MCQ
$K_2Cr_2O_7 + H_2O_2$ $\xrightarrow{{{H^ + }}}$
- A$K_2CrO_4$
- ✓$CrO_5$
- C$Cr_2(SO_4)_3$
- DNone
✓
Answer
Correct option: B.
$CrO_5$
b
Step $(i)$ $K _2 Cr _2 O _7+ H _2 SO _4 \rightarrow K _2 SO _4+ H _2 Cr _2 O _7$
Step $(i)$ $K _2 Cr _2 O _7+ H _2 SO _4 \rightarrow K _2 SO _4+ H _2 Cr _2 O _7$
Step $(ii)$ $\left[ H _2 O _2 \rightarrow H _2 O +( O )\right] 4$
Step $(iii)$ $H _2 Cr _2 O _7+4( O ) \rightarrow 2 CrO _5+ H _2 O$
Hence, $K _2 Cr _2 O _7+ H _2 SO _4+4 H _2 O _2 \rightarrow 2 CrO _5+ K _2 SO _4+5 H _2 O$
The formation of Chromium Pentaoxide leads to the formation of blue colour from orange (as potassium dichromate is orange in colour). Chromium Pentaoxide is blue, so we get a blue colour after the reaction.
In the reaction, Hydrogen peroxide acts as an oxidizing agent, because it gets reduced, and its oxidation number changes to $-2$ from $-1$
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