96,485 coulombs (1 Faraday) will deposit 1 equivalent weight of Al (27/3 = 9 grams).
Your problem has 4.5 g; therefore, there must have been 1/2 x 96,485 coulombs (1/2 F). That same quantity of electricity will release 1 equivalent weight of hydrogen or H. That is 1/2 of a mole (1/2 of H2) and a mole occupies 22.4L @ STP; therefore, 1/2 mole will occupy 1/2 x 22.4 = ?
Atoms and ions react as moles and not grams.
A quantity of electrical charge that brings about the deposition of 4.5 g Al from Al3+ at a cathode will also produce what volume (STP) of H2(g) from H+ at a cathode?
I know there are 3 e- difference b/w the two reactions. I took 4.5 g/3 and get 1.5 g H2. Then (1.5 g H2)(1 mol H2/2 g H2)(22.4 L/mol H2)=16.8 L. Answer in the book says 5.6 L ? What am I doing wrong? Thanks.
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