`18.97 g` fused `SnCl_(20` was electrolyzed using inert electrodes. `1.187g Sn` was deposited at cathode. If nothing is obtained during electrolysis, calculate the ration of weight of `SnCl_(2)` and `SnCl_(4)` in fused state after electrolysis.
Given`:`
Atomic weight of `Sn=118.7, Mw `of `SnCl_(2)=189.7, Mw` of `SnCl_(4)=260.7`

Explanation

Fused `SnCl_(2)overset(El ectrolysis)rarr Sn^(2+)+2Cl^(c-)`
At cathode `: Sn^(2+)+2e^(-)rarr Sn,,,(` Reduction `)`
At anode `: 2Cl^(c-)rarr Cl_(2)(g)+2e^(-),,,(` Oxidation `)`
Further , `Cl_(2)(g)` formed at anode reacts with left over `SnCl_(2)` to give `SnCl_(4)` .
`SnCl_(2) +Cl_(2) rarr SnCl_(4)`
During electrolysis`:`
Eq of `SncL_(2)` lost `=` Eq of `Cl_(2)` formed `=` Eq of `Sn` formed.
`Eq` of `Sn` formed `=(W_(Sn))/(Ew of Sn)`
`=(1.187)/(118.7//2) " "Sn^(2+)+2e^(-)rarrSn`
=`2xx10^(-2)" "n` factor of `Sn=2`
`Eq` of `SnCl_(2)` lost `=Eq` of `Cl_(2)` formed
`=Eq `of `SnCl_(4)` formed
`=Eq` of `Sn` formed
`=2xx10^(-2)`
Now total loss is equivalent of `SnCl_(2)` during complete course
`=Eq` of `SnCl_(2)` lost during electrolysis is `+ Eq` of `SnCl_(2)` lost during reaction with `Cl_(2)`
`=2xx10^(-2)+2xx10^(-2)=4xx10^(-2)`
`Eq` of `SnCl(` initially `)=(18.97g)/(189.7//2)=2xx10^(-1)`
`Eq` of `SnCl_(2)` left is molten solution
`=2xx10^(-1)-4xx10^(-2)=0.16`
`Eq` of `SnCl_(4)` formed `=2xx10^(-2)=0.02`
`("Weight of SnCl left")/("Weight of SnCl formed ")=("Weight of "SnCl_(2)"left" xx Ew of SnCl_(2))/("Weight of "SnCl_(4) "formed "xx Ew of SnCl_(4))`
`=((0.16xx189.7//2))/((0.02xx260.7//2))=5.82`