MCQ
When acidified $K_2Cr_2O_7$ react with $H_2O_2$ a blue colour solution is obtained. This blue colour is due to the formation of
- A$Cr_2O_3$
- B$CrO_3$
- C$Cr_2(SO_4)_3$
- ✓$CrO_5$
✓
Answer
Correct option: D.
$CrO_5$
d
Step
(i) $\quad K_{2} C r_{2} O_{7}+H_{2} S O_{4} \rightarrow K_{2} S O_{4}+H_{2} C r_{2} O_{7}$
Step
(ii) $\left[H_{2} O_{2} \rightarrow H_{2} O+(O)\right] 4$
Step
(iii) $\quad H_{2} C r_{2} O_{7}+4(O) \rightarrow 2 C r O_{5}+H_{2} O$
Hence, $K_{2} C r_{2} O_{7}+H_{2} S O_{4}+4 H_{2} O_{2} \rightarrow 2 C r O_{5}+K_{2} S O_{4}+5 H_{2} O$
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
Hence, the correct option is $D$
Step
(i) $\quad K_{2} C r_{2} O_{7}+H_{2} S O_{4} \rightarrow K_{2} S O_{4}+H_{2} C r_{2} O_{7}$
Step
(ii) $\left[H_{2} O_{2} \rightarrow H_{2} O+(O)\right] 4$
Step
(iii) $\quad H_{2} C r_{2} O_{7}+4(O) \rightarrow 2 C r O_{5}+H_{2} O$
Hence, $K_{2} C r_{2} O_{7}+H_{2} S O_{4}+4 H_{2} O_{2} \rightarrow 2 C r O_{5}+K_{2} S O_{4}+5 H_{2} O$
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
Hence, the correct option is $D$
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