In the following reaction, what is oxidized and what is reduced? Also include th

Question

In the following reaction, what is oxidized and what is reduced? Also include the change in oxidation number for what is oxidized and what is reduced. 6H2SO4 + 2AL --> AL2(SO4)3 + SO2 + 6 H2O

Explanation / Answer

Dissolving Al in NaOH will produce either insoluble Al(OH)3 or more likely, Al(OH)4^- ions along with hydrogen gas and some leftover sodium ions. Aluminum always has a coating of Al2O3. The NaOH dissolves the coating allowing Al to react with water to produce Al3+ and H2. It is essentially the same reaction as reacting sodium metal with water. But it only occurs in basic solutions where the aluminum is no longer protected by its oxide coating. You will notice that solid Al won't react with solid NaOH. Otherwise boxes of Drano would explode, which they don't. It's only when there is water present does the Al react. That's because it is reacting with the water. Here is the balanced equation for the reaction of aluminum metal with water in the presence of sodium hydroxide: 2Al + 2NaOH + 6H2O --> 2NaAl(OH)4(aq) + 3H2 By the way, the only way to produce Na3AlO3 would be to completely dehydrate NaAl(OH)4(aq) NaAlOHOHOHOH = NaAlO HOH O HOH = NaAlO2 + 2H2O This is why "sodium aluminate" is usaually written as NaAlO2. ===================================== As to the reaction of aluminum and H2SO4... In dilute solutions of sulfuric acid we get the formation of hydrogen gas, just as we might suspect. 2Al(s) + 3H2SO4(aq) --> Al2(SO4)3(aq) + 3H2(g) But in contact with metals, *concentrated H2SO4* will oxidize the metal and produce sulfur dioxide gas. By "concentrated" we generally mean liquid H2SO4, or H2SO4 diluted with very little water. 2Al(s) + 6H2SO4(conc) --> Al2(SO4)3(aq) + 3SO2(g) + H2O(l) Source(s): 35 years of teaching chemistry.

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