Asked by Chris
If a current of 5.0 Amps is passed through the electrolytic cell for .5 hour, how should you calculate the number of grams of bromine produced?
Answers
Answered by
DrBob222
5.0 amps x 0.5 hr x 60 min/hr x 60 min/s = 9,000 coulombs.
2Br^- ==> Br2 + 2e
You know that 96,485 coulombs will form 1 equivalent of Br2 which is 159.8/2 = 79.9 g Br2.
You have 9,000 coulombs; therefore, how much Br2 will be formed?
79.9 x (9,000/96,485) = ?
2Br^- ==> Br2 + 2e
You know that 96,485 coulombs will form 1 equivalent of Br2 which is 159.8/2 = 79.9 g Br2.
You have 9,000 coulombs; therefore, how much Br2 will be formed?
79.9 x (9,000/96,485) = ?
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