Exercise 8 Page 1 Illinois Central College CHEMISTRY 132 Laboratory Section: Redox Titration Name: Equipment 1-25.00 ml burette 0.100 N KMn 4 solution 2-50 ml beakers KHS 3 solution of unknown Normality bjectives The objectives of this experiment are to develop an understanding of oxidation-reduction titration and the use of the "eqivalent" concept as applied to oxidizing and reducing agents. Background xidation Reduction reactions are chemical reactions in which substances undergo changes in oxidation state. xidation is defined as the loss of electrons (or an increase in oxidation state) and reduction as the gain of electrons (or a decrease in oxidation state). In acid base titrations, equivalent amounts of acid and base must be used for exact neutralization at the titration endpoint. In oxidation-reduction reactions, there is a similar equivalence between oxidizing and reducing agents. In acid base reactions, one "equivalent" corresponded to 1 mole of H +1 (for an acid) or 1 mole of H -1 (for a base). In oxidation-reduction reactions, one "equivalent" refers to 1 mole of electrons either provided (by a reducing agent) or taken (by an oxidizing agent). So, just as with acids and bases, one equivalent of a reducing agent will reduce one equivalent of an oxidizing agent. At the endpoint of a redox titration; number of equivalents of oxidizing agent = number of equivalents of reducing agent. xidizing Agent: the substance which takes up electrons Reducing Agent: the substance which gives up electrons Alternatively, we can write: "milliequivalents" of oxidizing agent = "milliequivalents" of reducing agent where a "milliequivalent is 1/1000 of an equivalent.
Exercise 8 Page 2 Consider the reaction of potassium permanganate with oxalic acid in the presence of sulfuric acid. The balanced chemical equation and net ionic equations are; 2 KMn 4 + 5 H 2 C 2 4 + 3 H 2 S 4 10 C 2 + 2 MnS 4 + 8 H 2 (oxidation: loss of 2 e -1 ) x 5 = 10 e -1 2 Mn 4-1 + 5 H 2 C 2 4 + 6 H +1 10 C 2 + 2 Mn +2 + 8 H 2 (reduction: gain of 5 e -1 ) x 2 = 10 e -1 The molecular mass of KMn 4 is 158.0 g/mol Since one mole of KMn 4 actually removes 5 moles of electrons in this redox reaction, we can say that 1 mol KMn 4 = 5 equivalents of KMn 4 (at least, in this particular reaction). This allows us to define the "equivalent weight" of a compound, that is, the mass which would provide 1 equivalent of oxidizing power. For KMn 4, that would be; ( 158.0 grams KMn 4 1 mol KMn 4 )x( 1 mol KMn 4 5 equivalents ) = 31.60 grams/equivalent The molecular mass of oxalic acid is 90.0 grams. Since one mole of oxalic acid actually provides 2 moles of electrons in this redox reaction, we can say that 1 mol H 2 C 2 4 = 2 equivalents of H 2 C 2 4 (again, in this particular reaction). So, the equivalent weight of this reducing agent can also be calculated, that is, the mass of ocxalic acid that would provide 1 equivalent of reducing power. For oxalic acid, that would be; ( 90.0 grams H 2 C 2 4 1 mol H 2 C 2 4 )x( 1 mol H 2C 2 4 ) = 45.0 grams/equivalent 2 equivalents This leads us to the logical end of defing the "Normality" of oxidizing and reducing agents just as we did with acids and bases. Since different oxidizing and reducing agents can provide various numbers of equivalents, we simply redefine the strength of an oxidizing or reducing agent as "the number of equivalents per liter" (or milliequivalents per milliliter),or Normality. N oxidizing agent = milliequivalents of oxidizing agent milliliter N reducing agent = milliequivalents of reducing agent milliliter Since the endpoint of a redox titration demands that the equivalents of oxidizing agent equal the equivalents of reducing agent, then; N oxidizing agent x ml oxidizing agent = N reducing agent x ml reducing agent
Exercise 8 Page 3 Safety Precautions Safety goggles must be worn in the lab at all times. Any skin contacted by chemicals should be washed immediately. Procedure In this experiment, you will determine the normality of the reducing agent, potassium bisulfite (KHS 3 ) by titrating with a standard 0.100 N KMn 4 solution. 1. Set up a 25.00 ml burrette and fill it with your unknown concentration of reducing agent (KHS 3 ). Be sure to record the letter of your unknown on your Report Sheet. 2. Draw 15-25 ml of the 0.100 N KMn 4 solution (from the dispensing table) into a clean, dry 50 ml beaker. Carefully read the burette to the nearest 0.01 ml before and after the withdrawal and record these values on your Report Sheet. 3. Add approximately 5 ml of concentrated (12 M) H 2 S 4 to your permanganate solution. If the solution turns brown during the titration with your reducing agent, add additional acid.. 4. Dispense the reducing agent into your permanganate solution while stirring drop by drop as the endpoint is approached, until one drop of the reducing agent completely decolorizes the permanganate. (White paper or some other white background makes the color change more discernable.) Record your final burette reading on your Report Sheet. 5. Starting with a different volume of permanganate, repeat the titration for Trial 2. 6. Calculate the Normality of the KHS 3 and record it on your Report Sheet. 7. Report the average Normality on your Report Sheet. 8. Write a balanced oxidation-reduction equation for the reaction of KMn 4 with KHS 3. The unbalanced equation is: KMn 4 + KHS 3 + H 2 S 4 MnS 4 + KHS 4 + K 2 S 4 + H 2 9. Based on their Normalities, calculate the weights of KMn 4 and KHS 3 present in one liter of each solution and record these values on your Report Sheet.
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Exercise 8 Page 5 Illinois Central College CHEMISTRY 132 Laboratory Section: REPRT SHEET Redox Titration Name: Unknown Number xidizing Agent: KMn 4 Final burette reading Initial burette reading Volume used Normality of KMn 4 0.100 N 0.100 N Reducing Agent: KHS 3 Final burette reading Initial burette reading Volume used Normality of KHS 3 Average Normality 1. Write the balanced redox equation for the reaction of KMn 4 with KHS 3. (over)
Exercise 8 Page 6 2. Mass of KMn 4 per liter in a 0.100 N solution. g/l 3. Mass of KHS 3 per liter g/l (based on your experimental normality) Show All Calculations Below:
Exercise 8 Page 7 Illinois Central College CHEMISTRY 132 Laboratory Section Name: PRELAB: Exp. 8 Redox Titration 1. Balance the following Redox reactions. a) Zn + AgN 3 Zn(N 3 ) 2 + Ag b) Sn + HN 3 Sn 2 + N 2 + H 2 c) Mn + Pb 2 + HN 3 HMn 4 + Pb(N 3 ) 2 + H 2 d) Zn + NaN 3 + NaH Na 2 Zn 2 + NH 3 + H 2 e) K 2 Cr 2 7 + H 2 S + H 2 S 4 K 2 S 4 + Cr 2 (S 4 ) 3 + S + H 2 f) MnCl 2 + Pb 2 + HN 3 HMn 4 + Cl 2 + Pb(N 3 ) 2 + H 2
Exercise 8 Page 8