Question

For the given redox reaction
$K_{2}C{r}_2{O}_7+14HCl\to 2KCl+2CrC{l}_3+3C{l}_2+7H_{2}O$
What is the equivalent mass of $HCl$? [Given; $M_{HCl}=36.6{\text{g mol}}^{-1}$]

Answer 1

$K_{2}C{r}_2{O}_7+14HCl\to 2KCl+2CrC{l}_3+3C{l}_2+7H_{2}O\text{n - factor of}{K}_{2}C{r}_2{O}_7=6$
Again the oxidation state of $Cl$ is getting changed from $-1$ to $0$
$\therefore \text{n - factor of HCl =}\frac{6}{14}=\frac{3}{7}\therefore E_{HCl}=\frac{M_{HCl}}{\text{n-factor of HCl}}=\frac{7M}{3}=\frac{7\times 36.6}{3}=85.40$