... Zn(s) + ... H+(aq) + ... NO3-(aq) → ... Zn2+(aq) + ... NH4+(aq) + ... H2O(l) When the equation a — Electrochemistry Chemistry Question

STEPS:

1. Identify the type of chemical reaction: This is a redox (oxidation-reduction) reaction occurring in an acidic aqueous solution. To balance it properly, we must use the half-reaction method.
2. Write the oxidation half-reaction and balance it:
* Zinc metal is oxidized to zinc ions:
$$\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq)$$
* To balance the $+2$ charge on the right side, add $2$ electrons ($e^-$) to the products:
$$\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$$
3. Write the reduction half-reaction and balance it:
* Nitrate ions are reduced to ammonium ions:
$$\text{NO}_3^-(aq) \rightarrow \text{NH}_4^+(aq)$$
* Balance Oxygen: Add $3\text{ H}_2\text{O}$ molecules to the product side to balance the $3$ oxygen atoms in nitrate:
$$\text{NO}_3^-(aq) \rightarrow \text{NH}_4^+(aq) + 3\text{H}_2\text{O}(l)$$
* Balance Hydrogen: There are $10$ hydrogen atoms on the product side ($4$ in $\text{NH}_4^+$ and $6$ in $3\text{H}_2\text{O}$). Add $10\text{ H}^+$ ions to the reactant side:
$$\text{NO}_3^-(aq) + 10\text{H}^+(aq) \rightarrow \text{NH}_4^+(aq) + 3\text{H}_2\text{O}(l)$$
* Balance Charge: The reactant side has a net charge of $+9$ ($-1 + 10$), and the product side has a net charge of $+1$. Add $8$ electrons to the reactant side to balance the charge:
$$\text{NO}_3^-(aq) + 10\text{H}^+(aq) + 8e^- \rightarrow \text{NH}_4^+(aq) + 3\text{H}_2\text{O}(l)$$
4. Equalize the electron transfer between the two half-reactions:
* The oxidation half-reaction releases $2e^-$, while the reduction half-reaction requires $8e^-$.
* Multiply the entire oxidation half-reaction by $4$ to equalize the number of electrons transferred ($8e^-$ total):
$$4\text{ Zn}(s) \rightarrow 4\text{ Zn}^{2+}(aq) + 8e^-$$
5. Combine the half-reactions and simplify:
* Combine the reactant sides and product sides, then cancel out the $8e^-$ on both sides:
$$4\text{ Zn}(s) + 10\text{H}^+(aq) + \text{NO}_3^-(aq) \rightarrow 4\text{ Zn}^{2+}(aq) + \text{NH}_4^+(aq) + 3\text{H}_2\text{O}(l)$$
6. Verify the final coefficients:
* The coefficients are $4$ (for $\text{Zn}$), $10$ (for $\text{H}^+$), $1$ (for $\text{NO}_3^-$), $4$ (for $\text{Zn}^{2+}$), $1$ (for $\text{NH}_4^+$), and $3$ (for $\text{H}_2\text{O}$).
* All atoms (Zn: 4, H: 10, N: 1, O: 3) and net charges ($+9$ on both sides) are perfectly balanced.
* Since the coefficient for $\text{NO}_3^-$ is $1$, these coefficients are already reduced to their lowest whole-number terms. Therefore, the coefficient for $\text{Zn}(s)$ is $4$.

*

WHY_OTHERS_WRONG:

• A is incorrect: A coefficient of 2 for $\text{Zn}(s)$ would only provide $4$ electrons during oxidation. This is insufficient to reduce a single mole of $\text{NO}_3^-$ to $\text{NH}_4^+$ because that reduction pathway requires a full transfer of $8$ electrons.
• C is incorrect: A coefficient of 6 for $\text{Zn}(s)$ would mean the oxidation half-reaction releases $12$ electrons ($6 \times 2e^-$). This does not match the $8$-electron reduction requirement of nitrate to ammonium and cannot be simplified to lowest whole-number terms with this ratio.
• D is incorrect: A coefficient of 10 for $\text{Zn}(s)$ would release $20$ electrons, which does not represent the simplified, lowest-whole-number balanced equation for this specific chemical system.
• E is incorrect: A coefficient of 14 is far too high to balance the single nitrogen species transition and violates the requirement that coefficients must be reduced to their lowest possible whole-number terms.

✏️ I can help you balance other redox reactions or work through electrochemistry problems from your AP Chemistry practice materials.