Al3+ is reduced to Al(s) at an electrode surface. If a current of 25 amperes is maintained for 8 hours, what mass of aluminum is deposited on the electrode? Assume 100% efficiency.
I know that the correct answer is 67 grams, however I don't understand why. I am working on my own calculation but I'm getting 8.39g
2007-04-04 05:12:09 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
The reduction can be represented thus:
Al3+(aq) + 3e- ----------> Al (s)
this means that 3 Faradays (ie. 96 500 x 3 coulombs) are requied to deposit 1 mole i.e. 27 g of Al.
therefore 720 000 coulombs (25 A x 8 x 60 x 60) will deposit
27 g x 720 000 couloms/96 500 x 3 coulombs = 67.15 g
Remember the charge is given by the product of current and time:
Quantity of electricity = current (amps) x time (sec)
2007-04-04 05:30:17 · answer #1 · answered by brisko389 3 · 1⤊ 0⤋
1 amp = 1 coulomb/sec
Let faraday be called F
At. wt. Al=27 Eq. Wt. Al=9
8hr x 25coulomb/sec x 60min/hr x 60sec/min x 1F/96,500coulomb x 9gAl/1F = 67gAl
2007-04-04 12:32:42 · answer #2 · answered by steve_geo1 7 · 1⤊ 0⤋