A current of 0.75 A is passed through an acidic CuSO4 solution for 5 min. The volume of oxygen at STP liberated at Pt anode is

A current of 0.75 A is passed through an acidic CuSO4 solution for 5 min. The volume of oxygen at STP liberated at Pt anode is

  1. A

    13.05 mL 

  2. B

    26.1 mL 

  3. C

    39.15 mL

  4. D

     52.2 mL

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    Solution:

    Quantity of electricity passed , Q=It=(0.75×5×60)C

    Amount o electricity passed n=QF=0.75×5×6096500mol

    The reaction at anode is 2H2O(l)4H+(aq)+O2(g)+4e

    Thus, 4 mol of electrons causes the release of 1 mol of O2 (= 22.414 Lat STP). Hence, for the amount n of electricity, volume of O2 released will be

    V=140.75×5×6096500mol22.414 L mol1=0.01306 L=13.06 mL

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