Question

Calculate equivalent weight of $K_{2}C{r}_2{O}_7$ in acidic medium.

Answer 1

In acidic medium
$\underset{\text{+6}}{K_{2}C{r}_2{O}_7}\to \underset{\text{+3}}{C{r}_2(S{O}_4{)}_3}$
Change in the oxidation number of one $Cr$ atom $=6-3=3$.
Change in the oxidation number of two $Cr$ atoms $=2\times 3=6$.
The equivalent mass of $K_{2}C{r}_2{O}_7=\frac{\text{Formula mass of}{K}_{2}C{r}_2{O}_7}{\text{Total change in the oxidation number of Cr}}=\frac{294}{6}=49$