K_2Cr_2O_7 + H_2O_2 xrightarrow{H^{+}} | vedclass

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(B) Step $I$: $K_2Cr_2O_7 + H_2SO_4 \rightarrow K_2SO_4 + H_2Cr_2O_7$Step $II$: $H_2O_2 \rightarrow H_2O + [O]$Step $III$: $H_2Cr_2O_7 + 4H_2O_2 \righ

Vedclass · Accepted Answer

$CrO_5$

Vedclass · Answer

(B) Step $I$: $K_2Cr_2O_7 + H_2SO_4 \rightarrow K_2SO_4 + H_2Cr_2O_7$Step $II$: $H_2O_2 \rightarrow H_2O + [O]$Step $III$: $H_2Cr_2O_7 + 4H_2O_2 \rightarrow 2CrO_5 + 5H_2O$Overall reaction: $K_2Cr_2O_7 + H_2SO_4 + 4H_2O_2 \rightarrow 2CrO_5 + K_2SO_4 + 5H_2O$The formation of $CrO_5$ (Chromium pentoxide) results in a deep blue color. $CrO_5$ has a butterfly structure with two peroxide linkages.

$K_2Cr_2O_7 + H_2O_2 \xrightarrow{H^{+}} $

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