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Three electrolytic cells A, B, C containing solutions of ZnSO_4, AgNO_3 and CuSO_4, respectively are connected in series. A steady current of 1.50 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited ? [At. weights of Cu = 63.5, Zn = 65.3, Ag = 108] |
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Answer» Solution :`Ag^(+) + e^(-) to Ag` 1 mole or 108 G of Ag are deposited by 96500 C of electricity `therefore 1.45`g of Ag will be deposited by `(96500)/108 xx 1.45C = 1295.6 `C Time of FLOW of current is OBTAINED as under: `Q = l xx t` or `t = Q//l = 1295.6//1.50 = 863.7 s = 14` MIN 24 s Masses of Cu and Zn deposited can be obtained as under: `Cu^(2+) + 2e^(-) to Cu` `2 xx 96500 C ` deposit = 63.5 g Cu `therefore 1295.6` C will deposit `=63.5/(2 xx 96500) xx 1295.6 = 0.438 g`. |
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