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!!!
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)
Make that 6SO₄⁻² a 3SO₄⁻²
Thank you very much :)
To identify what has been oxidized and reduced in a reaction, it is necessary to assign oxidation numbers to each element in the reactants and products. The oxidation number is a way to keep track of the electrons gained or lost by an atom during a chemical reaction.
In the given equation: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
To determine the oxidation numbers, we need to know the rules:
1. The oxidation number of any uncombined element is zero.
2. The oxidation number of a monatomic ion is equal to its charge.
3. In most compounds, oxygen has an oxidation number of -2.
4. Hydrogen usually has an oxidation number of +1 when combined with non-metals.
5. The sum of oxidation numbers in a compound or polyatomic ion is equal to its charge.
Let's assign the oxidation numbers:
For 2Al: Since it is an uncombined element, its oxidation number is zero.
For 3H2SO4:
- Oxygen is usually -2, and there are four oxygen atoms, so the total oxidation number from oxygen is -8.
- Hydrogen usually has an oxidation number of +1 when combined with non-metals, and there are six hydrogen atoms, so the total oxidation number from hydrogen is +6.
- Sulfur is an unknown oxidation number.
Adding the oxidation numbers from oxygen and hydrogen gives us a total of -8 + 6 = -2. For the compound H2SO4 to be neutral (since it's not an ion), the total must be zero. So the oxidation number of sulfur must be +6.
For Al2(SO4)3:
- The oxidation numbers for oxygen and hydrogen remain the same as before: oxygen is -2, hydrogen is +1.
- The overall charge of Al2(SO4)3 is zero since it's not an ion.
- Since there are three SO4 groups, the total oxidation number from sulfur is (3 x -2) = -6.
- To make the compound neutral, the two aluminum atoms must collectively have an oxidation number of +6.
Therefore, the aluminum atoms are oxidized from an oxidation state of zero in Al to +6 in Al2(SO4)3, and the sulfur atoms in SO4 are reduced from +6 to -2 in Al2(SO4)3.
I hope this explanation helps you understand how to balance the oxidation numbers in the given equation.