QUANTITATIVE ANALYSIS [REDOX REACTION TITRATION]

 

Name of teacher:           

Week:                                                                   Date:                                                     Time:

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

Subject:                               CHEMISTRY                                                                         Class: SS THREE

Topic:                                    REDOX REACTION (TITRATION)

Sub topic:

Reference materials:

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

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

(3) INTERNET

Instructional materials: beaker, beam balance

Entry behavior: The students have been taught density.

Objective: At the end of the lesson the students should be able to:

1.       Define oxidation and reduction.

2.       Identify an oxidizing agents and reducing agents

3.       Solve problem involving redox titration

4.       Test for oxidizing agents and reducing agents.

CONTENT

REDOX REACTIONS (TITRATION)

In Redox titration, the solutions commonly used are the oxidizing agents and reducing agents. Examples of oxidizing agents include solution KMnO₄, K₂Cr₂O₇, iron (iii) compounds, Conc. HNO₃, Conc. H₂SO₄, Potassium trioxoiodate (v), KIO₃, iodine, powdered Manganese (iv) oxide, MnO₂ and gaseous O₂ and Cl₂ while example of reducing agents are solution of iron(ii)compound, Tin(ii)compounds, KI, Conc. HCl, Na₂S₂O₃, pure metals, carbon, H₂, H₂S and SO₂.

In acid-base titrations, the neutralization reaction are represented by molecular equations e.g

HCl  +  KOH                KCl  +   H₂O

But in redox titrations only the net ionic equation is represented e.g.

2MnO₄^-  +  5C₂O₄^2-   + 16H⁺                               2Mn²⁺ + 10CO₂ + 8H₂O

When an oxidizing agent (O.A) is titrated against a reducing agent (R.A) the concentration of their solutions can be determined as is done in acid-base titration.

Example

A is a solution of KMnO₄, B is 0.02moldm^-3 of ethanedioc acid and solution C is 1.0moldm^-3 of conc. H₂SO₄. Titrate solution A against 25cm³ of solution B which has been acidified with 5cm³ of solution C. Average volume of 16.5cm³ of A was required for complete oxidation of solution B from the result obtained calculate (a) Conc. of solution A in moldm^-3 (b) Conc. of solution A ingdm^-3.

Equation of reaction: 2MnO₄^-  +  5C₂O₄^2-   + 16H⁺                               2Mn²⁺ + 10CO₂ + 8H₂O   [K=39, Mn=55, O=16].

Solution

Co = ?, CR= 0.02M , Vo= 16.5cm3, VR= 25cm3 , no = 2, nR= 5

  =  ; Co =  = 0.012112M

(a)    Molar mass of KMnO₄ = K + Mn + 4O = 39 + 55 + (4x16) = 158gmol-1

Mass conc. of A = molar conc. of A x molar mass of A = 0.0121 moldm-3 x 158gmol-1 = 1.91gdm-3.

Note

In this redox reaction, add KMnO4 from the burette until the solution is permanently pink, that is the end point.

KMnO₄ is a useful oxidizing agent. However, it has the disadvantage that it is sometimes too strong. For example it will oxidize chlorine ions to chlorine.

2Cl^-      Cl₂ + 2e^-

And it must not be used in titration, if Chloride is present. In this case, a weaker oxidizing agent must be used and potassium heptaoxochromate (vi) is a suitable one.

Cr₂O₇^- + 14H⁺  + 6e^-                      2Cr³⁺ + 7H₂O

The chromate ion is yellow and chromate (iii) salt is green. So that there is not the sharp end point as with KMnO₄. In this case an indicator an indicator must be used and a common one is diphenylamine. The reduced form is colourless while the oxidized form is dark blue.

Experiment: To Determine the Concentration of a given solution of Sodium thiosulphate, using a standard solution of potassium iodate; then the thiosulphate solution is used to determine the concentration of a given solution of sulphuric acid.

Procedure: Prepare 0.01M KIO₃ solution by weighing 0.90g of KIO₃. Dissolve and transfer to a 250cm³ volumetric flask.

 If 250cm³ of the solution contains 0.90g

 1000cm³ (1dm³) of the solution will contain 0.90g x 4 = 3.6g

Molar mass of KIO₃ – K + I + 3O = 39 + 126.9 + (3x16) = 213.9 gmol-1

Molarity of the solution is 3.60/ 213.9 = 0.017M

This is the standard solution of KIO₃.

25cm³ portions, add about 10cm³ of 1M H₂SO₄ and 10cm³ of 10% KI solution, and shake the flask. Titrate the iodine librated with the sodium thiosulphate from the burette, adding about 2cm3 of starch solution near the end point.

Titration result: I₂ vs S₂O₃^2-

Burette Reading (cm3)	Rough or Trial(cm3)	1st Titre (cm3)	2nd Titre (cm3)	3rd Titre (cm3)
Final	33.20	33.10	33.10	33.00
Initial	0.00	0.00	0.00	0.00
Volume used	33.20	33.10	33.10	33.3

Indicator used: Starch

Average of accurate titres =  = 33.07cm3 of S₂O₃^2-

Reaction equation:

(1)    IO^3- + 5I- + 6H⁺              3I₂ + 3H₂O;    (2) 2S₂O3^2- + I₂                     S₄O₆^2- + 2I^-

FROM EQUATION (i) and (ii)

1 mol of IO₃^-  =     6 mol of S₂O₃^2-

  =  ; Co =  = 0.077M

Molarity of thiosulphate ion, S₂O₃^2- = 0.077M

 3.  KMnO₄ liberate O₂ from hydrogen peroxide according to the equation: MnO₄^- + H₂O₂            Mn²⁺ + O₂

(a) Balance this equation in acid solution.

(b) If 24.0cm³ of acidified KMnO₄ liberates 0.100cm³ of O₂ measured at S.T.P, calculate the molarity of KMnO₄ [ Molar gaseous volume at S.T.P is 22.4dm³]

Solution

(a)    2MnO₄^- + 5H₂O₂ + 6H⁺                2Mn²⁺ + 5O₂ + 8H₂O

(b)   From the above equation: 2 moles of MnO₄^- liberates 5 moles of O₂

0.100dm³ of O₂ at stp in moles 0.100/22.4 = 0.0045 mol

But 5mol of O₂ are liberated by 2 mol of MnO₄^-

0.0045 mol of O₂ = 0.0018 mol of MnO₄^-

If 0.0018 mol of MnO₄^- is present in 1000cm³ of solution, the molarity of the solution will be 0.0018M

If 0.0018 mol is present in 24.0 cm³ of solution, molarity = 0.0018 x 1000/24 = 0.075M

1.       KMnO₄ is used to titrate a sample containing an unknown % of Fe. The sample is dissolved in H₃PO₄/ H₂SO₄ mixture to reduce all of the iron to Fe²⁺ ions. The solution is then titrated with 0.01625M K₂Cr₂0₇, producing Fe³⁺ and Cr³⁺ ions in acidic solution. The titration requires 32.26cm³ of K₂Cr₂O­₇ for 1.2765g the sample (a) Balance the net ionic equation using the half-reaction method. (b) Determine the % Fe in the sample (c ) is the sample ferrous iodate, ferrous phosphate or ferrous acetate?

Solution

(a)    6Fe²⁺ + Cr₂O₇^2- + 14H⁺                                   6Fe³⁺ + 2Cr³⁺ + 7H₂O

(b)   Mole of Cr₂O₇^2- = molar conc. x volume (dm3) = 0.01625M x 32.26/1000 = 0.000524225mol.

From equation: 1 mol Cr₂O₇^2- reacts with 6 mol of Fe²⁺

0.000524225 mol Cr₂O₇^2- reacts with 6x 0.000524225 mol [ 0.00314535 mol] of Fe²⁺

Mass of Fe react = Amount x Molar mass of Fe = 0.00314535 x 55.85 = 0.175652g

% of iron in the sample = 0.175652/1.2765 x 100% = 13.76%

(c ) which compound contains 13.6% Fe? The only way to determine this is to calculate the % composition of the three substances.

Ferrous iodate [Fe(IO₃)₂] ; %Fe = 55.85/403.67 x 100 = 13.77%

Fe₃(PO₄)₂ = 46.87% ; Fe(C₂H₃O₂)₂ = 32.11%

TEST FOR OXIDANTS AND REDUCTANTS

TEST FOR OXIDISING AGENTS

EXPERIMENTS	OBSERVATION	INFERENCE
1 a. Solution + dilute H2SO4 + KI solution b. To the brown solution add starch	Solution turns brown, iodine liberated Starch turns blue-black	O.A suspected. O.A confirmed.
2 a. Solution + dil. H2SO4 + FeSO4 sol.     b. Pale-Yellow brown sol. + NaOH sol	The green colour of FeSO4 changes to pale Yellow brown indicating Fe2+ is oxidized to Fe3+ Reddish-brown precipitate forms	O.A suspected     O.A confirmed
3. a. Solid substance + carbon + heat   b. Gas pass through the lime water	A gas evolved which is faintly acidic to litmus. Probably CO2 gas. Lime water change to milky. White precipitate	O.A suspected   O.A confirmed
4.  H2S gas bubbled in the solution	Yellow precipitate appears, H2S gas oxidize to sulphur	O.A confirmed
5. Solution + MnO2, warm gently	A greenish Yellow gas evolved with irritating smell, the gas changes blue litmus paper red and then bleaches it white. The gas is chlorine.	O.A confirmed
6a. Solution + MnO2 and warm gently   b. Gas tested with glowing splint	A colourless, odourless gas evolved, the gas is neutral to litmus, possible oxygen gas Gas rekindled glowing splint	O.A suspected   O.A confirmed

TEST FOR REDUCING AGENTS

EXPERIMENTS	OBSERVATION	INFERENCE
1. Solution + dil. H2SO4 + KMnO4 sol	The purple colour of KMnO4 is decolourised	R.A confirmed
2. solution + dil. H2SO4 + k2Cr2O7 sol	The yellow colour of K2Cr2O7 turns green	R.A confirmed
3a. Solution + FeCl3 sol.   b. Pale green solution + NaOH sol.	The brown colour of FeCl3 turns to pale green. Fe3+ reduces to Fe2+. Green gelatinous precipitate observed	R.A suspected   R.A confirmed
4. Gaseous substances passed over heated CuO	The black CuO turns brown. CuO is reduced to metallic Cu.	R.A confirmed
5. Heat solid substance + CuO	The black CuO turn brown. CuO is reduced to metallic Cu.	R.A confirmed

PRESENTATION

Step I: The teacher define oxidizing agent and reducing agents.

Step II: The teacher explains how to balance a redox reaction .

Step III: The teacher leads the students in solving problem on redox titration.

Step IV: The teacher explains test for oxidizing and reducing agents.

Step V: The teacher allows the students to ask questions.

EVALUATION

The teacher asks the following questions:

1.       Define oxidation and reduction.

2.       Give two example each of an oxidizing agents and reducing agents

3.       Explain one each Test for oxidizing agents and reducing agents.

ASSIGNMENT

1.A solution  of I^-_3(aq) can be standardized by using heat to titrate AsO_6(aq). A titration of 0.102g of As₄O₆ dissolved in 30.00cm³ of water requires  36.55cm³ of I^-_3(aq). Calculate the molarity I^-_3(aq) solution. The unbalanced equation is:

                As₄O₆ + I^-3                         AsO₁₀ + I^-_(aq) . [ As= 74.9, O=16]