Question (3 marks): Copper reacts with concentrated nitric acid to give copper(II) nitrate, nitrogen dioxide and water. Use oxidation numbers to balance the equation:Cu + HNO3 → Cu(NO3)2 + NO2 + H2O Step 1: find the changes. Copper goes from 0 to +2, an increase of 2. In HNO3, nitrogen is +5 (1 + x + 3(−2) = 0). In NO2 it is +4 (x + 2(−2) = 0), a decrease of 1. The nitrogen in Cu(NO3)2 stays at +5.Step 2: match the changes. The increase of 2 for one Cu must equal the total decrease, so 2 NO2 must form:Cu + ? HNO3 → Cu(NO3)2 + 2 NO2 + ? H2O
Step 3: balance the other atoms. Nitrogen on the right is 2 + 2 = 4, so 4 HNO3 are needed. Their four hydrogen atoms form 2 H2O:Cu + 4 HNO3 → Cu(NO3)2 + 2 NO2 + 2 H2O
Step 4: check oxygen. Left: 4 × 3 = 12. Right: 6 + 4 + 2 = 12. Balanced.Examiner tip: only two of the four nitric acid molecules are reduced. The other two provide the nitrate ions in the salt, which is why you balance the atoms that change first and the rest last. Copper is the reducing agent and nitric acid is the oxidising agent.
Step 3: balance the other atoms. Nitrogen on the right is 2 + 2 = 4, so 4 HNO3 are needed. Their four hydrogen atoms form 2 H2O:
Step 4: check oxygen. Left: 4 × 3 = 12. Right: 6 + 4 + 2 = 12. Balanced.Examiner tip: only two of the four nitric acid molecules are reduced. The other two provide the nitrate ions in the salt, which is why you balance the atoms that change first and the rest last. Copper is the reducing agent and nitric acid is the oxidising agent.