3.31 Three electrolytic cells A, B, C containing solutions of ZnSO4 , AgNO3 and CuSO4 , respectively are connected in series. A steady current of 1.5 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? 

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    Vishal Baghel | Contributor-Level 10

    5 months ago

    Equivalent weight is Ag, EAg = 180/1  = 180

    Equivalent weight is Cu, ECu = 63.5 / 2 = 31.75

    Equivalent weight is Zn, EZn= 65/2 = 32.5

    Using Faraday’s second law of electrolysis, to find the mass of Cu and Zn, we use Equation 1,

    ∴ WZn = 0.436 g

    To find the time of current flow, using Faraday’s first law of electrolysis we get,

    M = Z ×I ×t ⇒ Equation 2

    ? Z = Equivalent Weight / 96487, Equation 2 becomes,

    M = 108 / 96487 X 1.5 X t

    t = 1.45 X 96487 / 108X 1.5

    t = 864 seconds.

    The time of current flow, t = 864 seconds, the mass of Cu is 0.426 g and mass of Zn is 0.436 g

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