GRADE 11 PHYSICAL SCIENCES

REDOX

REACTIONS

GRADE 11
WHAT LEARNERS MUST KNOW:

Explain the meaning of oxidation number

• Use rules of oxidation to assign oxidation numbers to atoms in a variety of
molecules and ions
E.g. H2O, CH4, CO2, H2O2, HOCℓ - use oxidation number guidelines or rules

• Determine the oxidation number from a chemical formula and electronegativities.

• Identify a reduction - oxidation reaction
• Apply the correct terminology
• Describe oxidation reduction reactions:
o Involving electron transfer
o Involving changes in oxidation number
• Balance redox reaction equations by using oxidation numbers via the ion-electron
method.

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CLASSIFICATION BASED ON MICROSCOPIC LEVEL CHANGES

A chemical reaction involving the transfer of one or more protons (hydrogen ions) between
molecules has already been discussed and is called a proton transfer (acid-base) reaction.

E.g. HCl(g ) + H 2 O(l) ® H 3 O + (aq) + Cl − (aq)

ELECTRON TRANSFER REACTIONS (REDOX REACTIONS)

• A chemical reaction involving the transfer of one or more electrons between chemical
species is called an electron transfer reaction.

E.g. Cu(s) + Zn 2+ (aq) ® Zn(s) + Cu 2+ (aq)

• If the oxidation numbers of atoms change during a chemical reaction, then the reaction is
classified as a redox reaction.

What is an oxidation number?

The oxidation number of an atom in a molecule or ion is the charge that the atom can have or
has, should all compounds be regarded as ionic.

Are oxidation numbers ‘real’?

• Apart from monatomic ions, e.g. Cl–, or ionic compounds, e.g. NaCl, the oxidation numbers
are not the actual charge on the atoms in a molecule or ion.
• Oxidation numbers assume that the atoms in a molecule are positive or negative ions. This
is not true.
• In H2O, the charge on the H-atom is not +1, but the oxidation number is.
• The H atom has a partial positive charge, but is not an ion.
• The oxidation number of the O atom in water is –2, but the O atom is not an O O2– ion.
What is the use of oxidation numbers?
• Oxidation numbers provide information on the movement of electrons during a chemical
reaction.
• An increase in the oxidation number, or charge of an atom, is an indication that the atoms
lost electrons (negative charges).
• In the same way, a decrease in the oxidation number will indicate that an atom gained
electrons.
• In this unit you will use oxidation numbers to classify reactions as electron transfer
reactions.

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Guidelines for awarding oxidation numbers

• Each atom in a pure element has an oxidation number of zero.

Example: The oxidation number of Fe in iron metal is 0. The oxidation number of I in
I2 and S in S8 is also 0.
• In monatomic ions, the oxidation number equals the charge on the ion.
Example: The oxidation number of the H+ ion is +1 and that of the Cl– ion is –1.
• Elements of Group I, II and III form positive ions of which the charges equal the
group number.
Example: Mg forms Mg2+ ions and consequently the oxidation number is +2.
• Fluorine always has an oxidation number of –1 in compounds.
• The oxidation number of H is always +1, except in the metal hydrides.
Example: In H2O the oxidation number of H equals +1. In NaH, the oxidation number
of Na equals + 1 and that of H equals –1.
• The oxidation number of O is always –2, except peroxides which have an oxidation
number of –1.
Example: The oxidation number of O in H2O is –2. In H2O2, the oxidation number of
O equals –1.
• The oxidation numbers of the halogens (Cl, Br and I) are always –1, except when
these atoms are bonded to atoms with a higher electronegativity (O or F).
Example: In HCl, the oxidation number of Cl equals –1. In ClO– the oxidation number
of Cl equals +1.
• In neutral compounds, the sum of the oxidation numbers of the constituent atoms
equals 0.
Example: The sum of the oxidation numbers of Na and Cl in NaCl is:
+1 – 1 = 0.
• In polyatomic ions, the sum of the oxidation numbers of the constituent atoms
equals the charge on the ion.
Example: The sum of the oxidation numbers of N and H in NH +4 is:
N + 4(+1) = +1. Consequently, N has an oxidation number of –3.

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Examples

The following examples show how these guidelines can be applied to determine the oxidation
numbers of atoms.

Example 1: Determine the oxidation number of N in NO.

Answer: NO(g) has O = -2, so the oxidation number of N = +2.

Example 2: Determine the oxidation number of N in NO2.

Answer: NO2(g) has O = -2, so (-2)2 + N = 0 ∴N = +4.

Therefore, the oxidation number of N equals +4.

Example 3: Determine the oxidation number of S in SO42-

Answer: SO42- has O = -2. Thus S + 4(-2) = -2.

Solving the equation gives S = -2 + 8 = +6.

Therefore, the oxidation number of S equals +6.

Example 4: Determine the oxidation number of Cr in K2Cr2O7

Answer: K2Cr2O7 has K = +1 and O = -2.

Thus 2(+1) + 2 Cr + 7(-2) = 0; 2 Cr = 12; ∴ Cr = +6.

2

Therefore, the oxidation number of Cr equals +6.

Example 5: Determine the oxidation number of Cr in the dichromate ion ( Cr2 O 7 -).

Answer: Cr2 O 72 - has O = -2

2 Cr + 7(–2) = –2 ∴2Cr –14 = –2 ∴Cr = +6

Therefore, the oxidation number of Cr equals +6.

Example 6: Determine the oxidation number of C in Na2CO3.

Answer: Na2CO3 has Na = +1 and O = -2

2 Na + C + 3O) = 0 ∴2(+1) + C + 3(–2) = 0 ∴C = +4

Therefore, the oxidation number of C equals +4.

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Electrochemical (Redox) Reactions:

Electrochemistry is the branch of chemistry that deals with the transformation between
electrical energy and chemical energy.

Electrochemical reactions are redox reactions. All redox reactions are characterised by transfer
of an electrons. The word ‘redox’ refers to reduction and oxidation.

Reduction and oxidation

In terms of electrons, reduction and oxidation can be defined as follows:

• Reduction is when electrons are gained.

• Oxidation is when electrons are lost.

A reduction always goes together with an oxidation and vice versa.

In terms of oxidation numbers, reduction and oxidation can be defined as follows:

• Reduction is a decrease in oxidation number.
• Oxidation is an increase in oxidation number.

Oxidation states and redox reactions

• The oxidation state of an atom signifies the number and sign of the charge the atom
would have in a molecule if bonding electrons were transferred completely.
o If the oxidation number of an element increases during a reaction, then that
element is oxidised during the reaction.
o If the oxidation number decreases, the element is reduced.
• During oxidation-reduction (redox) reactions:
o Electrons are transferred
o There are two simultaneous half-reactions
o The two half-reactions produce the total or net redox reaction
o Electrons are lost during the oxidation half-reaction
o Electrons are gained during the reduction half reaction
o The reducing agent is oxidized
o The oxidizing agent is reduced.

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Reducing agents and oxidising agents

• A substance that loses electrons, i.e. undergoes oxidation, is a reducing agent. Such a
substance causes the reduction of another substance.

• A substance that gains electrons, i.e. undergoes reduction, is an oxidising agent. Such a
substance causes the oxidation of another substance.

Balancing of redox reactions

When we balance a chemical reaction equation, the primary concern is to obey the
principle of conservation of mass - the total mass of the reactants must be equal the total
mass of the products. This is done by conserving and accounting for atoms.

In redox reactions we must obey a second principle as well, the conservation of charge. The
total number of electrons lost must equal the total number of electrons gained. In other words
the electrons cannot be left lying around.

There are TWO very important methods for balancing oxidation-reduction reactions:

1. Oxidation number method

2. Ion-electron method

However, redox reactions involve the transfer of electrons from one species to another, so
electrical charge must be considered explicitly when balancing redox equations. The key to
balancing complicated redox equations is to balance electron transfers.

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Oxidation Number Method

During a redox reaction, the total increase in oxidation number must be equal to total
decrease in oxidation number. This is the basic principle for balancing chemical equations.

In addition, the number of atoms of each kind on one side of the equation must be equal to
the number of atoms of the corresponding elements on the other side (the law of conservation
of mass should not be violated). The following steps should be followed.

Steps for balancing redox equations

1. Write the skeleton redox reaction.

2. Indicate the oxidation number of atoms in each compound above the symbol of the
element.
3. Identify the element or elements, which undergo a change in oxidation number, one
whose oxidation number increases (reducing agent) and the other whose oxidation
number decreases (oxidizing agent).
4. Calculate the increase or decrease in oxidation numbers per atom. Multiply this
number of increase/decrease of oxidation number, with the number of atoms, which
are undergoing change.
5. Equate the increase in oxidation number with decrease in oxidation number on the
reactant side by multiplying the formula of the oxidizing and reducing agents.
6. Balance the equation with respect to all other atoms except hydrogen and oxygen.
7. Finally, balance hydrogen and oxygen.
8. For reactions taking place in acidic solutions, add H+ ions to the side deficient in
hydrogen atoms.
9. For reactions taking place in basic solutions, add H2O molecules to the side deficient
in hydrogen atoms and simultaneously add equal number to OH- ions on the other side
of the equation.

Let us discuss the above method stepwise with the help of reaction between zinc and
hydrochloric acid.

Step 1

The skeleton equation is:

Zn + HCl → ZnCl2 + H2

Step 2

Oxidation number of various atoms involved in the reaction

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Step 3

The oxidation number of zinc has increased from 0 to +2 while that of hydrogen has
decreased from +1 to 0. However, the oxidation number of chlorine remains same on both
sides of the equation. Therefore, zinc is reducing agent while HCl is oxidizing agent in
reaction and the changes are shown as:

Step 4

The increase and decrease in oxidation number per atom can be indicated as: O.N. increases
by 2 per atom

Step 5

The increase in oxidation number of 2 per atom can be balanced with decrease in oxidation
number of 1 per atom if Zn atoms are multiplied by 1 and HCl by 2. The equation will be:

Zn + 2HCl → ZnCl2 + H2

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Ion Electron Method

The balancing of a chemical equation by ion-electron method (using half reactions) is done
according to the following steps.

• Find the elements whose oxidation numbers are changed. Choose the substance,
which acts as an oxidizing agent and one that acts as a reducing agent.

• Separate the complete equation into two half reactions, one for the change undergone
by the oxidizing agent and the other for the change undergone by the reducing agent.

• Balance half equations by the following steps:

1. Balance all atoms other than H and O.

2. Calculate the oxidation number on both sides of the equation and add electrons to
whichever side is necessary, to make up the difference.
3. Balance the half equation so that both sides get the same charge.
4. Add water molecules to complete the balancing of the equation.

• Add two balanced half equations. Multiply one or both half equations by suitable
numbers so that on adding the two equations, the electrons are balanced.

Redox reactions take place in all the three media acidic or basic or neutral.

v If H+ ions appear on either side of the equation, the reaction takes place in acidic medium.
v If OH- ions appear on either side of the equation, the solution is basic.
v If neither H+ nor OH- ions are present, the reaction occurs in neutral solution.

For balancing redox reactions involving acidic and basic media, the method has to be
modified slightly. The steps are summarized by the following flowchart.

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Balancing Redox Reactions Examples

Question: Metallurgy involves reducing the metal ions in ores to the elemental form. The
production of manganese from the ore pyrolusite, which contains MnO2, uses aluminum as
the reducing agent. Using oxidation states, balance the equation for this process.

MnO2(s) + Al(s) → Mn(s) + Al2O3(s)

Solution:

First assign oxidation states.

MnO2(s) + Al(s) → Mn(s) + Al2O3(s)

Left side:

Mn = +4
O = -2 (each O)
Al = 0

Right side:

Mn = 0
Al = +3 (each Al)
O = -2 (each O)

Each Mn atom undergoes a decrease in oxidation state of 4 (from +4 to O), whereas each Al
atom undergoes an increase of 3 (from 0 to +3).

Thus we need three Mn atoms for every four Al atoms to balance the increase and decrease in
oxidation states.

Increase = 4(3) = decrease = 3(4)

3MnO2(s) + 4Al(s) →Products

We balance the rest of the equation by inspection

3MnO2(s) + 4Al(s) → 3Mn(s) + 2Al2O3(s)

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Half Reactions

Redox reaction is nothing but both oxidation and reduction reactions taking place
simultaneously. During a redox reaction, when an oxidation or reduction reaction happens, it
is called as half reaction for the redox reaction. Change in oxidation states of individual
substances during a redox reaction denotes half reaction.

Balancing by means of half reactions

The balancing of redox reactions can be complicated, and balancing by means of inspection is
sometimes difficult. Half reactions provide an easy method for balancing redox reactions. For
this method an unbalanced redox reaction has to be broken up into two half reactions.
The following steps are handy during balancing with half reactions:
• Allocate oxidation numbers to all elements in the redox reaction to determine which
substance is oxidised and which substance is reduced.
• Now break the redox reaction up into an oxidation and a reduction half reaction.
• Multiply the half reactions by whole numbers so that the number of electrons that are
lost during oxidation is equal to the number of electrons that are gained during
reduction.
• Add the two half reactions together and cancel out electrons on both sides of the
equation. Also cancel out species that appear on both sides of the equation.
• Check if the number of atoms of each type on the left-hand side is equal to the number
of atoms of each type on the right-hand side.
• Check if the sum of the charges on the left-hand side is equal to the sum of the charges
on the right-hand side.

NB: More of this will be discussed in GRADE 12.

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Example of Balancing Half reaction

In this method the two half reactions are balanced separately and then added together to get
the balanced equation.

Let us take an example – Oxidation of Fe2+ ions to Fe3+ ions by Cr2O72- ion in acidic medium.

The following steps are involved in balancing a redox reaction.

Step 1:

Write the unbalanced equation in ionic form

Fe2+(aq) + Cr2O72-(aq) → Fe3+(aq) + Cr3+(aq)

Step 2:

Separate the equation into two half reactions, i.e., oxidation half reaction and reduction half
reaction.

Oxidation Half reaction: Fe2+(aq) → Fe3+(aq)

Reduction half reaction: Cr2O72- (aq) → Cr3+(aq)

Step 3:

Balance the two half reactions separately. First we balance the atoms other than O and H
atoms.
Oxidation half reaction Fe2+ (aq) → Fe3+(aq) is already balanced with respect to Fe atoms .

Reduction half reaction Cr2O72-(aq) → 2Cr3+(aq) we have balanced Cr atoms by multiplying

Cr3+ by 2.

Step 4:

Balance O atoms by adding H2O and balance H atoms by adding H+ ions if the reaction
occurs in acidic medium.

Oxidation half reaction is already balanced with respect to O and H

Reduction half reaction

Cr2O72-(aq) + 14H+(aq) → 2Cr3+(aq) + 7H2O

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Step 5:

Balance the charge of both the half reactions by adding electrons to one side of the half
reaction. To make the number of electrons in both the half reactions equal, we need to
multiply one or both the reactions by suitable coefficients.

The charge of oxidation half reaction can be balanced by adding a single electron on the right
hand side and can be written as

Fe2+(aq) → Fe3+(aq) + 1e-

To balance the reduction half reaction we will add six electrons on the left hand side and the
reaction can be written as

Cr2O72-(aq) + 14H+(aq) +6e- → 2Cr3+(aq) + 7H2O

To equalize the number of electrons in both the half reactions we have to multiply the
oxidation half reaction by 6 and can be written as

6Fe2+(aq) → 6Fe3+(aq) + 6e-

Step 6:

We add the two half reactions and cancel the electrons from both sides.

6Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O

This equation is fully balanced in terms of number of atoms of each kind and also in terms of
charge.

For the reaction in basic Medium

First balance the atoms as is done for acidic medium, and then for each H+ ion , add an equal
number of OH- ions to both sides of the equation. Where H+ and OH- appear on the same side
of the equation, combine these to form H2O.

Example : Oxidation of I- ion by MnO4- ion in basic medium.

Step 1:

Write the skeletal ionic equation

MnO4-(aq) + I-(aq) → MnO2(s) + I2(s)

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Step 2:

Divide the reaction into two half reactions

Oxidation half Reaction: I-(aq) → I2(s)

Reduction half reaction: MnO4-(aq) → MnO2(s)

Step 3:

Balance the Oxidation half reaction 2I-(aq) → I2(s)

Step 4:

To balance the reduction half reaction we will add two H2O molecules on the right hand side
to balance the O atoms

MnO4-(aq) → MnO2(s) + 2H2O

To balance the H atoms, we will add four H+ on the left side and since the reaction is taking
place in a basic medium, we will add four OH- on both sides .

MnO4-(aq) + 4H+(aq) + 4OH-(aq) → MnO2(s) + 2H2O + 4OH-(aq)

Replacing 4H+ and 4OH- by 4H2O, the resultant equation is

MnO4-(aq) + 2H2O → MnO2(s) + 4OH-(aq)

Step 5 :

We balance the charge of both the reactions by adding the required number of electrons
Oxidation half reaction 2I-(aq) → I2(s) + 2e-

Reduction half reaction

MnO4-(aq) + 2H2O + 3e- → MnO2(s) + 4OH-(aq)

Step 6 :

Now to equalize the number of electrons, we multiply the oxidation half reaction by 3 and the
reduction half reaction by 2 and then, add both the reactions to get the final balanced
equation.

6I-(aq) + 2MnO4-(aq) + 4H2O → 3I2(s) + 2MnO2(s) + 8OH-(aq)

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WORKSHEET 1

1. Explain the meaning of the term ‘oxidation number’.

2. Are oxidation numbers ‘real’?

3. Determine the oxidation numbers for the underlined atoms in the following
compounds or ions:
3.1 H3PO4
3.2 Fe2O3
3.3 H2SO4
3.4 Al3O3
3.5 H2SO3
3.6 HNO3
3.7 MnO4-
3.8 NO2+
4. Study the following chemical equation:

H2 S + FeCl3 → FeCl2 + S + HCl

4.1 Balance the equation.

4.2 Determine the oxidation numbers of the underlined atoms.
4.3 Which compound lost electrons during the reaction? Supply a reason for your answer.
4.4 Which compound gained electrons during the reaction? Supply a reason for
your answer.
4.5 What are oxidation numbers used for?

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WORKSHEET 2

1. State the Oxidation Number of each of the underlined elements.

a) NH3 _____ b) H2SO4 _____

c) ZnSO3 _____ d) Al(OH)3 _____

e) Na _____ f) Cl2 _____

g) AgNO3 _____ h) ClO4- _____

i) SO2 _____ j) K2Cr2O4 _____

k) Ca(ClO3)2 _____ l) K2Cr2O7 _____

m) HPO32- _____ n) HClO _____

o) MnO2 _____ p) KClO3 _____

q) PbO2 _____ r) PbSO4 _____

s) K2SO4 _____ t) NH4+ _____

u) Na2O2 _____ v) FeO _____

w) Fe2O3 _____ x) SiO44- _____

y) NaIO3 _____ z) ClO3- _____

. For each of the following reactants, identify: the oxidizing agent, the reducing agent, the
substance oxidized and the substance reduced.
a) Cu2+ (aq) + Zn (s) → Cu(s) + Zn2+ (aq)

Substance oxidized _____ Substance reduced _____

Oxidizing agent _____ Reducing agent _____

b) Cl2 (g) + 2 Na (s) → 2 Na+ (aq) + 2 Cl- (aq)

Substance oxidized _____ Substance reduced ____

Oxidizing agent _____ Reducing agent _____

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WORKSHEET 3

1. Explain each of the following concepts in your own words:

1.1 Oxidation
1.2 Reducing agent 1.3
Spectator ions 1.4
Half reaction 1.5
Redox reaction
2. For each reaction below, identify the substance oxidized, the substance reduced, the
oxidizing agent, the reducing agent, the oxidation half reaction, the reduction half
reaction, and then complete and balance the equation by the method of oxidation-
reduction showing all electrons transfers.
2.1 Al(s) + H+(aq) → Al3+(aq) + H2(g)
2.2 2
Cr2 O 7 − (aq) + Cl-(aq) → Cr3+(aq) + Cl2(g)

2.3 −
Cu(s) + NO 3 (aq) → Cu2+(aq) + NO2(g)

2.4 Cr2 O 7 − (aq)

2
+ H2S(g) → Cr3+(aq) + S(s)

2.5 MnO 4 + −SO2(g) → Mn2+(aq) + SO 4 − (aq)

2

2.6 −
Cu(s) + NO 3 (aq) → Cu2+(aq) + NO(g)

2.7 Cr2 O 7 − (aq)

2
+ SO2(g) → Cr3+(aq) + SO 2− (aq)
4

2.8 Pb + Fe3+ → Fe2+ + Pb2

2.9 Sn2+ + Br2 → Sn4+ + 2Br-
2.10 Cl2 + F- → F2 + 2Cl-

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 Half reactions
Substance Substance Oxidizing Reducing
Q2.
Oxidized Reduced Agent Agent

Oxidation Reduction

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

2.9

2.10

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