Question

In the following unbalanced redox reaction,
$Cu_{3}P+Cr_{2}O^{2-}_{7} \to Cu^{2+}+H_{3}PO_{4}+Cr^{3+}$
Equivalent weight of $H_{3}PO_{4}$ is

• A. $\frac{M}{3}$
• B. $\frac{M}{6}$
• C. $\frac{M}{7}$
• D. $\frac{M}{8}$

Answer 1

Answer: (D)

Change in oxidation number = $8$ units of phosphorus
$\therefore $ Equivalent weight of $H_{3}PO_{4}=\frac{M}{8}$