Questions 10-13 refer to the following information. Cu(s) + 4 HNO3(aq) → Cu(NO3)2(aq) + 2 NO2(g) + 2 — Electrochemistry Chemistry Question

STEPS:

1. Recall the definitions of chemical reaction types:
* Brønsted^-Lowry acid-base reactions involve the transfer of protons ($\text{H}^+$) from an acid to a base, with no change in the oxidation states of any elements.
* Reduction-oxidation (redox) reactions involve the transfer of electrons between chemical species, which is always accompanied by changes in the oxidation numbers of the atoms involved.
2. Assign oxidation numbers to each element on the reactant side:
* $\text{Cu}(s)$: As a pure element in its standard state, copper has an oxidation number of $0$.
* $\text{HNO}_3(aq)$:
* Hydrogen ($\text{H}$) has its standard oxidation number of $+1$.
* Oxygen ($\text{O}$) has its standard oxidation number of $-2$.
* To find Nitrogen ($\text{N}$) in the nitrate ion ($\text{NO}_3^-$):
$$\text{Oxidation Number of N} + 3(-2) = -1 \implies \text{N} - 6 = -1 \implies \text{N} = \mathbf{+5}$$
3. Assign oxidation numbers to each element on the product side:
* $\text{Cu(NO}_3)_2(aq)$: This ionic compound dissociates into $\text{Cu}^{2+}(aq)$ and $\text{NO}_3^-(aq)$ ions.
* The copper ion has a charge of $2+$, so the oxidation number of $\text{Cu}$ is $+2$.
* In the spectator nitrate ions ($\text{NO}_3^-$), $\text{N}$ remains at $+5$ and $\text{O}$ remains at $-2$.
* $\text{NO}_2(g)$:
* Oxygen ($\text{O}$) has an oxidation number of $-2$.
* To find Nitrogen ($\text{N}$) in this neutral molecule:
$$\text{Oxidation Number of N} + 2(-2) = 0 \implies \text{N} - 4 = 0 \implies \text{N} = \mathbf{+4}$$
* $\text{H}_2\text{O}(l)$: Hydrogen ($\text{H}$) remains $+1$ and oxygen ($\text{O}$) remains $-2$.
4. Compare the initial and final oxidation states to identify the redox behavior:
* Copper ($\text{Cu}$) increases in oxidation number from $0 \rightarrow +2$, meaning it loses electrons and is oxidized.
* Some Nitrogen ($\text{N}$) atoms decrease in oxidation number from $+5$ (in $\text{NO}_3^-$) $\rightarrow +4$ (in $\text{NO}_2$), meaning they gain electrons and are reduced.
* Hydrogen ($\text{H}$) remains at $+1$ on both sides of the equation and is not reduced.
5. Conclude the correct option: Because there is a transfer of electrons where $\text{Cu}(s)$ is oxidized and the nitrogen atom in $\text{NO}_3^-(aq)$ is reduced, this is a redox reaction. This directly corresponds to Option D.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This is not a Brønsted^-Lowry acid-base reaction. Furthermore, because a strong acid ($\text{HNO}_3$) is added to copper (which is not a Brønsted^-Lowry base), and because some unreacted strong acid remains in the solution, the final solution is highly acidic, not neutral.
• Option B is incorrect: While $\text{HNO}_3(aq)$ is indeed a strong acid, its presence does not automatically make every reaction it undergoes an acid-base reaction. In this reaction, nitric acid acts as a powerful oxidizing agent rather than a proton donor, facilitating a redox process instead.
• Option C is incorrect: Although it correctly identifies the reaction as redox and states that $\text{Cu}(s)$ is oxidized, it incorrectly claims that $\text{H}^+(aq)$ is reduced. As shown in the oxidation state analysis, the hydrogen atoms maintain a $+1$ oxidation state throughout the process in both $\text{HNO}_3$ and $\text{H}_2\text{O}$, meaning hydrogen is not reduced.