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InterviewSolution
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`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` |
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Answer» 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` |
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