Asked by Luke
If a current of 3.00 A is applied to a cell containing a 400g copper anode, determine the final mass of the copper anode after the cell runs for 48.0h
Answers
Answered by
DrBob222
coulombs = amperes x seconds
C = 3.00 x 48hr x (60 min/hr) x (60 s/min) = approx 518,000 C but you need to do that more accurately.
We know that 96,485 coulombs will use 1 equivalent of Cu (1 eq of Cu is 63.54/2 = approx 32 g). So how much Cu is used? That's approx
(63.54/2) x (518,000/96,485) = Cu used. Then Cu left is approx 400g-g Cu used = ?final mass Cu.
Remember to redo the estimates.
The only thing I've not disclosed is how do you know the Cu is being subtracted and not added to the 400 g Cu bar. If it's the anode it must be dissolving and that means subtracting.
C = 3.00 x 48hr x (60 min/hr) x (60 s/min) = approx 518,000 C but you need to do that more accurately.
We know that 96,485 coulombs will use 1 equivalent of Cu (1 eq of Cu is 63.54/2 = approx 32 g). So how much Cu is used? That's approx
(63.54/2) x (518,000/96,485) = Cu used. Then Cu left is approx 400g-g Cu used = ?final mass Cu.
Remember to redo the estimates.
The only thing I've not disclosed is how do you know the Cu is being subtracted and not added to the 400 g Cu bar. If it's the anode it must be dissolving and that means subtracting.
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