Answer:
a. Cu + 4 HNO₃ ⇄ Cu(NO₃)₂ + 2 NO₂ + 2 H₂O oxidation/reduction reaction
b. Cu + 4 H⁺ + 4 NO₃⁻ ⇄ Cu²⁺ + 2 NO₃⁻ + 2 NO₂ + 2 H₂O
c. NO₃⁻
d. Cu + 4 H⁺ + 2 NO₃⁻ ⇄ Cu²⁺ + 2 NO₂ + 2 H₂O
e. Cu
f. N
Explanation:
a. Write the overall balanced molecular equation for this reaction and classify it as a double replacement, decomposition, metathesis, or an oxidation/reduction reaction.
This is an oxidation/reduction reaction. So, to balance it, we will use the ion-electron method. We split the global equation in half-reactions of oxidation and reduction. Then we balance the mass (adding H⁺ and H₂O if needed) and the charges (electrons). Afterward, we multiply both half-reactions by numbers that secure that the number of electrons gained and lost is the same, and we add both half-reactions. Finally, we balance spectator ions if necessary.
Cu + HNO₃ ⇄ Cu(NO₃)₂ + NO₂
Oxidation: (Cu ⇒ Cu²⁺ + 2e⁻) × 1
Reduction: (1e⁻ + 2H⁺ + NO₃⁻ ⇒ NO₂ + H₂O) × 2
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Cu + 4 H⁺ + 2 NO₃⁻ ⇄ Cu²⁺ + 2 NO₂ + 2 H₂O
We take these numbers to the global equation.
Cu + 4 HNO₃ ⇄ Cu(NO₃)₂ + 2 NO₂ + 2 H₂O
We realize that the only NO₃⁻ balanced by the ion-electron method are the 2 that reduce to NO₂, while the others have to be balanced manually.
b. Write the total ionic equation for this reaction.
In the total ionic equation we include all ions and species that do not dissociate in water.
Cu + 4 H⁺ + 4 NO₃⁻ ⇄ Cu²⁺ + 2 NO₃⁻ + 2 NO₂ + 2 H₂O
c. Identify any spectator ions from part b.
Some NO₃⁻ are spectator ions
d. Write the net ionic equation for this reaction.
The net ionic equation includes only ions that participate in the reaction and species that do not dissociate in water.
Cu + 4 H⁺ + 2 NO₃⁻ ⇄ Cu²⁺ + 2 NO₂ + 2 H₂O
e. Which element was oxidized?
Cu was oxidized. Its oxidation number goes from 0 to 2.
f. Which element (not compound) was reduced?
N was reduced. Its oxidation number goes from 5 to 4.
