Question

In the standardization of $N{a}_2{S}_2{O}_3$ using $K_{2}C{r}_2{O}_7$ by iodometry , the equivalent weight of $K_{2}C{r}_2{O}_7$ is :
(Given : Molecular weight of $K_{2}C{r}_2{O}_{7}is294g$)

Answer 1

Oxidation number of $Cr$ in $K_{2}C{r}_2{O}_7$ is $+6$
$\text{Equivalent weight}=\frac{Molecularweight}{Changeinoxidationnumber}$
In iodometry, $K_{2}C{r}_2{O}_7$ liberates $I_2$ from iodides $\left(NaIorKI\right)$. Thus, it is titrated with $N{a}_2{S}_2{O}_3$ solution
$2N{a}_2{S}_2{O}_3+I_2\to 2NaI+N{a}_2{S}_4{O}_6$
Oxidation number of $Cr$ changes from $+6\left(in{K}_{2}C{r}_2{O}_7\right)$ to $+3$ i.e., $+3$ change for each $Cr$ atom
$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$
Thus, one mole of $K_{2}C{r}_2{O}_7$ accepts $6$ mole of electrons
$\therefore ,\text{Equivalent weight}=\frac{Molecularweight}{6}=\frac{294}{6}=49$