers
The transition metals each have a number of
oxidation numbers (states) in their compounds
Element
Chromium
Manganese
Iron
Copper
Oxidation
Number
2, 3, 6
2, 3, 4, 6, 7
2, 3, 6
1, 2
Example 4:
What is the systematic name of MnCl2?
Solution:
- Let the O.N. of Mn
- The O.N. of Cl is
=x
= -1
So,
x + 2(-1) = 0
x -2 = 0
x = 2 → The O.N. of Mn atom = +2
Therefore when naming this molecule since Mn is a
transition metal need to identify its oxidation state in
its name:
Manganese(II) Chloride
Using oxidation numbers give the systematic names
of the following TRANSITION METAL compounds:
a) CuCl
b) CuCl2
c) FeO
d) Fe2O3
Quiz 1
What is the oxidation number of
Hydrogen in H2O
Hydrogen in H2O2
Oxygen in H2O2
Oxygen in Cu2O
Calcium in CaCl2
Nitrogen in NH3
Quiz 2
What is the oxidation number of:
Carbon in CO32Hydrogen in OHSulfur in SO42Hydrogen in MgH2
Carbon in C6H12O6
Manganese in MnO4-
Complete the following:
Book - pg 190 14.3 – 14.5
Workbook – W14.1 – W14.8
Oxidation and Reduction in terms of
Oxidation Numbers
Oxidation numbers may be used to find out what is
oxidised and what is reduced in a redox reaction
Oxidation, in terms of oxidation numbers, is an
increase in oxidation number
Reduction, in terms of oxidation numbers, is a
decrease in oxidation number
Example 1:
Using oxidation numbers, indicate the species oxidised
and reduced in the reaction:
2H2O + O2 → 2H2O
Example 2:
What is i) oxidised ii) reduced iii) the oxidising agent iv)
the reducing agent in the following redox reaction:
As2O3 + 2I2 + 2H2O → As2 O5 + 4I- + 4H+
Please leave space for answers – 1 page
Try the following:
What is a) oxidised b) reduced c) the oxidising agent d)
the reducing agent in the following redox reactions?
a) 2SO2 + O2 → 2SO3
b) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
c) 2MnO4- + 16H+ + 5C2O42- → 2Mn2+ + 10CO2 + 8H2O
Now review study of reducing agents in industry:
i) CO
ii) SO2
Homework: write out a full list of equations for all
the redox reactions that were investigated when
completing experiments on oxidation-reduction
reactions. Using oxidation numbers label the species
that have been oxidised and reduced.
You should be able to complete the
following questions:
Book – pg190 14.6 – 14.7
Workbook – W14.9 – 14.13
Balancing Redox Equations Using
Oxidation Numbers
We have previously learned how to balance equations
by inspection. This method is generally only appropriate
for fairly simple equations. So, for more complicated
equations it is useful to have a more systematic
method. In the case of oxidation-reduction reactions
we will use oxidation numbers to balance the
equations.
Step 1: Assign oxidation numbers to all the atoms in
the equation
Step 2: Identify clearly any elements which show a
change in oxidation number.
Step 3: Show the number of electrons lost and gained
Step 4: Work out the ratio of oxidising agent to
reducing agent
Step 5: Balance the remaining items in the equation
using the inspection method (As a general rule
leave oxygen and hydrogen atoms as the last
items to be balanced.
Example 1:
Using oxidation numbers, balance the following
equation:
Fe2+ + Cl2 → Fe3+ + Cl-
Example 2:
Using oxidation numbers balance the following
equation:
Cr2O72- + Fe2+ + H+ → Cr3+ + Fe3+ + H2O
Example 3:
Using oxidation numbers, balance the following
equation:
MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O
Try the following:
a) Cu + HNO3 + H+ → Cu2+ + NO + H2O
b) MnO4- + CH3OH + H+ → Mn2+ + HCHO + H2O