CHEM-LEC-LESSON-4.pdf

CHEM LEC Lesson 4
Redox Reactions
Corrosion
One of the most familiar redox reactions is the corrosion of a metal. In some instances
corrosion is limited to the surface of a metal, in other instances the corrosion goes
deeper which eventually compromises the structural integrity of the metal.
Examples of corrosion are the following:
•
o green coating on the copper metal which contains Cu2+Cu2+ ions combined
with carbonate and hydroxide anions.
•
o rust containing Fe3+Fe3+ combined with oxide and hydroxide anions.
•
o silver tarnish containing Ag+Ag+ combined with sulfide ions.

Based on the examples above, we define corrosion as the conversion of a metal into a
metal compound by a reaction between a metal and some substance in its environment.
When a metal corrodes, each metal loses electrons and so forms a cation, which can
combine with an anion to form an ionic compound.
Redox Reactions
A redox reaction is a type of chemical reaction wherein there is a transfer of electrons
from one substance to another causing a change in their oxidation states.
The term redox is used because the reaction is composed of two simultaneous
chemical reactions - reduction and oxidation. In these two simultaneous reactions, we
must remember that when one loses an electron another substance must gain that
electron.
The Nuclear Atom
We recall also that there are three primary subatomic particles: protons, electrons and
neutrons.
• Protons and neutrons make up the nucleus of an atom. Collectively, they are called
the nucleons.

• Protons are positively-charged and neutrons are uncharged making the charge of the
nucleus positive.
• Negatively charged electrons outside the nucleus occupy most of the volume of the
atom.
• All atoms of the same element have the same number of protons in their nuclei. The
number of protons in the nucleus of an atom is called the atomic number.
Calcium has an atomic number of 20. This means that there are 20 protons in the
nucleus of a Ca atom.
• A neutral atom has no net electrical charge because the number of electrons outside
the nucleus equals the number of protons inside the nucleus.
A neutral calcium atom has 20 protons and 20 electrons.
Ions
Electrons are considered as the most important subatomic particles by chemists
because they are the first part of the atom to contact another atom. When one atom
comes into contact with another atom, it is the electrons of both atoms that interact. This

means that electrons can be transferred from one atom to another. When this happens,
an ion is formed.
An ion is an atom, or a group of atoms, that have lost or gained one or more electrons
so that they are no longer electrically neutral.
• When one or more electrons are removed from one atom, a positive ion is
formed called a cation.
A Na+ ion is formed when a neutral sodium ion loses an electron. The charge is +1
because there is 1 more proton compared to the number of electrons.

▪ A Mg2+Mg2+ ion is formed when a neutral magnesium ion loses two electrons. The
charge is +2because there are 2 more protons compared to the number of electrons.
• The removed electrons are transferred to another atom forming a negative ion called
an anion.
A F− ion is formed when a neutral fluorine ion gains an electron. The charge is -1 because there
is 1 more electron compared to the number of protons.
An O2−O2− ion is formed when a neutral oxygen ion gains two electrons. The charge is
-2 because there are 2 more electrons compared to the number of protons.
An ion is formed when a neutral atom loses/gains a/an
Answer: electrons
When a neutral atom loses an electron, a/an ____ is formed.
Answer: cation
When a neutral atom gains an electron, a/an ____ is formed.
Answer: anion

Oxidation Number
An oxidation number is the total number of electrons that an atom will either gain or
lose in order to form a chemical bond with another atom.
We can predict the oxidation number of an element based on its position in the periodic
table.
Rules for Assigning Oxidation Numbers
1. For an atom in its elemental form, the oxidation number is always zero.

• I2 is the elemental form of iodine and has an oxidation number of 0.
Na is the elemental form of sodium and has an oxidation number of 0.
2. For any monoatomic ion, the oxidation number equals the ionic charge.
o Mg2+
is a monoatomic ion and has an oxidation number of +2.
3. Nonmetals usually have negative oxidation numbers, although they can
sometimes be positive:
• (a) The oxidation number of oxygen is usually -2 except in peroxides.
• (b) The oxidation number of hydrogen is usually +1 when bonded to nonmetals
and -1 when bonded to metals
• (c) The oxidation number of fluorine is -1 in all compounds.
• (d) The other halogens have an oxidation number of -1 in most binary
compounds but have positive oxidation numbers when combined with oxygen.
4. The sum of the oxidation numbers of all atoms in a neutral compound is zero.
• Mg3(AsO4)3 is a neutral compound and has an oxidation number of 0.
• This is not a peroxide so the oxidation number of oxygen is –2.
• Magnesium is an alkaline earth metal (group II) so its oxidation number is +2.
• Solve for the oxidation number of arsenic algebraically:

3Mg + 2As +8O = 0
3(+2) + 2As + 8(-2) = 0
6 + 2As - 16 = 0
As = 5
• Ca is an alkaline earth metal so its oxidation number is +2.
• CaH2 is a neutral compound and has an oxidation number of 0.
• Hydrogen is bonded to a metal so its oxidation number is –1.
This is not a peroxide so the oxidation number of O is – 2.
• MnFO3 is a neutral compound and has an oxidation number of 0.
• The oxidation number of fluorine in all compounds is -1.
• Solve for the oxidation number of manganese algebraically:
Mn + F + 3O = 0
Mn + (-1) + 3(-2) = 0
Mn -1 - 6 = 0
Mn = 7
5. The sum of the oxidation numbers of all atoms in a polyatomic ion equals the charge
of the ion.
• This is not a peroxide so the oxidation number of oxygen is -2.

• • ClO-
is a polyatomic anion with a charge of -1 so the summation of all charges
of all atoms is -1.
• • Solve for the oxidation number of chlorine algebraically:
•
Cl + O = -1 Cl + (-2) = -1
Cl = -1 + 2
Cl = 1
• This is not a peroxide so the oxidation number of oxygen is -2.
• • Cr2O7
2-
is a polyatomic anion with a charge of -2 so the summation of all charges of all atoms
is -2.
• • Solve for the oxidation number of chromium algebraically:
2Cr + 7O = -2
2Cr + 7(-2) = -2
2Cr - 14 = -2
2Cr = 14 - 2
Cr = 6
Give the name of the compound Al(NO3)3
Answer: Aluminum nitrate
How many nitrogen atoms are there in the compound Al(NO3)3?
Answer: 3
How many oxygen atoms are there in the compound Al(NO3)3?
Answer: 9
Give the oxidation number of aluminum in the compound Al(NO3)3?
Answer: +3
Give the oxidation number of oxygen in the compound Al(NO3)3?
Answer: -2

Give the oxidation number of nitrogen in the compound Al(NO3)3?
Answer: +5
What is the oxidation number of H in H2(g)
Answer: 0
What is the oxidation number of Cl in LiClO3?
Answer: +5
What is the oxidation number of Cr in Au2(CrO4)3?
Ans: +6
What is the oxidation number of C in C2O2−4?
Answer: +3
Reduction and Oxidation Reactions 1
Now that we have reviewed about the neutral atom, how ions are formed by the
movement of electrons and how oxidation numbers are assigned, we will now be able to
understand reduction and oxidation better.
As stated in the previous pages, reduction and oxidation reactions are two simultaneous
reactions which involve the transfer of electrons. We can, therefore, identify if a reaction
is a redox reaction by assigning oxidation numbers to the elements involved to keep
track of the electrons transferred
Reduction
1. This is a chemical reaction that involves the gaining of electrons by one of the atoms
involved in the chemical reaction between two substances.
2. When the atom gains electrons, it becomes more negatively charged and its oxidation
number decreases.
3. The substance which gained electrons has been reduced and the substance where the
electron comes from is called the reducing agent.
•

RDG
(Reduction, Decrease in oxidation number, Gain of e-)
Oxidation
1. This is a chemical reaction that involves the losing of electrons by one of the atoms
involved in the chemical reaction between two substances.
2. When the atom loses electrons, it becomes more positively charged and its oxidation
number increases.
3. The substance which lost electrons has been oxidized and the substance where the
electron goes is the oxidizing agent.
OIL
(Oxidation, Increase in oxidation number, Loss of -)
• The oxidized substance, the substance that loses electrons, is the reducing agent.
• The reduced substance, the substance that gains electrons, is the oxidizing agent.
Reduction and Oxidation Reactions 2
• We can see in the figure above that particle AA lost an electron to form A+A+.
• The lost electron was gained by particle BB to form B−B−.
It is important to remember that electrons are negatively charged. So
• When particle AA lost an electron, its oxidation number was increased to form
the A+A+ ion.
• When particle BB gained an electron, its oxidation number was decreased to form
the B−B− ion.
1. Particle AA was oxidized. It is the reducing agent causing particle BB to undergo
reduction.

2. Particle BB was reduced. It is the oxidizing agent causing particle AA to undergo
oxidation.
Balancing Redox Equations
Just like any other chemical reactions, redox reactions are represented using chemical
equation. We will refer to these equations as redox equations.
Whenever we balance a chemical equation, we must obey the law of conservation of
mass: the amount of each element must be the same on both sides of the equation. In
addition to this, the gains and losses of electrons must also be balanced in redox
equations.
In many simple chemical equations, balancing the electrons is balanced automatically in
the sense that we can balance the equation without explicitly considering the transfer of
electrons.
An example of this is the thermite reaction wherein ferric oxide and aluminum metal
reacts to give iron metal and aluminum oxide.
Fe2O3(s)+Al(s)→Fe(s)+Al2O3(s)Fe2O3(s)+Al(s)→Fe(s)+Al2O3(s)
We can identify whether the reaction is a redox reaction or not by assigning oxidation
numbers and looking for changes from the reactants to the products side.
• Oxidation numbers for reactants:
1. O = -2
2. Fe = +3
3. Al = 0
• Oxidation numbers for products:
1. O = -2
2. Fe = 0
3. Al = +3
Therefore,
1. The oxidation number of iron decreased, while the oxidation number of aluminum
increased.
2. The Fe3+Fe3+ in Fe2O3Fe2O3 was reduced (+3 to 0). Consequently, Fe2O3Fe2O3 is
the oxidizing agent.
3. The AlAl was oxidized to Al3+Al3+ ( 0 to +3). Al is the reducing agent.
To balance the equation, we can simply put a coefficient of 22 for
both Al(s)Al(s) and Fe(s)Fe(s)

Fe2O3(s)+2Al(s)→2Fe(s)+Al2O3(s)
Half-equation Method
To see more clearly how electrons are transferred, a redox equation can be thought of
as the result of two simultaneous half-reactions. A half reaction shows the reactants,
products and electrons transferred for either an oxidation or a reduction.
Let us consider the equation:
CuCl2(aq)+Al(s)→Cu(s)+AlCl3(aq)CuCl2(aq)+Al(s)→Cu(s)+AlCl3(aq)
• • Oxidation numbers on reactants side: Cu = +2, Cl = -1, Al = 0
• • Oxidation numbers on products side: Cu = 0, Cl = -1, Al = +3
• • Cu2+
is reduced to Cu and Al is oxidized to Al3+
as
Since CuCl2 and AlCl3 are dissociated into ions in solutions, we can rewrite the equation
into:
Cu2+(aq)+Cl−(aq)+Al(s)→Cu(s)+Al3+(aq)+Cl−(aq)Cu2+(aq)+Cl−(aq)+Al(s)→Cu(s)+
Al3+(aq)+Cl−(aq)
• • The chloride ions, Cl−(aq)Cl−(aq), appear on both sides of the equation (without
changes in their oxidation states). They are called spectator ions and can be omitted
from the equation.
• • Therefore,
Cu2+(aq)+Al(s)→Cu(s)+Al3+(aq)Cu2+(aq)+Al(s)→Cu(s)+Al3+(aq)
• • At first look it might appear that the equation is already balanced but it is actually not
because the total charge on the left is +2 and that on the right is +3, which violates the
principle of the conservation of charge.

To balance the charge on both sides of an oxidation half reaction, appropriate
number of electrons must be added to the ____ side of the reaction.
Answer: Products
Balance the equation representing reaction between solid copper and nitric acid
to yield aqueous copper(II) ions and nitrogen monoxide gas.
Cu(s)+HNO3(aq)→Cu2+(aq)+NO(g)Cu(s)+HNO3(aq)→Cu(aq)2++NO(g)
Which element undergoes oxidation?
Answer: Cu
Balance the equation representing reaction between solid copper and nitric acid to yield
aqueous copper(II) ions and nitrogen monoxide gas.
Cu(s)+HNO3(aq)→Cu2+(aq)+NO(g)Cu(s)+HNO3(aq)→Cu(aq)2++NO(g)How
many H+H+ ions are in the balanced equation?
Answer: 6

8H+(aq)+MnO−4(aq)+5Fe2+(aq)→5Fe3+(aq)+Mn2+(aq)+4H2O(l)8H+(aq)+
MnO4−(aq)+5Fe2+(aq)→5Fe3+(aq)+Mn2+(aq)+4H2O(l)
On the reactants side, give the oxidation number of H.
Answer: +1
Given the reaction
On the reactants side, give the oxidation number of Mn.
Answer: +7
Given the reaction
On the reactants side, give the oxidation number of O.
Answer: -2
On the reactants side, give the oxidation number of Fe.
Answer: +2
On the products side, give the oxidation number of Fe.
Answer: +3
On the products side, give the oxidation number of Mn.
Answer: +2

On the products side, give the oxidation number of H.
Answer: +1
On the products side, give the oxidation number of O.
Answer: -2
The reducing agent is
Answer: Fe2+(aq)
The oxidizing agent is
Answer: MnO−4(aq)
In balancing the equation for the reaction taking place in an acidic solution
FeS(s)+NO−3(aq)→NO(g)+SO2−4(aq)+Fe3+(aq)FeS(s)+NO3−(aq)→NO(g)+SO
42−(aq)+Fe3+(aq)
the coefficients of the H+(aq)H+(aq) and H2O(l)H2O(l) are x,y , respectively.
(separate the two coefficients with a comma without space. example: 3,4)
Answer: 4,2

In balancing the equation for the reaction taking place in a basic solution
H2O2(aq)+Fe2+(aq)→Fe3+(aq)H2O2(aq)+Fe2+(aq)→Fe3+(aq)
the coefficients of the OH−(aq)OH−(aq) and H2O(l)H2O(l) are x,y , respectively.
Answer: 2,0

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