BALANCE OF REDOX EQUATION

 

Week: ONE                        Date:     6-10/05/2019                                                     Time:

Period:                                 Duration: 1 HR 20 MIN.                                                 Average age of learners: 16YEARS

Subject:                               CHEMISTRY                                                                        Class: SS TWO

Topic:                                    OXIDATION-REDUCTION [REDOX] REACTION

Sub topic:  Balancing of Redox equation

Reference materials:

(1) ESSENTIAL CHEMISTRY, TONALD PUBLISHERS, I. O ODESINA

(2) NEW SCHOOL CHEMISTRY, AFRICAN FIRST PUBLISHERS, OSEI YAW ABABIO

(3) INTERNET

Instructional materials:

Entry behavior: The students have been taught chemical reaction

Behavioural objective: At the end of the lesson the students should be able to:

i.                     Recognize oxidizing and reducing agent in a redox reaction

ii.                   State the rules that guides balancing redox equation

iii.                  Balanced redox equation

CONTENT

OXIDIZING AND REDUCING AGENTS

An oxidizing agent is a substance that is reduced in a chemical reaction thereby bringing about oxidation while a reducing agent is oxidized thereby bringing about reduction e.g

CuO + CO              Cu  + CO2

In this equation:

The CuO losses oxygen to CO to become metallic copper and therefore reduced. The CO gain oxygen to become CO2, and therefore it is oxidized.

The CuO is reduced by the CO and thus CO is the reducing agent. The CO is oxidized by the CuOand thus CuO is the oxidizing agent.

Common Oxidising and Reducing Agents

Oxidising agent

Effective Change

Decrease in Oxidation Number

KMnO4 in acid solution

MnO4 -  → Mn2+

5

KMnO4 in alkaline solution

MnO4 - →  MnO2

3

K2Cr2O7 in acid solution

Cr2O72- →  Cr3+

3

dilute HNO3

NO3- →  NO

3

concentrated HNO3

NO3- →  NO2

1

concentrated H2SO4

SO42- →   SO2

2

manganese (IV) oxide

MnO2 → Mn2+

2

Chlorine

Cl →  Cl-

1

chloric (I) acid

ClO-  →  Cl-

2

KlO3 in dilute acid

IO3-  →  I

5

KlO3 in concentrated acid

IO3-   →  I-

4

Reducing agent

Effective Change

Increase in Oxidation Number

iron (II) salts (acid)

Fe2+ →  Fe3+

1

tin (II) salts (acid)`

Sn2+ →  Sn4+

2

ethanedioates (acid)

C2O42- →  CO2

1

sulphites (acid)

SO32-  →SO42-

2

hydrogen sulphide

S2- → S

2

iodides (dilute acid)

I- → I

1

iodides (concentrated acid)

I- → I+

2

metals, e.g. Zn

Zn → Zn2+

2

Hydrogen

 

 

Some other oxidizing and reducing agent

Oxidizing agent

Reducing agent

Oxygen

Hydrogen peroxide (H2O2)

Sulphur

 

Carbon

Sulphur(iv)oxide SO2

Ammonia NH3

 

Note: oxidizing agents give oxygen to another substance or remove hydrogen from it and gain electrons while a reducing agents remove oxygen from another substance or give hydrogen and loss electrons.

BALANCING REDOX EQUATIONS

Half-cell reaction method :- The equation is separated into two half – equations. One for Oxidation and  one for Reduction. Each equation is balanced by adjusting coefficients and adding H2O, H+ and e- in this other:

Balance elements in the equation other than O and H.

Balance the oxygen atoms by adding the appropriate number of water molecules to the opposite side of the equation.

Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation.

Add up the charges on each side. Make them equal by adding enough electrons to the more positive side ( Rule of thumb: e- and H+ are almost always on the same side)

The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.

The half equations are added together, cancelling out the electrons to form one balance equation.

Example

Write a balance ionic equation for the redox reaction below: Cu+aq  + Fes        Fe3+aq  + Cus

Solution

   Step I

Cu+aq is the oxidizing agent. It is reduced to metallic Cus

Cu+aq            Cus

 Fes is the reducing agent. It is oxidized to Fe3+aq ion

 Fes          Fe3+aq

Step  II

Reducing half-equation

Cu+aq  + e-           Cus

Oxidation  half-equation

Fes          Fe3+aq + 3e-

Balance the number of charges

Cu+aq  + e-           Cus           X3

Fes          Fe3+aq + 3e-           X1

3Cu+aq  +3 e-           3Cus

Fes          Fe3+aq + 3e-

3Cu+aq  + Fes          Fe3+aq  + 3Cus

 

 

 

 

 

Write a balance ionic equation for the redox reaction below: Cr2O72-aq  + HNO2         Cr3+aq  + NO3-aq

Solution

   Step I

Reduction half-equation

Cr2O72-aq           Cr3+aq

Cr2O72-aq + 14H+       2Cr3+aq  +  7H2O

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

Oxidation half-equation

HNO2          NO3-aq

HNO2  + H2O           3H+ + NO3-aq  + 3e-

Balance the number of charges

Cr2O72-aq + 14H+ + 6e-           2Cr3+aq  +  7H2O     X1

HNO2  + H2O           3H+ + NO3-aq  + 2 e-              X3

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

3HNO2  +3 H2O            9H+ +3 NO3-aq  + 6e-             

Cr2O72-aq  + HNO2  +5e-              2Cr3+aq  + 3NO3-aq  +  4H2O

PRESENTATION

Step I : The teacher defines oxidizing and reducing agents with relevant examples

Step II: The teacher states the rule guiding balancing of redox equation

Step III: The teacher leads the students in balancing redox equation

Step IV: The students are allow to ask questions

EVALUATION

The teacher evaluates the lessons by asking the following questions

1.       Give three example each of oxidizing  and reducing agents

2.       State the property exhibited by NO2 in each of the following equation:

i.                     4Cu + 2NO2          4CuO   +    N2    [O.A]

ii.                   H2O + 2NO2       v HNO3   +   HNO2   [O.A & R.A]

Assignment

MnO-4  + H2O2               Mn2+    +   O2

 

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