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Question

A volume of 500 mL of 0.1 M CuSO4 solution is electrolysed for 5 min at a current of 0.161 A. If Cu is produced at one electrode and oxygen at the other, the approximate pH of the final solution is:

A
3.0
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B
3
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C
3.00
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Solution

Cathode: Cu2++2eCu
Anode: 2H2OO2+4H++4e
Moles of electrons used=QF=0.161×5×60965005×104
Equivalents of Cu2+ present =500×0.11000×2=0.1(excess)
Since, Cu2+ is in excess, H+ will not be consumed at cathode.
Moles of H+ produced=Moles of electrons=5×104
[H+]final=5×104500×1000=103M
pH=log[H+]
pH=3.0


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