Question

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Answer 1

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)

For aluminum (Al):

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.

For copper (Cu):

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.

For nitrogen (N):

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

For oxygen (O):

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).