The value of n in the following processes: AO₄ⁿ⁻ + 2e → HAOₙ²⁻ is

Balance the half-reaction with H⁺ and H₂O: AO₄ⁿ⁻ + aH⁺ + 2e → HAOₙ²⁻ + bH₂O. Charge: -n + a - 2 = -2 ⇒ a = n. H: a = 1 + 2b. O: 4 = n + b ⇒ b = 4 - n. So n = 1 + 2(4-n) ⇒ 3n = 9 ⇒ n = 3. Ans: 3.

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Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsProblem involving Equivalent ConceptMedium2 minQB

ChemistryMediumnumerical

The value of n in the following processes: AO₄ⁿ⁻ + 2e $\to$ HAOₙ²⁻ is

Answer:

3.00

Solution:

Balance the half-reaction with H⁺ and H₂O: AO₄ⁿ⁻ + aH⁺ + 2e $\to$ HAOₙ²⁻ + bH₂O. Charge: -n + a - 2 = -2 ⇒ a = n. H: a = 1 + 2b. O: 4 = n + b ⇒ b = 4 - n. So n = 1 + 2(4-n) ⇒ 3n = 9 ⇒ n = 3. Ans: 3.

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Problem involving Equivalent Concept

⏱ 2mℹ️ Source: QB

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