Question

19 g fused $SnC{l}_2$ was electrolysed using inert electrodes. 0.119g $Sn$ was deposited at cathode. If nothing was given out during electrolysis, calculate the ratio of mass of $SnC{l}_2$ and $SnC{l}_4$ in fused state after electrolysis.(Round off to nearest integer)

Answer 1

The Redox changes during electrolysis of $SnC{l}_2$ are:
Anode: $2C{l}^{-}\to C{l}_2+2e$
Cathode:$S{n}^{2+}+2e\to Sn$
Also $C{l}_2$ formed at anode reacts with $SnC{l}_2$ to give $SnC{l}_4$
$SnC{l}_2+2e\to SnC{l}_4$
Now Eq. of $SnC{l}_2$ lost during elecrolysis = Eq. of $C{l}_2$ formed during electrolysis=Eq. of Sn formed during electrolysis
$=\frac{0.119}{0.119/2}=2\times {10}^{-3}$
$\therefore$ Eq. of $SnC{l}_4$ formed = $2\times {10}^{-3}$
Total loss in eq. of $SnC{l}_2$ during complete course=Eq. of $SnC{l}_2$ lost during electorlysis + Eq. of $SnC{l}_2$ lost during reaction with $C{l}_2$
= $2\times {10}^{-3}$+$2\times {10}^{-3}$=$4\times {10}^{-3}$
Initial eq. of $SnC{l}_2$ =\frac { 19 }{ 190/2 } =2\times { 10 }^{ -1 }
$\therefore$ Eq.of $SnC{l}_2$ left in solution
=$2\times {10}^{-1}-4\times {10}^{-3}$ =0.196.
Also Eq. of $SnC{l}_4$ formed= $2\times {10}^{-3}$=0.002
$\therefore$ $=\frac{Massof{SnCl}_{2}left}{Massof{SnCl}_{4}formed}=\frac{0.196\times \left(190/2\right)}{0.002\times \left(261/2\right)}=71.34$
Hence, the ratio of mass of $SnC{l}_2$ and $SnC{l}_4$ in fused state after electrolysis is 71.34.