Here is a reaction: 2Al+3H2SO4--->Al2(SO4)+3H2

determine the oxidation numbers of each element in both the rectants and the products of this oxidation reduction reaction of aluminum and sulfuric acid?

Which element is oxidized in this reaction?
Which element is rduced in this reaction?

http://www.woodrow.org/teachers/chemistry/links/chem1/Chapter15.html

To determine the oxidation numbers of each element in the reactants and products of the reaction 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2, we need to understand the rules for assigning oxidation numbers:

1. The oxidation number of an element in its elemental form is always zero.
2. In compounds, Group 1 elements (e.g., Al) have an oxidation number of +1, Group 2 elements have an oxidation number of +2, and oxygen has an oxidation number of -2, unless otherwise specified.
3. The sum of the oxidation numbers in a compound is always zero.
4. Hydrogen usually has an oxidation number of +1, except when it is bonded to metals.

Using these rules, let's determine the oxidation numbers for each element:

1. Aluminum (Al):
- In the reactants (2Al), aluminum is in its elemental form, so the oxidation number is zero.
- In the product (Al2(SO4)3), aluminum is in a compound. Since each sulfur atom in Al2(SO4)3 has an oxidation number of +6, and each oxygen atom has an oxidation number of -2, we can calculate the oxidation number for aluminum:
2x + 3(-2) = 0
2x - 6 = 0
2x = 6
x = +3
Therefore, the oxidation number of aluminum in Al2(SO4)3 is +3.

2. Sulfur (S):
- In the reactants (3H2SO4), sulfur is in a compound. Since each hydrogen atom has an oxidation number of +1 and each oxygen atom has an oxidation number of -2, we can calculate the oxidation number for sulfur:
3(+1) + 4x + 3(-2) = 0
3 + 4x - 6 = 0
4x - 3 = 0
4x = 3
x = +3/4 or +0.75 (approximately)
Therefore, the oxidation number of sulfur in H2SO4 is approximately +0.75.
- In the product (Al2(SO4)3), each sulfur atom has an oxidation number of +6, based on the compound's formula.

3. Hydrogen (H):
- In the reactants (3H2SO4), hydrogen is in a compound. Each hydrogen atom has an oxidation number of +1.
- In the product (3H2), hydrogen is again in a compound. Each hydrogen atom still has an oxidation number of +1.

Now, let's determine which element is oxidized and reduced in this reaction:

Oxidation is defined as the loss of electrons, while reduction is defined as the gain of electrons.

- In this reaction, aluminum goes from an oxidation number of 0 in the reactants to +3 in the product. Therefore, aluminum has been oxidized (its oxidation number increased), meaning it loses electrons.

- On the other hand, sulfur goes from an oxidation number of approximately +0.75 in the reactants to +6 in the product. Therefore, sulfur has been reduced (its oxidation number decreased), meaning it gains electrons.

In summary:
- Aluminum is oxidized.
- Sulfur is reduced.

For further information and examples, you can refer to the link you provided at http://www.woodrow.org/teachers/chemistry/links/chem1/Chapter15.html.

To determine the oxidation numbers of each element in a chemical reaction, you need to know the set of rules for assigning oxidation numbers. Here is a simplified version of the rules:

1. The oxidation number of an element in its uncombined state is zero.
2. The oxidation number of a monatomic ion is equal to its charge.
3. In compounds, the oxidation number of hydrogen is +1.
4. In compounds, the oxidation number of oxygen is typically -2.
5. The sum of the oxidation numbers in a neutral compound is zero.
6. The sum of the oxidation numbers in an ion is equal to the charge of the ion.
7. Fluorine always has an oxidation number of -1 in compounds.
8. Group 1 elements (alkali metals) always have an oxidation number of +1 in compounds.
9. Group 2 elements (alkaline earth metals) always have an oxidation number of +2 in compounds.
10. In binary compounds, the more electronegative element is assigned the negative oxidation number equal to the charge it would have as an anion.

Let's apply these rules to determine the oxidation numbers in the given reaction:

2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2

In aluminum (Al), the oxidation number is 0 since it is an element in its uncombined state.

In hydrogen (H2), the oxidation number is +1 based on rule 3.

In sulfur (S) in H2SO4, we need to determine its oxidation number. Since hydrogen always has an oxidation number of +1, and we have 3 hydrogens in H2SO4, the total oxidation number of hydrogen is +3. Given that the sum of the oxidation numbers in a neutral compound is zero, the oxidation number of sulfur in H2SO4 is +6.

In oxygen (O) in H2SO4, we use rule 4 and assign an oxidation number of -2. Given that there are 4 oxygens in H2SO4, the total oxidation number of oxygen is -8. Again, given that the sum of the oxidation numbers in a neutral compound is zero, we can calculate the oxidation number of sulfur as +6.

Now, let's shift our focus to the products of the reaction:

In aluminum sulfate (Al2(SO4)3), we need to determine the oxidation number of aluminum. Since the overall compound has no net charge, we know that the sum of the oxidation numbers must be zero. Given that there are two aluminum atoms, the oxidation number of aluminum must be +3.

In sulfate (SO4), we already know the oxidation number of sulfur from before, which is +6. Since the overall compound has a charge of -2, we can calculate the oxidation number of each oxygen as -2.

For hydrogen (H2), the oxidation number is +1 based on rule 3.

Now that we have determined the oxidation numbers, we can identify which element is oxidized and which is reduced.

In this reaction, aluminum (Al) goes from an oxidation number of 0 to an oxidation number of +3, which indicates that it has undergone oxidation. Therefore, aluminum is oxidized.

In sulfur (S), the oxidation number changes from +6 to +6 in the reaction and remains the same. Since the oxidation number does not change, sulfur is not oxidized or reduced.

In summary:
Aluminum (Al) is oxidized.
Sulfur (S) is not oxidized or reduced.

For further information, you can refer to the link you provided, which discusses oxidation-reduction reactions and their principles.

in the equation 2 Al + 8 H2O yields 2 H+1 + 3 H2 + 2 Al(OH)4-1 which element was reduced and which element was oxidized?