Question

Acidified solution of $K_{2}C{r}_2{O}_7$ on treatment with $H_2{O}_2$ yields:

• A. $Cr{O}_3+H_{2}O+O_2$
• B. $C{r}_2{O}_2+H_{2}O+O_2$
• C. $Cr{O}_5+H_{2}O+K_{2}S{O}_4$
• D. $H_{2}C{r}_2{O}_7+H_{2}O+O_2$

Answer 1

Answer: (A)

$Cr{O}_3+H_{2}O+O_2$

Acidified solution of K2Cr2O7{$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$ K_{2}Cr_{2}O_{7}$$K_{2}C{r}_2{O}_7$ on treatment with $H_2{O}_2$H2O2 yields chromium trioxide, water and oxygen.

The equation is as follows :

$K_{2}C{r}_2{O}_7+H_2{O}_2\to Cr{O}_3+H_{2}O+O_2$

Hence, the correct option is $\text{A}$