The objective of this experiment was to determine the percent composition of hydrogen peroxide by analyzing the hydrogen peroxide through an oxidation-reduction reaction. A buret was filled with 0.025 M of potassium permanganate and zeroed. Exactly 1 mL of hydrogen peroxide was poured into an erlenmeyer flask using a pipet and diluted with 25 mL of deionized water and then 5 mL of 6 M sulfuric acid was added as well. The purpose of including the sulfuric acid is to catalyze the reaction between potassium permanganate and hydrogen peroxide. The solution was then carefully titrated with potassium permanganate until it immediately turned into a light pink color. This shows that the equivalence point was met and the redox reaction is complete. The …show more content…
Reduction half-reaction: 2(MnO4-+ 8 H+ + 5 e- → Mn2+ + 4 H2O) Oxidation half-reaction: 5(H2O2 → O2 + 2 H+ + 2 e-) Balanced redox reaction: 2 MnO4-+ 6 H+ + 5 H2O2 → 2 Mn2+ + 8 H2O + 5 O2
Example Calculation Using Trial 1:
Final Volume of KMnO4 - Initial Volume of KMnO4 = Volume of KMnO4 Used
43.31 mL - 28 mL = 15.31 mL
Determining mol MnO4- titrated
Molarity KMnO4 x Volume of KMnO4 = mol MnO4- 0.025 M x 15.31mL x (1 L/ 1000 mL) = 3.83 x 10-4 mol MnO4-
Determining mol H2O2 mol MnO4- x [ 5 mol H2O2 / 2 mol MnO4- ]= mol H2O2
3.83 x 10-4 mol MnO4- x [ 5 mol H2O2 / 2 mol MnO4- ] = 9.57 x 10-4 mol H2O2
Determining mass of H2O2 mol H2O2 x Molar Mass of H2O2 = mass of H2O2
9.57 x 10-4 mol H2O2 x 34.016 g/mol = 0.0326 g H2O2
Determining the volume of H2O2 mass of H2O2 x Density of H2O2 = volume of H2O2
0.0326 g/mol x 1.00 L/g = 0.0326 L H2O2
Determining the percent composition of H2O2:
(volume of solute / volume of solution) x 100% = percent composition
(0.0326 mL / 1.00 mL) x 100% = 3.26%
Average Percent Composition of H2O2 (3.26 % + 3.20 % + 3.36 %) / 3 = 3.27 %
Trial
Percent Composition of H2O2
1
3.26 %
2
3.20 %
3
3.36 %
Average
3.27 %
Balanced Redox Reaction:
2 MnO4- + 6 H+ + 5 H2O2 → 2 Mn2++ 8 H2O + 5