Question

oxidation state of n in reactant: +5
in product:+2

oxidation state of o in reactant: -2
in product:-2

explain why this is a redox reaction.
this is a redox reaction because cu is oxidized (0 to +2) and n is reduced (+5 to +2)

i
b. write the half - reactions for the reaction. label the oxidation reaction and the reduction reaction. (3 points)

Explanation:

Step1: Identify Oxidation Half - Reaction

In a redox reaction, oxidation is the loss of electrons (increase in oxidation state). For copper (Cu), its oxidation state changes from 0 (in elemental Cu) to +2. The half - reaction for oxidation (Cu losing electrons) is:
$Cu
ightarrow Cu^{2+}+2e^-$
This is the oxidation reaction because the oxidation state of Cu increases from 0 to +2, meaning it loses electrons.

Step2: Identify Reduction Half - Reaction

Reduction is the gain of electrons (decrease in oxidation state). For nitrogen (N), its oxidation state changes from +5 (in the reactant, e.g., in $NO_3^-$ or similar species) to +2. Let's assume the nitrogen - containing species is $NO_3^-$ (a common species with N in +5 oxidation state) and the product has N in +2 (e.g., $NO$). The half - reaction for reduction (N gaining electrons) is:
$NO_3^- + 4H^++3e^-
ightarrow NO + 2H_2O$ (assuming acidic conditions, and balancing O and H). The key here is that the oxidation state of N decreases from +5 to +2, so it gains electrons. This is the reduction reaction.

Answer:

Oxidation reaction (loss of electrons, increase in oxidation state): $Cu
ightarrow Cu^{2+}+2e^-$
Reduction reaction (gain of electrons, decrease in oxidation state): $NO_3^- + 4H^++3e^-
ightarrow NO + 2H_2O$ (or a similar reduction half - reaction depending on the specific reactants and products, but the general principle is N going from +5 to +2 by gaining electrons and Cu going from 0 to +2 by losing electrons)