Which of the following is the correctly balanced redox reaction for Cu(s) + NO3^-(aq) → Cu^{2+}(aq) + NO2(g) in acidic solution?

• A. Cu(s) + 2 NO3^-(aq) + 4 H^+(aq) → Cu^{2+}(aq) + 2 NO2(g) + 2 H2O(l)
• B. Cu(s) + NO3^-(aq) → Cu^{2+}(aq) + NO2(g)
• C. 2 Cu(s) + 2 NO3^-(aq) + 4 H^+(aq) → 2 Cu^{2+}(aq) + 2 NO2(g) + 4 H2O(l)
• D. Cu(s) + NO3^-(aq) + 2 H^+(aq) → Cu^{2+}(aq) + NO2(g) + H2O(l)

Answer: (D)

Verified step by step guidance

1. Identify the oxidation and reduction half-reactions. Copper (Cu) is oxidized from Cu(s) to Cu^{2+}(aq), and nitrate ion (NO_3^-) is reduced to nitrogen dioxide (NO_2).
2. Write the oxidation half-reaction: \(\mathrm{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-}\).
3. Write the reduction half-reaction for nitrate to NO_2 in acidic solution, balancing atoms other than H and O first, then balance O by adding H_2O, and H by adding H^+, and finally balance charge by adding electrons.
4. Combine the two half-reactions by equalizing the number of electrons lost and gained, then add the half-reactions together, canceling electrons and any species that appear on both sides.
5. Verify that the final equation is balanced in terms of atoms (Cu, N, O, H) and charge, ensuring the number of atoms and total charge are the same on both sides.