Question

When an aIkaline solution of $K_{2}Cr{O}_4$ is treated with 3% $H_2{O}_2$ solution, red brown paramagnetic peroxochromate is obtained as per following equation:
$2K_{2}Cr{O}_4+7H_2{O}_2+2KOH\to 2K_{3}Cr{O}_8+8H_{2}O$
The equivalent weight of $K_{2}Cr{O}_4$, for above transformation must be:

(Assuming $M$ is the molar mass of $K_{2}Cr{O}_4$)

• A. $\frac{M}{16}$
• B. $M$
• C. $\frac{M}{12}$
• D. $\frac{M}{2}$

Answer 1

Answer: (B)

$M$

$\stackrel{+6}{K_{2}C{r}_2{O}_7}⟶\stackrel{+5}{K_{3}Cr{O}_8}$

$[2+2x-14=0$ $[3+x-8\times 1=0$

$x=+6]$ $x=+5]$

Since $K_{3}Cr{O}_8$ is a peroxide, the oxidation number of $O$ is $-1$ in $K_{3}Cr{O}_8$ .

Hence, a net change of electrons is $1$

Therefore equivalent weight of $K_{2}Cr{O}_4=M/1=M$