Question

The equivalent weight of $K_{2}C{r}_2{O}_7$ in acidic medium is expressed in terms of its molecular weight $\left(M\right)$ as :

• A. $M/3$
• B. $M/4$
• C. $M/6$
• D. $M/7$

Answer 1

The correct option is C $M/6$

Potassium dichromate in acidic medium is a strong oxidizer. It means it gains electrons during redox reaction. Potassium dichromate in acidic solution results in

$K_{2}C{r}_2{O}_7+14H^{+}+6e^{-}\to 2K^{+}+2C{r}^{3+}+7H_{2}O$

There is overall gain of 6 electrons. Therefore the equivalent weight is $=\frac{M}{6}$.