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Redox Reactions

In the standa...

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 $K_{2} C r_{2} O_{7} i s 294 g$ )

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Solution

Oxidation number of $C r$ in $K_{2} C r_{2} O_{7}$ is $+ 6$
$Equivalent weight = \frac{M o l e c u l a r w e i g h t}{C h a n g e i n o x i d a t i o n n u m b e r}$
In iodometry, $K_{2} C r_{2} O_{7}$ liberates $I_{2}$ from iodides $(N a I o r K I)$ . Thus, it is titrated with $N a_{2} S_{2} O_{3}$ solution
$2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}$
Oxidation number of $C r$ changes from $+ 6 (i n K_{2} C r_{2} O_{7})$ to $+ 3$ i.e., $+ 3$ change for each $C r$ atom
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$
Thus, one mole of $K_{2} C r_{2} O_{7}$ accepts $6$ mole of electrons
$∴, Equivalent weight = \frac{M o l e c u l a r w e i g h t}{6} = \frac{294}{6} = 49$

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