Analytical Chemistry One - Section Seven PDF

MEDICINAL CHEMISTRY DEPARTMENT ANALYTICAL CHEMISTRY ONE SECTION SEVEN  Balancing Redox Equations H2C2O4 + MnO4-→ CO2 + Mn2+ Ionic Form Mass balance H2C2O4 + MnO4- → CO2 + Mn2+ H2C2O4 + MnO4- → 2CO2 + Mn2+  Balancing Redox Equations...

MEDICINAL CHEMISTRY DEPARTMENT ANALYTICAL CHEMISTRY ONE SECTION SEVEN  Balancing Redox Equations H2C2O4 + MnO4-→ CO2 + Mn2+ Ionic Form Mass balance H2C2O4 + MnO4- → CO2 + Mn2+ H2C2O4 + MnO4- → 2CO2 + Mn2+  Balancing Redox Equations -2 e - ×5 H2C2O4 2×+1 +2C+4×-2=0 +6 +8 2+2C-8=0 H2C2O4 + MnO4- → 2CO2 + Mn2+ Electron +7 +2 2C=+6 balance +5 e - ×2 H2C2O4 + 2MnO4- → 2CO2 + 2Mn2+ 5H2C2O4 + 2MnO4- → 10CO2 + 2Mn2+  Balancing Redox Equations Charge 5H2C2O4 + 2MnO4- → 10CO2 + 2Mn2+ balance Charge LHS RHS zero -2 zero +4 Total Charge -2 +4 5H2C2O4 + 2MnO4- +6H+ → 10CO2 + 2Mn2+ 5H2C2O4 + 2MnO4- +6H+ → 10CO2 + 2Mn2++ 8H2O  Balancing Redox Equations 5H2C2O4 + 2MnO4- +6H+ → 10CO2 + 2Mn2++ 8H2O  Redox titrations These are titrations in which the reaction between the titrant and the sample involves oxidation reduction process. Oxidation: it is a process of loss of electrons and accompanied by an increase in the oxidation number. Example: Fe2+ → Fe3+ + e- Reduction: it is a process of gain of electrons and accompanied by a decrease in the oxidation number. Example: Ce4+ + e- → Ce3+  Potassium Permanganate (KMnO4) As a titrant It is a strong oxidizing agent and has a lot of applications in both acidic and alkaline media. It is a self indicator as it has a dark purple color (First drop excess will produce a faint pink color) At e.p  1- Determination of Oxalic Acid Principle: Oxalic acid is determined by direct titration with standard KMnO4 in acidic medium. 5H2C2O4 + 2MnO4- + 6H+ → 10CO2 + 2Mn2+ + 8H2O purple colorless Mn+2 Procedure: +5 es 1. Pipette 10 mL sample of oxalic acid MnO4- 2. Add 20 mL dil. H2SO4 +3 es MnO2 3. Heat to about 70◦C (to accelerate the rate of reaction) 4. Titrate while hot with 0.05 M KMnO4 (the first few drops of KMnO4 should be added slowly till a sufficient amount of Mn2+ (autocatalyst) is formed otherwise a brown ppt o MnO2 f MnO2 is formed due to gain of 3 electrons instead of 5 electrons). Color change: Colorless → first faint pink Calculations: 5H2C2O4 + 2MnO4- + 6H+ → 10CO2 + 2Mn2+ + 8H2O At e.p. No. mmoles of analyte = mmoles of Titrant × R R 5/2 5 mmoles H2C2O4 = 2 mmoles MnO4- R = no. ofmmoles of Analyte /no. of mmoles of titrant Wt.(mg) = mmol x M.wt Wt.(g) = Wt.(mg) / 1000 % w/v =(Wt.(g) / V (mL) ) x 100