Question

In acidic medium, the equivalent weight of $K _{2} Cr _{2} O _{7}$ (Mol. wt. $=M$ ) is

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

Answer 1

Answer: (B)

Key Idea Equivalent weight

$=\frac{\text { Molecular mass }}{\text { Change in oxidation number per atom }}$

Potassium dichromate, $K _{2} Cr _{2} O _{7}$ is a powerful oxidising agent. In acidic medium, its action can be represented as

$K _{2} \overset{+6}{Cr _{2}} O _{7}+4 H _{2} SO _{4} \longrightarrow K _{2} SO _{4}+ \overset{+3}{Cr _{2}}\left( SO _{4}\right)_{3}+4 H _{2} O +3[ O ]$

or $Cr _{2} O _{7}^{2-}+14 H ^{+}+6 e^{-} \longrightarrow 2 Cr ^{3+}+7 H _{2} O$

Eq. wt. of $K _{2} Cr _{2} O _{7}$

$=\frac{\text { molecular weight of } K _{2} Cr _{2} O _{7}}{2 \times \text { change in oxidation number }}$

$=\frac{M}{2 \times 3}\,\,[\because$ Two Cr atoms are involved. $]$

$\therefore $ Eq. wt. of $K _{2} Cr _{2} O _{7}=\frac{M}{6}$