A current of 96.5 A is passed for 18 minutes through 500 mL solution of 2 M CuSO4 using Pt electrodes. The molarity of Cu2+ ion after electrolysis would be (Consider volume constant)
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Answer:
Q =i t
Charge =96.5 x18 x60 = 104220 C
2 x96500 C charge deposite 1 mole of Zn+2
104220 C charge Deposite fu of much mole of Zn2+?
Explanation:
104220/2x96500 = 0.54 moles
Intial moles = MV =2 x500 Milli moles
= 1 mole
No of moles in solution = 1 -0.54 =046 moles
Concentration equal 0.46moles present in 500ml
M =n/v X 1000
M =0.46/500 x1000
= 0.46x2= 0.92 M
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