Sulfuric acid exists in solution as H⁺ and HSO₄⁻ ions, but both react with Al metal to produce Al⁺³ ions and SO₄⁻² ions. The most accurate representation of what's going on is:
2Al(s) + 3H⁺ + 3HSO₄⁻ --> 3H₂(g) + 2Al⁺³ + 6SO₄⁻²
or as ...
2Al(s) + 3H₂SO₄(aq) --> 3H₂(g) + Al₂(SO₄)₃(aq)
Hi, the question is: in each reaction, identify what have been oxidised and reduced.
The equation is: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
I am confused about the Al2(SO4)3 I know what oxides and reductions us I just don't understant how Al2(SO4)3 balances out to O with the oxidation numbers. Please help!!!
3 answers
Make that 6SO₄⁻² a 3SO₄⁻²
Thank you very much :)
0 answers