In the equation ${\mathrm{SnCl}}_2+2{\mathrm{HgCl}}_2⟶{\mathrm{Hg}}_2{\mathrm{Cl}}_2+{\mathrm{SnCl}}_4$ The equivalent weight of stannous chloride (molecular weight= 190) will be

${\mathrm{SnCl}}_2+2{\mathrm{HgCl}}_2⟶{\mathrm{Hg}}_2{\mathrm{Cl}}_2+{\mathrm{SnCl}}_4$

$E_{{\mathrm{SnCl}}_2}=\frac{\text{ Molecular wt. }}{\begin{array}{c}\text{ No. of electron gain or loss by per }\\ \text{ molecule of oxidant or reductant }\end{array}}$

$\begin{array}{l}=\frac{190}{2}=95\\ {\mathrm{Sn}}^{2+}⟶{\mathrm{Sn}}^{4+}+2e^{-}\end{array}$