Answer:
Oxidation state of Al in reactant: 0
. In product: +3
Oxidation state of Cu in reactant: +2
. In product: 0
Oxidation state of N in reactant: +5
. In product: +5
Oxidation state of O in reactant: -2
. In product: -2
Explanation:
The balanced redox equation is the following:
2 Al(s) + 3 Cu(NO₃)₂(aq) → 3 Cu(s) + 2 Al(NO₃)₃(aq)
<u>For aluminum (Al):</u>
In reactants, it is in the metallic solid-state (Al(s)), so it is the pure element. Pure elements has an oxidation number equal to 0.
In products, it is combined with N and O and forms an ionic compound, which can dissociate into ions as follows:
Al(NO₃)₃ → Al³⁺ + 3 NO₃⁻
So, aluminum is a monoatomic ion, and its oxidation number is equal to the charge of the ion. Thus, the oxidation number is +3.
<u>For copper (Cu):</u>
In reactants, it is present as a nitrate salt. So, it dissociates into ions as follows:
Cu(NO₃)₂ → Cu²⁺ + 2 NO₃⁻
Since Cu is in the form of a monoatomic ion, its oxidation number is equal to the charge of the ion (+2).
In products, it is in the metallic solid form (pure element). Thus, the oxidation number is 0.
<u>For nitrogen (N):</u>
Both in reactants and products, N is present in the nitrate ion (NO₃⁻) which is derived from the nitric acid (HNO₃). The oxidation number is calculated from the addition of the oxidation numbers of N and O multiplied by their subscripts, which must be equal to the charge of the compound (-1):
(N x 1) + (O x 3) = -1
In oxide compounds, the oxygen atom has an oxidation number of -2. So, the oxidation number of O is -2 because it derives from an oxide (nitric acid is obtained from nitric oxide). Thus, we calculate the oxidation number of N:
N + (-2 x 3) = -1
N - 6 = -1
N = - 1 + 6 = +5
<u>For oxygen (O):</u>
Both in reactants and products, O is present in the nitrate ion (NO₃⁻) which is derived from the nitric acid (HNO₃).
As we saw previously, the oxidation number of O is -2 because it derives from an oxide (nitric acid is obtained from nitric oxide).
