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
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.
2 answers
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