A solution of KHC 2 O 4 ⋅ H 2 C 2 O 4 ⋅ 2 H 2 O is 0.2 N as an acid. Then it is:

KHC₂O₄·H₂C₂O₄·2H₂O has (i) three ionizable H⁺. If it is 0.2 N, then it is: = 3 0.2 M = 0.067 M as an acid. (ii) Two C₂O₄²⁻ are oxidized to CO₂: C 2 O 4 2 − → 4 C O 2 + 4 e − Thus, 0.067 M = 0.067 × 4 = 0.267 N as a reducing agent. Therefore, (a), (c), and (d) are true.

Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsEquivalent Weight and n-factorMedium2 minQB

ChemistryMediummultiple choice

A solution of $KHC_{2} O_{4} \cdot H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is 0.2 N as an acid. Then it is:

Options:(select one or more)

Answer:

A, C, D

Solution:

KHC₂O₄·H₂C₂O₄·2H₂O has

(i) three ionizable H⁺. If it is 0.2 N, then it is:

$= \frac{0.2}{3} M = 0.067 M$ as an acid.

(ii) Two C₂O₄²⁻ are oxidized to CO₂:

$C_{2} O_{4}^{2 -} \to 4 C O_{2} + 4 e^{-}$

Thus, 0.067 M = $0.067 \times 4 = 0.267$ N as a reducing agent.

Therefore, (a), (c), and (d) are true.

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Equivalent Weight and n-factor

⏱ 2mℹ️ Source: QB

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