HCHO disproportionates to HCO O − and C H 3 OH in the presence of O H − (Cannizzaro's reaction). 2HCHO + O H − → HCO O − + C H 3 OH . If the equivalent weight of HCHO is E, then the value of E/6 is

One C goes 0 → +2 (loses 2e) and the other 0 → -2 (gains 2e): 2 e − transferred per 2 mol HCHO , so n-factor = 1. E = M/1 = 30/1 = 30. E/6 = 5. Ans: 5.

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Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsEquivalent Weight and n-factorEasy2 minQB

ChemistryEasynumerical

$HCHO$ disproportionates to $HCO O^{-}$ and $C H_{3} OH$ in the presence of $O H^{-}$ (Cannizzaro's reaction). 2HCHO + $O H^{-}$ $\to$ $HCO O^{-}$ + $C H_{3} OH$ . If the equivalent weight of $HCHO$ is E, then the value of E/6 is

Answer:

5.00

Solution:

One C goes 0 $\to$ +2 (loses 2e) and the other 0 $\to$ -2 (gains 2e): $2 e^{-}$ transferred per 2 mol $HCHO$ , so n-factor = 1. E = M/1 = 30/1 = 30. E/6 = 5. Ans: 5.

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

⏱ 2mℹ️ Source: QB

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