The n factor of KI in the following reaction is- CuSO 4 + KI → Cu 2 I 2 + K 2 SO 4 + I 2

(1) 4 I − → I 2 2 − + I 2 + 2 e − ⇒ n factor of kI = 4 2 = 2 1

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Chemistry - Redox Reactions Question with Solution | TestHub

ChemistryRedox ReactionsEquivalent Weight and n-factorMedium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

In a balance redox reaction, net gain of electron(s) is equal to net loss of electrons(s). n_factor is a reaction specific parameter and for inter molecular redox reaction n_factor of oxidising or reducing agent is the no. of moles of electron gained / loss by one mole of compound, respectively.

The $n_{factor}$ of KI in the following reaction is- $CuSO_{4} + KI \to Cu_{2} I_{2} + K_{2} SO_{4} + I_{2}$

Options:

Answer:

Solution:

(1) $4 I^{-} \to I_{2}^{2 -} + I_{2} + 2 e^{-}$

$\Rightarrow n_{factor}$ of $kI = \frac{2}{4} = \frac{1}{2}$

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

⏱ 2mℹ️ Source: QB

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