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An electric current is passed through three cells in series containing respectively solutions of copper sulphate, silver nitrate and potassium iodide. What weights of silver and iodine will be liberated while 1.25 g of copper is being deposited ? |
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Answer» Solution :`("Wt. of COPPER")/("Wt. of IODINE")=("Eqvt. Wt. copper")/("Eqvt. Wt. of Iodine")` `"or"(1.25)/(x)=(31.7)/(127),""x=(1.25xx127)/(31.7)` Hence, Wt. of Iodine x = 5.0 g of iodine `"Also,"("Wt. of Copper")/("Wt. of Silver")=(1.25)/(y)` `=("Eqvt. wt. of Cu(= 31.7)")/("Eqvt. wt. of Silver( = 108)")` `"Wt. of silver (y) "=(108xx1.25)/(31.7)` = 4.26g |
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