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Three electrolytic cells A,B and C containing electrolytes ZnSO_(4),AgNO_(3) and CuSO_(4) respectively were connected in series. A steady current of 1.50 ampere was passed through them until 1.45g of Ag were deposited at the cathode of cell B. how long did the current flow? what mass of copper and zinc were deposited? (At. wts. of Cu=63.5,Zn=65.3,Ag=108) |
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Answer» Solution :`Ag^(+)+e^(-)TOAG,` i.e., 108 g of Ag are deposited by 1F=`96500C` `therefore1.45g` of Ag will be deposited by `(96500)/(108)xx1.45C=1295.6C` `Q=Ixxt` or `t=Q//I=1295.6//1.50=863.7s=14`min, 24sec. `Cu^(2+)+2E^(-)TOCU` i.e., `2xx96500C` deposit Cu=63.5g `therefore1295.6C` will deposit `Cu=(63.5)/(2xx96500)xx1295.6=0.426g` `Zn^(2+)+2e^(-)TOZN. therefore` Zn deposited`=(65.3)/(2xx96500)xx1295.6=0.438g`, |
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