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$H_2O_2$ reacts with acidified $K_2Cr_2O_7$ the product formed is

$\begin{array}{1 1}(a)\;K_2CrO_4\\(b)\;CrO_3\\(c)\;Cr(OH)_3\\(d)\;CrO_5\end{array}$

Can you answer this question?

$CrO_4^{-2}+2H^++2H_2O_2\rightarrow CrO_5+3H_2O$
But $CrO_5$ is unstable in acidic medium.
$4CrO_5+12H^+\rightarrow 4Cr^{+3}+6H_2O+7O_2$
Hence (d) is the correct answer.
answered Jan 24, 2014