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19 g molten SnCl_(2) was electrolysed by two inert electrodes. On electrolysis, 0.119g of Sn was deposited at cathode. Find the ratio of weights of SnCl_(2) and SnCl_(4) after electrolysis, assuming, there is no loss of materíal during electrolysis. |
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Answer» Solution :`Sn^(2+)+2e rarrSn,` electricity required to produce `"0.11 g Sn "=(2xx96500)/(119)xx0.199=193C` `2SnCl_(2)rarrSn+SnCl_(4)or, 2Sn^(2+)rarr Sn+Sn^(4+)` `Sn^(2+)rarrSn^(4+)+2e" HENCE, "2xx96500C "-="1 mol "Sn^(4+)` `therefore""193C-=10^(-3)" mol "Sn^(4+)` `therefore` AMOUNT of `SnCl_(4)` PRODUCED `=10^(-3)xx(119+142)=0.261g` `"0.119 g Sn "=10^(-3)" mol Sn "-=10^(-3)" mol "SnCl_(2) and 10^(-3)" mol "SnCl_(4)-=10^(-3)" mol "SnCl_(2)` `therefore` To produce `10^(-3)` mol Sn and `10^(-3)` mol `SnCl_(2)2xx10^(-3)` mol `SnCl_(2)` is required. `therefore""SnCl_(2)` LEFT after electrolysis `=(19-2xx10^(-3)xx190)=18.62g, SnCl_(4)` produce `=10^(-3)xx(119+4xx35.5)=0.261g.` `therefore"Ratio of the AMOUNTS of "SnCl_(2)" to "SnCl_(4)=(18.62)/(0.261)=71.34` |
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