Chemistry - Redox Reactions Question with Solution | TestHub
ChemistryRedox ReactionsDetermination of Oxidation StateEasy2 minQB
ChemistryEasysingle choice
A certain amount of a reducing agent reduces x mole of MnO₂ and y mole of K₂CrO₄ in different reactions in acidic medium. If the changes in oxidation states of reducing agent in the reactions are in 1:2 ratio, respectively, then the ratio of x and y is
Options:
Answer:
C
Solution:
MnO₂: Mn +4 +2 takes 2 e⁻; K₂CrO₄: Cr +6 +3 takes 3 e⁻. With reductant n-factors n₁:n₂ = 1:2 and the same amount of reductant: x = Rn₁/2, y = Rn₂/3. So x/y = (3n₁)/(2n₂) = (3 × 1)/(2 × 2) = 3/4. Hence (c).
Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Determination of Oxidation State
⏱ 2mℹ️ Source: QB
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