Asked by Wesley
Calculate the volume of gas liberated at the anode at STP during the electrolysis of a CuSO4 solution by a current of 1 A passed for 16 minutes and 5 seconds.
Answers
Answered by
DrBob222
Cu is deposited at the negative electrode (the cathode) and oxygen is liberated at the positive electrode (anode in an electrolysis cell) by the equation 2H2O(l) ===> 4H+(aq) + O2(g) + 4e–
coulombs = amperes x seconds
C = 1 A x 16 min x (60 sec/min) + 5 sec = 965
96,485 C will liberate (22.4/4 = 5.6 ) L O2 gas @ STP; therefore, 965 C will liberate.
5.6 L x (965/96,485) = ? L
NOTE: I suspect the author of the problem used 96,500 instead of the more exact 96,485 so the final calculation comes out easier as
5.6 L x (965/96,500) = ? L
coulombs = amperes x seconds
C = 1 A x 16 min x (60 sec/min) + 5 sec = 965
96,485 C will liberate (22.4/4 = 5.6 ) L O2 gas @ STP; therefore, 965 C will liberate.
5.6 L x (965/96,485) = ? L
NOTE: I suspect the author of the problem used 96,500 instead of the more exact 96,485 so the final calculation comes out easier as
5.6 L x (965/96,500) = ? L
Answered by
Anonymous
Bsdk answer to likh deya kro site g9 mrane ki leye banaye hai
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.