MCQ
When $Cl_2$ is passed through cold dil. $NaOH$ the products are
- ✓$NaCl, \,NaOCl$
- B$NaCl, \,NaClO_2$
- C$ NaCl,\, NaClO_3$
- D$ NaCl,\, NaClO_4$
✓
Answer
Correct option: A.
$NaCl, \,NaOCl$
a
$I _{2( s )}+ H _2 O _{2( aq )}+2 OH _{\text {(aq) }}^{-} \rightarrow 2 I _{\text {(aq) }}^{-}+2 H _2 O _{\text {(l) }}+ O _{2( g )}$
$I _{2( s )}+ H _2 O _{2( aq )}+2 OH _{\text {(aq) }}^{-} \rightarrow 2 I _{\text {(aq) }}^{-}+2 H _2 O _{\text {(l) }}+ O _{2( g )}$
Therefore, the product of the reaction of $I _2 I _2$ with $H _2 O _2. H _2 O _2$ in the basic medium is Iodine.
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