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Q.

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

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a

13.05 mL 

b

26.1 mL 

c

39.15 mL

d

 52.2 mL

answer is A.

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Detailed 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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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