Copper sulphate solution(250) ml was electrolysed using a platinum anode and a copper cathode. A constant current of 2mA was passed for 16 minutes. It was found that after electrolysis the absorbance of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate in the solution to begin with.
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Soln:-
Given I=2×10^-3A, t=(16×60)=960 sec
Charge passed=I×t=(2×10^-3×960)C=1.92 coulomb=19.2/96500 faraday
The g-equivalent of CuSO4 decomposed=1.92/96500 g-eq. CuSO4
It is equal to half of the initial amount of CuSO4 in soln.
CuSO4 originally present in 250ml=1.92×2/96500 g-eq = 1.92/96500mole
Concentration of CuSO4 solution=1.92×1000/96500×250M =7.96×10^-5M
Hope it helps
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Given I=2×10^-3A, t=(16×60)=960 sec
Charge passed=I×t=(2×10^-3×960)C=1.92 coulomb=19.2/96500 faraday
The g-equivalent of CuSO4 decomposed=1.92/96500 g-eq. CuSO4
It is equal to half of the initial amount of CuSO4 in soln.
CuSO4 originally present in 250ml=1.92×2/96500 g-eq = 1.92/96500mole
Concentration of CuSO4 solution=1.92×1000/96500×250M =7.96×10^-5M
Hope it helps
plzz mark
Anonymous:
nice answer bhai
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