Chemistry - Electrochemistry Question with Solution | TestHub

At cathode: ${\mathrm{Sn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Sn}$

At anode : $2{\mathrm{Cl}}^{-}\to {\mathrm{Cl}}_{2(\mathrm{g})}+2{\mathrm{e}}^{-}$

${\mathrm{Cl}}_2$ formed at anode reacts with left ${\mathrm{SnCl}}_2$ to form ${\mathrm{SnCl}}_4$

During electrolysis Eq. of ${\mathrm{SnCl}}_2$ lost $=$ Eq. of ${\mathrm{Cl}}_2$ formed $=$ Eq. of Sn formed

Eq. of Sn formed $=\frac{{\mathrm{W}}_{\mathrm{Sn}}}{\text{ Eq. }\mathrm{Wt}.\text{ of }\mathrm{Sn}}=\frac{1.187}{118.7/2}=2\times 10^{-2}$

Eq. of ${\mathrm{SnCl}}_2$ lost $=2\times 10^{-2}=\mathrm{Eq}$. of ${\mathrm{SnCl}}_4$ formed

Total loss in Eq. of ${\mathrm{SnCl}}_2$ during complete course $=$ Eq. of ${\mathrm{SnCl}}_2$ lost during electrolysis + Eq. of ${\mathrm{SnCl}}_2$ lost during reaction with ${\mathrm{Cl}}_2$

$=2\times 10^{-2}+2\times 10^{-2}$

$=4\times 10^{-2}$

Eq. of ${\mathrm{SnCl}}_2$ initially $=\frac{18.97}{189.7/2}=2\times 10^{-1}$

Eq. of ${\mathrm{SnCl}}_2$ left in molten solution

$\begin{array}{rlr}=2\times 10^{-1}-4\times 10^{-2}& =0.16& \end{array}$

$\frac{wt.ofSnC{l}_{2}left}{wt.ofSnC{l}_{4}formed}=\frac{0.16\times 189.7/2}{0.02\times 260.7/2}=5.82$