JEE Challenger
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ChemHardAdv 2018 • Paper 2

Determine Amount of Manganese Chloride in Redox Reaction with Oxalic Acid

To determine the quantity of dissolved MnCl2\text{MnCl}_2 in an aqueous solution, it was converted entirely into KMnO4\text{KMnO}_4 via the reaction: MnCl2+K2S2O8+H2O→KMnO4+H2SO4+HCl(equation unbalanced)\text{MnCl}_2 + \text{K}_2\text{S}_2\text{O}_8 + \text{H}_2\text{O} \rightarrow \text{KMnO}_4 + \text{H}_2\text{SO}_4 + \text{HCl} \quad \text{(equation unbalanced)} A few drops of concentrated HCl\text{HCl} were added to the solution and gently warmed. Oxalic acid (225 mg225\text{ mg}) was then added incrementally until the characteristic colour of the permanganate ion disappeared completely. Find the mass of MnCl2\text{MnCl}_2 (in mg\text{mg}) originally present in the solution. (Given atomic weights in g mol−1\text{g mol}^{-1}: Mn=55\text{Mn} = 55, Cl=35.5\text{Cl} = 35.5)

Official Numerical Answer126

Topics & Concepts

Redox ReactionsStoichiometryBalancing Redox Reactions

Step-by-Step Solution

To determine the mass of MnCl2\text{MnCl}_2 originally present, we analyze the stoichiometry of the two sequential chemical transformations.

Step 1: Conversion of MnCl2\text{MnCl}_2 to KMnO4\text{KMnO}_4

All the manganese originally present as MnCl2\text{MnCl}_2 is completely oxidized to KMnO4\text{KMnO}_4. Based on the conservation of manganese atoms: Moles of MnCl2=Moles of KMnO4\text{Moles of } \text{MnCl}_2 = \text{Moles of } \text{KMnO}_4

Step 2: Redox reaction between KMnO4\text{KMnO}_4 and Oxalic Acid

In an acidic medium, permanganate ions (MnO4−\text{MnO}_4^-) oxidize oxalic acid (H2C2O4\text{H}_2\text{C}_2\text{O}_4) to carbon dioxide (CO2\text{CO}_2): 2MnO4−+5H2C2O4+6H+→2Mn2++10CO2+8H2O2\text{MnO}_4^- + 5\text{H}_2\text{C}_2\text{O}_4 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}

Valency Factor (nn-factor) calculation:

  1. For KMnO4\text{KMnO}_4: Mn+7+5e−→Mn+2  ⟹  nKMnO4=5\text{Mn}^{+7} + 5e^- \rightarrow \text{Mn}^{+2} \implies n_{\text{KMnO}_4} = 5

  2. For Oxalic Acid (H2C2O4\text{H}_2\text{C}_2\text{O}_4): C2+3→2C+4+2e−  ⟹  nH2C2O4=2\text{C}_2^{+3} \rightarrow 2\text{C}^{+4} + 2e^- \implies n_{\text{H}_2\text{C}_2\text{O}_4} = 2

Step 3: Molar Calculations

  • Molar Mass of Oxalic Acid (H2C2O4\text{H}_2\text{C}_2\text{O}_4): MH2C2O4=2(1)+2(12)+4(16)=90 g mol−1M_{\text{H}_2\text{C}_2\text{O}_4} = 2(1) + 2(12) + 4(16) = 90\text{ g mol}^{-1}

  • Moles of Oxalic Acid added: Moles of H2C2O4=225 mg90 g mol−1=2.5 mmol\text{Moles of } \text{H}_2\text{C}_2\text{O}_4 = \frac{225\text{ mg}}{90\text{ g mol}^{-1}} = 2.5\text{ mmol}

  • Equivalents Law: Equivalents of KMnO4=Equivalents of H2C2O4\text{Equivalents of } \text{KMnO}_4 = \text{Equivalents of } \text{H}_2\text{C}_2\text{O}_4 nKMnO4×(Moles of KMnO4)=nH2C2O4×(Moles of H2C2O4)n_{\text{KMnO}_4} \times (\text{Moles of } \text{KMnO}_4) = n_{\text{H}_2\text{C}_2\text{O}_4} \times (\text{Moles of } \text{H}_2\text{C}_2\text{O}_4) 5×(Moles of KMnO4)=2×2.5 mmol5 \times (\text{Moles of } \text{KMnO}_4) = 2 \times 2.5\text{ mmol} Moles of KMnO4=5.05=1.0 mmol\text{Moles of } \text{KMnO}_4 = \frac{5.0}{5} = 1.0\text{ mmol}

Step 4: Mass of MnCl2\text{MnCl}_2

Since Moles of MnCl2=Moles of KMnO4=1.0 mmol\text{Moles of } \text{MnCl}_2 = \text{Moles of } \text{KMnO}_4 = 1.0\text{ mmol}:

  • Molar Mass of MnCl2\text{MnCl}_2: MMnCl2=55+2×35.5=126 g mol−1M_{\text{MnCl}_2} = 55 + 2 \times 35.5 = 126\text{ g mol}^{-1}

  • Mass of MnCl2\text{MnCl}_2: Mass=1.0 mmol×126 g mol−1=126 mg\text{Mass} = 1.0\text{ mmol} \times 126\text{ g mol}^{-1} = 126\text{ mg}

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