Question

In the standardization of ${Na}_2{S}_2{O}_3$ using $K_2{Cr}_2{O}_7$ by iodometry, the equivalent weight of $K_2{Cr}_2{O}_7$ is:

• A. (molecular weight)$/2$
• B. (molecular weight)$/6$
• C. (molecular weight)$/3$
• D. same as molecular weight

Answer 1

The correct option is B (molecular weight)$/6$

Chromium is being reduced by $N{a}_2{S}_2{O}_3$ from $C{r}^{6+}$ in $K_{2}C{r}_2{O}_7$ to $C{r}^{3+}$ according to the following equation:

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

Since there is an exchange of 6 electrons, the equivalent weight will be ((molecular weight)/6).