How does $H_{2} O_{2}$ reacts with $K_{2} C r_{2} O_{7}$ in acidic medium in ether cold conditions?

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Solution

When $H_{2} O_{2}$ is treated with acidified $K_{2} C r_{2} O_{7}$ in ether cold conditions, then blue colored chromic acid is formed.
$4 H_{2} O_{2} + K_{2} C r_{2} O_{7} + H_{2} S O_{4} \to 2 C r O_{5} + K_{2} S O_{4} + 5 H_{2} O$

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How does H2O2 reacts with K2Cr2O7 in acidic medium in ether cold conditions?

When H2O2 is treated with acidified K2Cr2O7 in ether cold conditions, then blue colored chromic acid is formed.4H2O2+K2Cr2O7+H2SO4→2CrO5+K2SO4+5H2O

How does $H_{2} O_{2}$ reacts with $K_{2} C r_{2} O_{7}$ in acidic medium in ether cold conditions?

When $H_{2} O_{2}$ is treated with acidified $K_{2} C r_{2} O_{7}$ in ether cold conditions, then blue colored chromic acid is formed.
$4 H_{2} O_{2} + K_{2} C r_{2} O_{7} + H_{2} S O_{4} \to 2 C r O_{5} + K_{2} S O_{4} + 5 H_{2} O$