Oxidizing And Reducing Agents

CONTENT

• Identification of Oxidizing and Reducing agents
• Balancing of Redox equation in Acidic and Alkaline medium

OXIDIZING AND REDUCING AGENTS

An oxidizing agent is defined as a substance which loses oxygen or electronegative element to another substance. Or an oxidizing agent is a substance which gains hydrogen from a substance. Or an oxidizing agent is a substance which gains electron from a substance. Consider the reaction below

            C_(s)        +         ZnO_(s) → CO_2(g) +       Zn_(s)

ZnO is the oxidizing agent because it loses oxygen to C.

A reducing agent is defined as substance, which removes and accepts oxygen from other substances. Or a reducing agent is defined as a substance, which removes and accepts electronegative element from another substance. Or a reducing agent is defined as a substance which loses and donates electron to another substance. From the reaction above, C is the reducing agent because it removes and accepts oxygen from ZnO.

In an oxidation and reduction reaction, the oxidizing agent is the reduced species while the reducing agent is the oxidized species.

NOTE: An oxidizing agent accepts electron, is reduced and its oxidation number decreases while a reducing agent donate electron, is oxidized and its oxidation number increases.

IDENTIFICATION OF OXIDIZING AND REDUCING AGENTS

TESTS FOR OXIDIZING AGENTS: The presence of an oxidizing agent can be detected using any of the following reagents.

1. Acidified potassium iodide, KI with starch
2. Sulphur (IV) oxides, SO₂ with acidified Barium trioxonitrate (V) solution
3. Iron (II) Chloride solution (FeCl₂)
4. Hydrogen sulphide gas (H₂S)

SUMMARY OF TEST

S/NO	TEST	OBSERVATION	INFERENCE
1	O.A + FeCl2(aq)	Green colour of Fe2+ solution turns to reddish- brown of Fe3+	O.A is present
2	O.A + H2S(g)	Formation of yellow deposits of sulphur	O.A is present
3a.     b.	O.A + acidified KI   Red- brown solution + starch	Reddish – brown coloration produced. Iodine is liberated. Reddish- brown turns dark blue. The iodine reacts with the starch	O.A is present
4	O.A + SO2(g) +                  dilute HNO3(aq) + Ba(NO3)2(aq)	White precipitate of insoluble BaSO4 is formed	O.A is present

TEST FOR REDUCING AGENTS:  Reducing agent is detected in the laboratory using any of the following reagents.

1.     Acidified Potassium tetraoxomanganate (VII)

2.     Acidified Potassium heptaoxodichromate (VI)

S/N	TEST	OBSERVATION	INFERENCE
1	R.A + acidified KMO4	Purple solution of KMnO4 turns colorless on addition of R.A	R.A is present
2	R.A + acidified K2Cr2O7	Orange solution of K2Cr2O7 turns green solution addition of R.A	R.A is present

Common oxidizing agents are: concentrated HNO₃, H₂SO₄, KMnO₄, K₂Cr₂O₇, O₂, Cl₂ etc.

Common reducing agents are: concentrated HCl, pure metals, carbon, H₂, SO₂, H₂S, etc.

EVALUATION

1.    Define Oxidizing agent and Reducing agent in terms of electron transfer

2.    Describe one test each for identifying an Oxidizing agent and a Reducing agent

BALANCING OF REDOX EQUATIONS

Redox equations are balanced by first considering the two half equations involved in such reaction. Steps involved are

1.       Identify the oxidizing and reducing agents and deduce expected products.

2.       Write the half equations for oxidation and reduction. Balance the atoms and charges for each equation.

3.       Make sure that the electrons loss in the oxidation half equation is balanced by the electrons gain in the reduction half equation.

4.       Combine the halves equations to eliminate the electrons and get the overall redox equation.

EXAMPLE 1: Write a balanced ionic equation for the redox reaction between potassium tetraoxomanganate(VII) and Iron (II)tetraoxosulphate(VI)in acidic medium.

SOLUTION:   

O.A → MnO₄^–

R.A → Fe²⁺

OXIDATION HALF EQUATION

         Fe²⁺ →      Fe³⁺    +    e^–

REDUCTION HALF EQUATION

        MnO₄^–     +      H⁺ → Mn²⁺    + H₂O

BALANCED HALF EQUATIONS

        5Fe²⁺ →  Fe³⁺      +    5e^–

        MnO₄^–  +     8H⁺+     5e^–   → Mn²⁺    +     4H₂O

COMBINED EQUATION

  5Fe²⁺    +       MnO₄^–     +8H⁺                +   5e^–   →  Fe³⁺+  5e^– +    Mn²⁺ +4H₂O

The electrons on both sides of the equation cancel out and the overall equation is

5Fe²⁺     +       MnO₄^–     + 8H⁺ → Fe³⁺    +    Mn²⁺  +  4H₂O

EXAMPLE 2: Write a balanced equation for the following reaction in basic medium

                   Cr³⁺ +    BrO^– → CrO4^2-  +   Br^–

SOLUTION:

O.A       → BrO^–

R.A       →         Cr³⁺

OXIDATION HALF EQUATION

         Cr³⁺ → CrO₄^2-

Balancing of atoms:  Cr³⁺  + 8OH^– → CrO₄^2-  +  4H₂O

Balancing of charges: Cr³⁺  + 8OH^– → CrO₄^2-  +  4H₂O + 3e^–

REDUCTION HALF EQUATION

Balancing the atoms: BrO^–  +  H₂O → Br^–  +  2OH^–

Balancing the charges: BrO^–  +  H₂O  +  2e^–  →  Br^–  +  2OH^–

But electron lost in the oxidation half must equal electron gained in the reduction half equation.

Multiplying the oxidation half equation by 2 and the reduction half equation by 3 gives

BALANCED HALF EQUATIONS

2Cr³⁺  +16OH^–                   → 2CrO₄^2-  +  8H₂O + 6e^–

3BrO^–  +  3H₂O  +  6e^– → 3Br^–  +  6OH^–

COMBINED EQUATION

2Cr³⁺  +  16OH^–  3BrO^–  +  3H₂O  +  6e^–  →      2CrO₄^2-  +  8H₂O + 6e^– 3Br^–  +  6OH^–

The electrons on both sides of the equation cancel out and the overall equation is

2Cr³⁺  +  16OH^–    3BrO^–  +  3H₂O → 2CrO₄^2-  +  8H₂O + 3Br^–  +  6OH^– 

GENERAL EVALUATION/REVISION

1. Determine the oxidation number of

(a) Fe in  Fe₂O₃      (b) Cu in [Cu(NH₃)₄]²⁺

2.    Name the following compounds

• H₂CO₃          (b) KMnO₄

3.The compound Na₂S is called ————–

4.The IUPAC name of NaHSO₄ is————-

READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O. Y. Ababio, pages 193-196