Question

In the standardisation of $N{a}_{2}S{O}_3$ using $K_{2}C{r}_2{O}_7$ by iodometry, the equivalent weight of $K_{2}C{r}_2{O}_7$ is :

• A. $\frac{Mw}{2}$
• B. $\frac{Mw}{6}$
• C. $\frac{Mw}{3}$
• D. same as the molecular weight

Answer 1

The correct option is B $\frac{Mw}{6}$

The reduction half reaction for this is:

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

As $1Cr$ from $C{r}_2{O}_7^2-$ gains $3$ electrons to give $1C{r}^{3+}$,

So $2Cr$ from $C{r}_2{O}_7^2-$ gains $6$ electrons to give $2$ $C{r}^{3+}$.

Here oxidation state changes by $3$,

$+6$ is in $C{r}_2{O}_7^2-$ which changes to $+3$ in $C{r}^{3+}$

Since $equivalentweight=\frac{molecularweight}{totalelectronsgained}$

so $E=\frac{M_w}{6}$ is the right answer.