Question

In acidic medium, equivalent weight of $K_2{Cr}_2{O}_7(Mol.wt.=M)$ is:

• A. $\frac{M}{3}$
• B. $\frac{M}{4}$
• C. $\frac{M}{6}$
• D. $\frac{M}{2}$

Answer 1

The correct option is C $\frac{M}{6}$

The oxidation state of chromium in $K_2{Cr}_2{O}_7$ is +6.

Balanced chemical reaction of $K_2{Cr}_2{O}_7$ in acidic medium will be-

$K_2{Cr}_2{O}_7+14H^{+}+6e^{-}⟶2K^{+}+2{Cr}^{3+}+7H_{2}O$

In the above reaction, oxidation state chromate ion is changing from +6 to +3, i.e., the transfer of 6 electrons is taking place.

$\therefore$ Equivalent weight = $\frac{M}{6}$