Question

b. write the half - reactions for the reaction. label the oxidation reaction and the reduction reaction. (3 points)
cu> cu2+ + 2e
no - 3 + 4h + 3e> no + 2h2o
i
c. in this equation, what was oxidized and what was reduced? (3 points)

Explanation:

Part b

Step 1: Analyze the first half - reaction ($\ce{Cu -> Cu^{2+} + 2e^-}$)

In this reaction, the oxidation state of copper (Cu) changes from 0 (in elemental Cu) to +2 (in $\ce{Cu^{2+}}$). Since the oxidation state of an element increases when it loses electrons, this is an oxidation reaction. The loss of electrons is characteristic of oxidation.

Step 2: Analyze the second half - reaction ($\ce{NO_3^- + 4H^+ + 3e^- -> NO + 2H_2O}$)

First, calculate the oxidation state of N in $\ce{NO_3^-}$: Let the oxidation state of N be $x$. We know that the oxidation state of O is - 2 and the overall charge of $\ce{NO_3^-}$ is - 1. So, $x+3\times(- 2)=-1$, which gives $x = + 5$. In NO, the oxidation state of N is +2. Since the oxidation state of N decreases from +5 to +2, and the reaction involves the gain of electrons (3 electrons are gained), this is a reduction reaction. The gain of electrons is characteristic of reduction.

• For oxidation: A substance is oxidized when it loses electrons and its oxidation state increases. In the half - reaction $\ce{Cu -> Cu^{2+} + 2e^-}$, the oxidation state of Cu goes from 0 to +2. So, Cu is oxidized.
• For reduction: A substance is reduced when it gains electrons and its oxidation state decreases. In the half - reaction $\ce{NO_3^- + 4H^+ + 3e^- -> NO + 2H_2O}$, the oxidation state of N in $\ce{NO_3^-}$ (oxidation state +5) goes to +2 in NO. So, $\ce{NO_3^-}$ (specifically the N in $\ce{NO_3^-}$) is reduced.

Answer:

Oxidation reaction: $\ce{Cu -> Cu^{2+} + 2e^-}$ (Cu loses electrons, oxidation state increases from 0 to +2)
Reduction reaction: $\ce{NO_3^- + 4H^+ + 3e^- -> NO + 2H_2O}$ (N gains electrons, oxidation state decreases from +5 to +2)

Part c