Assign oxidation numbers for the following chemical equation:
2Al + 3H2SO4 ---> Al2(SO4)3 +3 H2
What was oxidized?
What was reduced?
My answer:
Oxidation numbers going from the left side of the chemical eq to the right side..
Al: 0
H: +1
S: +6
O : -2
Al: +3
S: +6
O: -2
H: 1
The Al was oxidized.
I don't know what is reduced because the oxidation numbers seem to be the same for S.. Where is my error??
you err on the right side. H is zero (free element). H was reduced, it gained electrons.
Your analysis of the oxidation numbers is correct. The oxidation number of sulfur remains at +6 on both sides of the equation. However, in order to determine what is reduced, you need to compare the oxidation numbers of the hydrogen in H2SO4 before and after the reaction.
Before the reaction, the oxidation number of hydrogen in H2SO4 is +1. After the reaction, the oxidation number of hydrogen in H2 is 0. This means that hydrogen has been reduced from an oxidation number of +1 to 0. Therefore, hydrogen is the species that has been reduced in this chemical equation.
To determine what was oxidized and what was reduced in a chemical equation, you need to compare the oxidation numbers of the elements before and after the reaction.
In the given chemical equation:
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
Let's determine the oxidation numbers for each element:
Al (on the left side): Since it is an element, its oxidation number is 0.
H (in H2SO4): Hydrogen usually has an oxidation number of +1 when it is bonded to a nonmetal (as in H2SO4).
S (in H2SO4): Since there are four oxygen atoms in H2SO4, we can assign a -2 oxidation number to each oxygen ion. Considering the overall charge balance of the sulfate ion, sulfur must have an oxidation number of +6 in H2SO4.
O (in H2SO4): It usually has an oxidation number of -2.
So, the oxidation numbers on the left side of the equation are:
Al: 0
H: +1
S: +6
O: -2
Now, let's determine the oxidation numbers on the right side of the equation:
Al (in Al2(SO4)3): Since there are two aluminum atoms, they will each have an oxidation number of +3 for the overall compound. This is because the sulfate ion has an overall charge of -2, and there are three of them.
S (in Al2(SO4)3): The sulfate ion has an overall charge of -2, and since there are three of them, the sulfur atom must have an oxidation number of +6 in Al2(SO4)3.
O (in Al2(SO4)3): It will have an oxidation number of -2.
H (in H2): Hydrogen usually has an oxidation number of +1 when it is bonded to a nonmetal.
So, the oxidation numbers on the right side of the equation are:
Al: +3
S: +6
O: -2
H: +1
Now, let's determine what was oxidized and what was reduced based on the changes in oxidation numbers:
Al went from an oxidation number of 0 in Al to +3 in Al2(SO4)3. Therefore, Al was oxidized.
S stayed at an oxidation number of +6 both in H2SO4 and Al2(SO4)3. Therefore, S did not change its oxidation number and was neither oxidized nor reduced.
So, the correct answer is that Al was oxidized, and there was no reduction in this reaction.