Experiment 7 - Redox Titration PDF

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EX PERIMENT 7:REDOX TITRATIONNAME : NUR SHAHIRA BINTI SHARIMANSTUDENT ID : 2020837258CL ASS : AS2463BDATE OF EXPERIMENT : 11 / 11/ 2020DATE OF SUBMITION : 17 / 11/ 2020L ECTURE NAME : DR HAIRUL AMANI BINTI ABDUL HAMIDPURPOSEThe purpose of this experiment is to standardize the potassium permanganate ...

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EXPERIMENT 7: REDOX TITRATION

NAME

: NUR SHAHIRA BINTI SHARIMAN

STUDENT ID

: 2020837258

CLASS

: AS2463B

DATE OF EXPERIMENT

: 11 / 11/ 2020

DATE OF SUBMITION

: 17 / 11/ 2020

LECTURE NAME

: DR HAIRUL AMANI BINTI ABDUL HAMID

PURPOSE The purpose of this experiment is to standardize the potassium permanganate solution and to determine the composition of metal by titration method. PROCEDURE

DATA

CALCULATION 1. Molarity of potassium permanganate 5Na2C2O4 + 2KMnO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 10CO2 + 5Na2SO4 + 8H2O TRIAL 1 Moles of Na2C2O4:   

Moles of Na2C2O4 =     . 

= [ ()()()]/ =

. 

 /

= 1.54 x 10-3 mol From equation: 5 mol of Na2C2O4 → 2 mol KMnO4 1.54 x 10-3 mol of Na2C2O4 → X mol KMnO4

X=

.   

x2

= 6.18 x 10-4 mol KMnO4 Molarity of KMnO4: Molarity of KMnO4 = =

  

  

.    . 

= 0.088 M TRIAL 2 Moles of Na2C2O4 Moles of Na2C2O4 =

  

   

. 

= [ ()()()]/ . 

=  / = 1.51 x 10-3 mol From equation: 5 mol of Na2C2O4 → 2 mol KMnO4 1.51 x 10-3 mol of Na2C2O4 → X mol KMnO4 X=

.   

x2

= 6.06 x 10-4 mol KMnO4 Molarity of KMnO4: Molarity of KMnO4 = =

     

.    . 

= 0.089 M Average molarity of KMnO4: Average molarity of KMnO4 =

.   .  

= 0.0885 M 2. Concentration of iron (II) in iron ore solution Moles of iron (II):    ()

Moles of iron (II) =     () =

.   /

= 3.6 x 10-3 mol Iron(II) in 100 mL Iron(II) in 100 mL =

=

  () 󰇡

 󰇢 

.    . 

= 0.036 mol/L 3. Percentage purity of iron(II) in unknown sample

MnO4- + 8H+ + 5Fe2+ → 2Mn2+ + 4H2O + 5Fe3+ TRIAL 1 Moles of KMnO4: Moles of KMnO4 = Molarity x Volume of KMnO4 .

= 0.1 M x 󰇡 󰇢 𝐿 = 3.0 x 10-5 mol From equation: 1 mol MnO4- → 5 mol Fe2+ 3.0 x 10-5 mol MnO4- → X mol Fe2+ X = 3.0 x 10-5 mol x 5 = 1.5 x 10-4 mol Fe2+ 1.5 x 10-4 mol Fe3+ in 25 mL mixed solution

Fe2+ in 100 mL: Fe2+ in 100 mL =

.    

𝑥 100 𝑚𝐿

= 6.0 x 10-4 mol Mass of Fe2+: Mass of Fe2+= Molar mass x moles of Fe2+ = 56 g/mol x 6.0 x 10-4 mol = 0.0336 g Percentage of Fe2+ in 100 mL: Percentage of Fe2+ in 100 mL =

. 

. 

𝑥 100

= 16.66% TRIAL 2 Moles of KMnO4: Moles of KMnO4 = Molarity x Volume of KMnO4 .

= 0.1 M x 󰇡 󰇢 𝐿 = 4.0 x 10-5 mol From equation: 1 mol MnO4- → 5 mol Fe2+ 4.0 x 10-5 mol MnO4- → X mol Fe2+ X = 4.0 x 10-5 mol x 5 = 2.0 x 10-4 mol Fe2+ ‫؞‬2.0 x 10-4 mol Fe2+ in 25 mL mixed solution Fe2+ in 100 mL: Fe2+ in 100 mL =

.    

= 8.0 x 10-4 mol

𝑥 100 𝑚𝐿

Mass of Fe2+: Mass of Fe2+= Molar mass x moles of Fe2+ = 56 g/mol x 8.0 x 10-4 mol = 0.0448 g Percentage of Fe2+ in 100 mL: Percentage of Fe2+ in 100 mL =

.  . 

𝑥 100

= 22.21% TRIAL 3 Percentage of Fe2+ in Trial 3 is similar to Trial 1 because the volume of KMnO4 used is similar which is 0.3 mL. Percentage purify of Fe2+: Average of percentage purify of Fe2+ =

( .  .  . )% 

= 18.51% DISCUSSION Redox or oxidation-reduction reaction are considered electron-transfer reaction. It occur simultaneously. Redox reaction involved changes in oxidation number of elements in reactant and product. The term oxidation reaction refers to the half-reaction that involves loss of electrons. A reduction reaction is half-reaction that involves gain of electrons. Redox is use in standardizing solution and determine an element in reactant. Standardization is very important. It used to determine their exact concentration of a solution. In this experiment, sodium oxalate (Na2C2O4) is used to standardize potassium permanganate (KMnO4). Na2C2O4 is act as reducing reagent by losing electrons to MnO4- meanwhile KMnO4 is act as oxidizing agent due to gain electron from Na2C2O4, according to the equation below: 5Na2C2O4 + 2KMnO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 10CO2 + 5Na2SO4 + 8H2O From this experiment, the molarity of KMnO4 can be determined. The average molarity of KMnO4 is 0.0885M. In the experiment, potassium permanganate (KMnO4) will react with iron ore with the presence of sulfuric acid (H2SO4). KMnO4 is act as oxidizing agent. MnO4- converted to Mn2+ because it accepts electrons from Fe2+. The conversion of the permanganate ion cause the change of colour from light

purple to pink and from that, KMnO4 also act as indicator. Iron ore act as reducing agent because it donates electrons to oxygen and converted Fe2+ to Fe3+. The presence of H2SO4 is use to increase acidity of the solution to prevent MnO4 reduced to MnO2. Hence, the acid is stable in the presence of strong oxidizing agent. Furthermore, acid is used to provide H + necessary for the reaction to occur quickly. The reaction can be seen equation as follow: MnO4- + 8H+ + 5Fe2+ → 2Mn2+ + 4H2O + 5Fe3+ Based on the calculation, percentage purity of Fe2+ in Trial 1 and Trial 3 is 16.66% and the percentage in Trial 2 is 22.21%. The average of the percentage is calculated because of the difference in the percentage result. The average percentage purity of Fe2+ is 18.51%. From this percentage, it shows that some errors had been occur such as misreading volume, error in concentration and using equipment incorrectly. CONCLUSION Potassium permanganate (KMnO4) can be standardize with 0.0885M and the composition of iron(II) can be determine with the percentage purify of 18.51% by using redox titration method. QUESTION 1. Calculate following: a) Molarity of potassium permanganate solution from the titration reaction: 2MnO4- + 5C2O42- + 16H+ → 2Mn2+ + 10CO2 + 4H2O TRIAL 1 Moles of Na2C2O4:   

Moles of Na2C2O4 =     =

.  [ ()()()]/

=

. 

 /

= 1.54 x 10-3 mol From equation: 5 mol of C2O42- → 2 mol MnO41.54 x 10-3 mol of C2O42- → X mol MnO4X=

.   

x2

= 6.18 x 10-4 mol MnO4Molarity of MnO4-: Molarity of MnO4- = =

  

   

.    . 

= 0.088 M TRIAL 2 Moles of Na2C2O4 Moles of Na2C2O4 = =

  

   

.  [ ()()()]/ . 

=  / = 1.51 x 10-3 mol

From equation: 5 mol of C2O42- → 2 mol MnO41.51 x 10-3 mol of C2O42- → X mol MnO41.52 X =

.   

x2

= 6.06 x 10-4 mol MnO4-

Molarity of MnO4-: Molarity of MnO4- =

=

   

  

.    . 

= 0.089 M

b) Concentration of iron(II) in iron ore solution (initial volume 100mL) Moles of iron (II): Moles of iron (II) = =

   ()

    ()

. 

 /

= 3.6 x 10-3 mol

Iron(II) in 100 mL Iron(II) in 100 mL =

=

  () 󰇡

 󰇢 

.    . 

= 0.036 mol/L

c) Percent purity of iron(II) in unknown sample MnO4- + 8H+ + 5Fe2+ → 2Mn2+ + 4H2O + 5Fe3+ TRIAL 1 Moles of KMnO4: Moles of KMnO4 = Molarity x Volume of KMnO4 .

= 0.1 M x 󰇡 󰇢 𝐿 = 3.0 x 10-5 mol From equation: 1 mol MnO4- → 5 mol Fe2+ 3.0 x 10-5 mol MnO4- → X mol Fe2+ X = 3.0 x 10-5 mol x 5 = 1.5 x 10-4 mol Fe2+

1.5 x 10-4 mol Fe3+ in 25 mL mixed solution Fe2+ in 100 mL: Fe2+ in 100 mL =

.  

𝑥 100 𝑚𝐿

 

= 6.0 x 10-4 mol Mass of Fe2+: Mass of Fe2+= Molar mass x moles of Fe2+ = 56 g/mol x 6.0 x 10-4 mol = 0.0336 g Percentage of Fe2+ in 100 mL: Percentage of Fe2+ in 100 mL =

. 

. 

𝑥 100

= 16.66% TRIAL 2 Moles of KMnO4: Moles of KMnO4 = Molarity x Volume of KMnO4 = 0.1 M x 󰇡

.

󰇢𝐿



= 4.0 x 10-5 mol From equation: 1 mol MnO4- → 5 mol Fe2+ 4.0 x 10-5 mol MnO4- → X mol Fe2+ X = 4.0 x 10-5 mol x 5 = 2.0 x 10-4 mol Fe2+ ‫؞‬2.0 x 10-4 mol Fe2+ in 25 mL mixed solution Fe2+ in 100 mL: Fe2+ in 100 mL =

.    

= 8.0 x 10-4 mol

𝑥 100 𝑚𝐿

Mass of Fe2+: Mass of Fe2+= Molar mass x moles of Fe2+ = 56 g/mol x 8.0 x 10-4 mol = 0.0448 g Percentage of Fe2+ in 100 mL: Percentage of Fe2+ in 100 mL =

.  . 

𝑥 100

= 22.21% TRIAL 3 Percentage of Fe2+ in Trial 3 is similar to Trial 1 because the volume of KMnO4 used is similar which is 0.3 mL. Percentage purify of Fe2+: Average of percentage purify of Fe2+ =

( .  .  . )% 

= 18.51% 2. No indicator needed in this experiment because potassium permanganate, KMnO4 act as indicator itself. The light purple colour of KMnO4 change to pink when the reaction is completed due to the changes of oxidation number of MnO4- to Mn2+. 3. Iron ore needed to dry before conducting the experiment because the water composition and other impurities in the iron can affect the whole experiment. This can cause an error in data and also calculation. REFERENCES 1. https://www.academia.edu/33355154/COURSE_CODE_CHM_138_EXPERIMENT_6_RED OX_TITRATION_DETERMINATION_OF_THE_MOLARITY_AND_CONCENTRATION _OF_IRON_II_SULPHATE_FeSO_4_SOLUTION_BY_TITRATION_WITH_POTASSIU M_PERMANGANATE 2. https://www.enotes.com/homework-help/why-kmno4-solution-must-standardized-before-use299421

3. http://www.titrations.info/permanganate-titration-standardization...