Balance the following redox equation in acidic solution and determine the coefficient placed in front of H₂O: ___ Cu + ___ NO₃^- + ___ H⁺ ___ Cu²⁺ + ___ NO + ___ H₂O

Question

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1. Assign oxidation states: Cu = 0 NO = -2 S in H₂O = +6 S in NO₃^- = +6 H in H⁺ = +1 Cu²⁺ = +2 2. Write half reactions: Oxidation: H₂O → H⁺ Reduction: Cu → Cu²⁺ 3. Balance elements other than H and O: Oxidation: H₂O → H⁺ + 4H2O Reduction: 2Cu → Cu²⁺ 4. Balance O by adding H2O: Oxidation: H₂O → H⁺ + 4H+ + 3H2O Reduction: 2Cu → Cu²⁺ 5. Balance H by adding H+: Oxidation: H₂O + 8H+ → H⁺ + 4H2O Reduction: 2Cu + 2H+ → Cu²⁺ 6. Balance charge by adding electrons: Oxidation: H₂O + 8H+ + 11e- → H⁺ + 4H2O Reduction: 2Cu + 2H+ + 2e- → Cu²⁺ 7. Multiply half reactions by coefficients to make the electrons equal and add them together: H₂O + 8H+ + 11e- → H⁺ + 4H2O 2Cu + 2H+ + 2e- → Cu²⁺ multiply reduction by 11 and add with oxidation reaction to get final equation 11Cu + NO₃^- + 8H₂O → Cu²⁺ + 3NO + 4H2O +11H⁺ The coefficient placed in front of H₂O is 8, so the answer is (D).

Balance the Following Redox Equation in Acidic Solution and Determine