Consider the following equations :
2Fe²⁺ + H₂O₂ → xA + yB
(in basic medium)
2MnO₄⁻ + 6H⁺ + 5H₂O₂ → x'C + y'D + z'E
(in acidic medium)
The sum of the stoichiometric coefficients x, y, x', y', and z' for products A, B, C, D and E, respectively, is ______.

Solution

Chemistry ma’am · 4 steps

H₂O₂ plays two roles here: oxidant with Fe²⁺, reductant with MnO₄⁻.

1. First reaction. H₂O₂ oxidises Fe²⁺ to Fe³⁺ and is itself reduced. In basic medium oxygen ends up as OH⁻. Two Fe²⁺ each lose one electron, and one H₂O₂ takes those two electrons.

   2Fe²⁺ + H₂O₂ → 2Fe³⁺ + 2OH⁻

2. Check and read off. Charge is +4 on both sides, and O and H balance. So x = 2 and y = 2.
3. Second reaction. Here MnO₄⁻ is the stronger oxidant, so H₂O₂ is oxidised to O₂. Two Mn go from +7 to +2 and gain 10 electrons. Five H₂O₂ lose 10 electrons. This gives 2MnO₄⁻ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂. It checks out: 16 H, 18 O and charge +4 on each side.
4. So. x' = 2, y' = 8, z' = 5. The sum is 2 + 2 + 2 + 8 + 5 = 19.

The idea to keep. H₂O₂ is an oxidant towards easily oxidised ions like Fe²⁺. It is a reductant giving O₂ towards strong oxidants like MnO₄⁻. Each H₂O₂ that gives O₂ loses exactly 2 electrons.

Answer from the NTA final answer key. Solution by PrepWiser.