Question

For given reaction:

${Cr}_2{O}_7^{2-}\underset{}{\overset{H^{+}}{\to }}{Cr}^3$

$Eq.wtof{Cr}_2{O}_7^{2-}$ is:

• A. mol. wt.$/6$
• B. mol. wt.$/3$
• C. mol. wt.$/4$
• D. mol. wt.$/1$

Answer 1

The correct option is A mol. wt.$/6$

Here, the oxidation state of chromium is changed from +6 to +3

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

There is an exchange of 6 electrons.

Hence the equivalent weight of $C{r}_2{O}_7^{2-}$ is $\frac{molecularweight}{6}$.