19 g of molten SnCl_(2) is electrolysed for same tine, using inert electrodes. 0.119 g of Sn is deposited at the cathode. No substance is lost during the electrolysis. Find the ratio of the weight of SnCl_(2) and SnCl_(2) after electrolysis (Sn = 119)

19 g of molten SnCl_(2) is electrolysed for same tine, using inert ele

1.	19 g of molten SnCl_(2) is electrolysed for same tine, using inert electrodes. 0.119 g of Sn is deposited at the cathode. No substance is lost during the electrolysis. Find the ratio of the weight of SnCl_(2) and SnCl_(2) after electrolysis (Sn = 119)
Answer» Solution :According to this problem `SnCl_(2)` should first ionise into the IONS of tin and chloride. Sn gets depositedat the cathode and as `Cl_(2)`, PRODUCED at anode, is not lost, it combines with the remaining `SnCl_(2)` to give `SnCl_(4)`. `SnCl_(2) rarr Sn + Cl_(2)` `SnCl_(2) + Cl_(2)rarr SnCl_(4)` Equivalent of Sn deposited `= (0.119)/(119//2) = 0.002 ""("eq. wt. of Sn " = (119)/(2))` `therefore` equivalent of `SnCl_(2)` decompose = 0.002. Equivalent of CHLORINE produced = 0.002. Equivalent of `SnCl_(4)` = formed = 0.002. `therefore` weight of `SnCl_(4)` formed = no. of eq. `xx` eq. wt. `= 0.002 xx (261)/(2) = 0.261` gram. `(("eq. wt. of " SnCl_(4) = (261)/(2)" , in " SnCl_(2) + Cl_(2), rarr SnCl_(4)),("+2",+4))` Further eq. of chlorine produced = initial wt. of `SnCl_(2)` - (wt. of `SnCl_(2)` decomposed+ wt. of `SnCl_(2)` combined with `Cl_(2)` to form `SnCl_(4)`) = 19 - (eq. of `SnCl_(2)` decomposed `xx` eq. wt. of `SnCl_(2)` + eq. of `SnCl_(2)` which combined with `Cl_(2) xx` eq. wt. of `SnCl_(2)`). `= 19 - [0.002 xx (190)/(2) + 0.002 xx (190)/(2)]` `= 19 - 0.38` `= 18.62g`. `therefore SnCl_(2) : SnCl_(4) = 18.62 : 0.261`. (weight ratio)