2Al+3H2SO4--> Al2(SO4)3+3H2
a.)determine the oxidation number of each element in both the reaction and products of this oxidation-reduction reaction of aluminum and sulfuric acid.
b.) what element is oxidized in this reaction?
c.) what element reduces in the reaction?
a.) Oxidation number of Al in the reactants: +3; Oxidation number of Al in the products: +3; Oxidation number of H in the reactants: +1; Oxidation number of H in the products: +1; Oxidation number of S in the reactants: +6; Oxidation number of S in the products: +6.
b.) Aluminum is oxidized in this reaction.
c.) Hydrogen is reduced in the reaction.
a.) To determine the oxidation number of each element, we need to assign oxidation numbers to each element based on the following rules:
- The oxidation number of an element in its elemental form is always zero.
- In a compound, the sum of the oxidation numbers of all the elements is equal to zero.
- The oxidation number of hydrogen is +1, and when it is combined with a nonmetal, it is -1.
- The oxidation number of oxygen is -2, except in peroxides where it is -1.
- The sum of the oxidation numbers in a polyatomic ion is equal to its charge.
In the reaction:
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
We can assign the oxidation numbers as follows:
- Aluminum (Al) has an oxidation number of 0 on the reactant side and +3 on the product side.
- Hydrogen (H) has an oxidation number of +1 on both sides of the reaction.
- Sulfur (S) has an oxidation number of +6 on the reactant side and +6 on the product side.
- Oxygen (O) has an oxidation number of -2 on the reactant side and -2 on the product side.
b.) The element that is oxidized in this reaction is aluminum (Al). It goes from an oxidation state of 0 in the reactants to +3 in the product, which means it loses electrons and is oxidized.
c.) The element that reduces in this reaction is hydrogen (H). It goes from an oxidation state of +1 in the reactants to 0 in the products, which means it gains electrons and is reduced.
a) To determine the oxidation number of each element in the reaction and products, we need to assign oxidation states to each atom.
In aluminum (Al), it is always assigned an oxidation state of +3.
In sulfuric acid (H2SO4), hydrogen (H) is always assigned an oxidation state of +1. The oxidation state of oxygen (O) is -2. Since we have four oxygen atoms in H2SO4, the overall charge from oxygen is -8. To balance this, the sulfur (S) must have an oxidation state of +6, since the total oxidation numbers of all atoms in a compound must sum up to zero.
In the product compound, aluminum sulfate (Al2(SO4)3), we already know that aluminum has an oxidation state of +3. The overall charge of the sulfate (SO4) ion is -2. To balance the charges, sulfur (S) must have an oxidation state of +6, and the oxygen (O) atoms have an oxidation state of -2.
b) To identify which element is oxidized, we look for the element that increases its oxidation state. In this reaction, the oxidation state of aluminum (Al) increases from 0 to +3. Therefore, aluminum is oxidized.
c) To identify which element is reduced, we look for the element that decreases its oxidation state. In this reaction, the oxidation state of sulfur (S) decreases from +6 in sulfuric acid to +6 in aluminum sulfate. Therefore, sulfur is reduced.