Chemistry - Redox Reactions Question with Solution | TestHub

Equivalent volume 5.6 L at STP means one mole of gas corresponds to n = 22.4/5.6 = 4 g-eq. (a) CH₄ combustion: C goes -4 $\to$ +4, n = 8, so 2.8 L. (b) O₃ as oxidant (O₃ $\to$ O₂ + O²⁻): accepts 2 e⁻ per molecule, n = 2, so 11.2 L. (c) H₂S as reductant (S²⁻ $\to$ S): n = 2, so 11.2 L. (d) C $\to$ CO₂: C goes 0 $\to$ +4, n = 4, so 22.4/4 = 5.6 L. Only (d).