Identify the oxidation and reduction half-reactions. Copper (Cu) is oxidized to Cu²⁺, and nitrate (NO3⁻) is reduced to nitrogen dioxide (NO2).
Write the oxidation half-reaction: Cu(s) → Cu²⁺(aq) + 2e⁻. This shows the loss of electrons by copper.
Write the reduction half-reaction: NO3⁻(aq) + 2e⁻ + 2H⁺(aq) → NO2(g) + H2O(l). This shows the gain of electrons by nitrate, forming nitrogen dioxide and water.
Balance the electrons transferred in both half-reactions. Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to balance the electrons.
Combine the balanced half-reactions, ensuring that the number of electrons lost equals the number gained, and balance the remaining atoms. The final balanced equation is: 3Cu(s) + 2NO3⁻(aq) + 8H⁺(aq) → 3Cu²⁺(aq) + 2NO2(g) + 4H2O(l).