4.1 CHEMICAL EQUATION

FTF :13 HOUR NFTF : 13 HOURS

SUMMARY

4.1 Chemical 4.2 Stoichiometry
equation



4.1 Chemical Equation

Learning Outcomes

At the end of the lesson, students should be able to:
(a)Determine the oxidation number of an element in

a chemical formula

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Redox Reaction

Redox reaction is a reaction that involves
both losing and gaining of electron which

occur simultaneously.

Oxidation (Zn lost 2 e-)
Reduction (Cu gain 2 e-)

https://byjus.com/jee/redox-reactions/

+1 +1 No lost or gain of
+1 +1 electron!!

What are oxidation numbers?

 Chemists use oxidation numbers (or oxidation
states) to keep track of how many electrons an
atom has.

 Oxidation numbers don’t always correspond to
real charges on molecules, and we can calculate
oxidation numbers for atoms that are involved in
covalent (as well as ionic) bonding.

Determining the oxidation number
(rules…)

For :

 1.Free element

 2.Neutral molecule

 3.Monoatomic ion

 4.Polyatomic ions

 5.Halogens

 6.Hydrogen Oxidation numbers of any atoms can
 7.Oxygen be determined by applying the

following rules :

1.In a free element, as an atom or a
molecule the oxidation number is zero

https://www.avensonline.org/blog/importance-of-
chemistry.html

2. In neutral molecule, the SUM of the
oxidation number of all atoms that made up the
molecule is equal to zero.

3. For monoatomic ion, the oxidation
number is equal to the charge on the ion.

4. For polyatomic ions, the total oxidation
number of all atoms that made up the
polyatomic ion must be equal to the nett
charge of the ion.

Example

Assign the oxidation number of Cr in
Cr2O72-.

Solution

Cr2O72- = - 2
2Cr + 7(-2) = - 2
2 Cr = + 12
Cr = + 6

5. Halogens (F, Cl, Br, I) always have oxidation
number of -1 in its compound. Only have a positive
number when combine with oxygen.

6. Hydrogen has an oxidation number of +1 in its
compound EXCEPT in metal hydrides which
hydrogen has an oxidation number of -1.

7. Oxygen has an oxidation number of -2 in most
of its compound.

Oxygen has two exceptional cases…

Oxygen - Two Exceptional Cases

Exercise

1. Assign the oxidation number of Mn in the following chemical compounds.
i. MnO2
ii. MnO4-

2. Assign the oxidation number of Cl in the following chemical compounds.
 i. KClO3

ii. Cl2O72-
3. Assign the oxidation number of following:
 i. Cr in Cl2O72-

ii. U in UO22+
iii. C in C2O42-

Answer

1. i. Mn = +4
ii. Mn = +7

2. i. Cl = +5
ii. Cl = +6

3. i. Cr = +6
ii. U = +6
iii. C = +3

4.1 Chemical Equation

Learning Outcomes

At the end of the lesson, students should be able to:
(b) Write a balance : chemical equation.

i. Chemical equation by inspection method
ii. Redox equation ion-electron method

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Chemical Equation

CH4 (g) + 2O2 (g) ⭢ CO2 (g) + 2H2O (l)

CH4 & O2 : reactants
CO2 & H2O : products
1, 2, 1, 2 : stoichiometric coefficients

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Balanced Chemical Equation

CH4 (g) + 2O2 (g) ⭢ CO2 (g) + 2H2O (l)

A chemical equation must be balanced. Why?

A balance equation follows mass conservation law i.e.

TOTAL MASS OF REACTANTS = TOTAL MASS OF

PRODUCTS 25

Balancing Chemical Equation

Two methods

Inspection Method Ion-electron Method
(for chemical equation) (for redox equation)

Balance all the elements. Acidic Medium
Basic Medium

Balance all the elements
AND

charges.

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Steps in Inspection

1. Balance the metallic
element, followed by
non-metallic atoms.

2. Then balance the
hydrogen and oxygen
atom

3. Check

Example

Balance the chemical equation by applying the
inspection method.

Balance the chemical equation below by
applying inspection method.
a. Fe(OH)3 + H2SO4 → Fe2(SO4)3 +H2O

b. C6H6 + O2 → CO2 + H2O

c. Y2O + Y2S → Y + SO2

Answer

a. 2Fe(OH)3 + 3H2SO4 → Fe2(SO4)3 + 6H2O
b. C6H6 + 15/2O2 → 6CO2 + 3H2O
c. 2Y2O + Y2S → 6Y + SO2

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Steps in Ion-electron Method

In Acidic Medium Basic Medium
general

Step 1 Write two half-reactions (oxidation & reduction)
Step 2 Balance both of the half-reactions EXCEPT for O and H.

Balance the oxygen atom by adding H2O and hydrogen by adding H+.
Balance the charge by adding electrons to the side with the greater overall
positive charge.

Step 3 Multiply each half-reaction by an interger (so that number of electron lost in
one half-reaction = the number of electrons gained in the other half
Step 4 reaction).
Step 5
Step 6 Add the two half-reactions and simplify where possible
(by canceling identical species).

Check (to make sure that there are balance in terms of elements/atoms and
also charges).

5 steps only. Add OH- to both sides of the
No step number 6. equation (so that it can be combined
with H+ to form H2O).

The number of hydroxide ions (OH-) 31
added is equal to the number of
hydrogen ions (H+) in the equation.

Acidic Medium – Example 1 Balance the equation.

Fe2+ + MnO4- → Fe3+ + Mn2+

Step 1 :

i. Fe2+ → Fe3+
ii. MnO4- → Mn2+

Step 2 :

i. Fe2+ → Fe3+ + 1e
ii. MnO4- + 8H+ → Mn2+ + 4H2O

MnO4- + 8H+ + 5e → Mn2+ + 4H2O 32

Step 3 :

i. 5Fe2+ → 5Fe3+ + 5e
ii. MnO4- + 8H+ + 5e → Mn2+ + 4H2O

Step 4 :

5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O

Step 5 :

5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O

Total charge reactant Total charge product

= 5 (+2) + (-1) + 8 (+1) = 5 (+3) + (+2) + 4 (0)

= + 10 - 1 + 8 = + 15 + (+2) 33
= + 17 = + 17

Acidic Medium – Example 2 Balance the equation.

C2O42- + MnO4- + H+ → CO2 + Mn2+ + H2O

Step 1:

i. C2O42- → CO2

ii.
Step 2 :

i.
ii.

Step 3 :

i. C2O42- → 2 CO2 + ____
ii. MnO4- + __H+ + ___ → Mn2+ + ___H2O

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Step 4 :

i.___ x (C2O42- → 2CO2 + 2e)

ii. ___ x (MnO4- + 8H+ + 5e → Mn2+ + 4H2O)
2MnO4- + 16H+ + 10e → 2Mn2+ + 8H2O

Step 5 :

i. 5C2O42- → 10CO2 + _____
ii. 2MnO4- + 16H+ + _____ → 2Mn2+ + 8H2O
______________________________________

5C2O42- +2MnO4- + 16H+→10CO2+ 2Mn2++ 8H2O

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Basic Medium – Example 1 Balance the equation.

Cr(OH)3 + IO3- + OH- → CrO32- + I- + H2O

Step 1: i. Cr(OH)3 → CrO32-
ii. IO3- → I-

Step 2 : i. Cr(OH)3 → CrO32- + 3H+
ii. IO3- + 6H+ → I- + 3H2O

Step 3 : i. Cr(OH)3 → CrO32- + 3H+ + 1e-
ii. IO3- + 6H+ + 6e- → I- + 3H2O

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MATTER 36

Step 4 : i. 6 (Cr(OH)3 → CrO32- + 3H+ + 1e)
6Cr(OH)3 → 6CrO32- + 18H+ + 6e

ii. IO3- + 6H+ + 6e → I- + 3H2O

Step 5 : i. 6Cr(OH)3 → 6CrO32- + 18H+ + 6e
ii. IO3- + 6H+ + 6e → I- + 3H2O

6Cr(OH)3 + IO3- → 6CrO32- + I- + 12H+ + 3H2O

Step 6 :6Cr(OH)3+IO3- +_____→6CrO32-+ I- +12H+ +3H2O+ ____

6Cr(OH)3 + IO3- + 12OH- → 6CrO32- + I- + 15H2O

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Exercise 1

Balance the following redox equations:

a. In Acidic Solution

i. Cu + NO3- + H+→ Cu2+ + NO2 + H2O
ii. MnO4- + H2SO3 → Mn2+ + SO42- + H2O + H+
iii. Zn + SO42- + H+ → Zn2+ + SO2 + H2O

b. In Basic Solution HINT
NO2 → NO3-
i. ClO- + S2O32- → Cl- + SO42-
ii. Cl2 → ClO3- + Cl- NO2 → NO
iii. NO2 → NO3- + NO
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