Question

In the standardisation of Na$_2S_2O_3$ using K$_2Cr_2O_7$ by
iodometry, the equivalent weight of K$_2Cr_2O_7$ is

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

Answer 1

Answer: (B)

The following reaction occur between $S_2O_3^{2-} \,$ and $ Cr_2O_7^{2-}:$
$26H^+ +3S_2O_3^{2-}+4Cr_2O_7^{2-}\rightarrow 6SO_4^{2-}+8Cr^{3+}+13H_2O$
Change in oxidation number of $Cr_2O_7^{2-}$ per formula unit is $6$ (it is always fixed for $Cr_2O_7^{2-}$
Hence equivalent weight of $K_2Cr_2O_7 =\frac{\text{Molecular weight}}{6}$