JEE Main 2022 · 27 July · morning · Numerical answer · +4 / −0
In the titration of KMnO₄ and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is ___________.
Answer and solution
Answer 1
Solution
Chemistry ma’am · 4 steps
Oxalic acid carbon goes from +3 to +4 in CO₂. Count the change per carbon, not per molecule.
- The reaction. In acidic medium, MnO₄⁻ oxidises oxalic acid to carbon dioxide and is itself reduced to Mn²⁺.
2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
- Carbon in oxalic acid. In H₂C₂O₄, let each carbon be x. Then 2(+1) + 2x + 4(-2) = 0, so 2x = 6 and x = +3.
- Carbon in CO2. In CO₂, x + 2(-2) = 0, so x = +4.
- So. Each carbon atom goes from +3 to +4, so the change in oxidation number of carbon is 1.
The common slip: 2. Counting both carbons of one oxalic acid molecule gives a total change of 2, but the question asks for the change in the oxidation number of carbon, which is 1 per atom.
The idea to keep. Oxidation number belongs to one atom. When a molecule has two identical atoms, find the change for one atom, then multiply only if the question asks for the whole molecule.
Answer from the NTA final answer key. Solution by PrepWiser.