Metallic copper is oxidized by dilute nitric acid to form Cu2+ and NO

Question

Metallic copper is oxidized by dilute nitric acid to form Cu²⁺ and NO.

 

i)Write the two half equations associated with this reaction?

 

ii) Identify the oxidation and reduction half- equations. YOu must show how you arrived at your answer.

 

a)Oxidation half equation

b)Reduction half equation

 

iii) Write the ionic equation for the reaction.

 

Net ionic equation: _______

 

iv)Write the molecular equation for the reaction.

Answer 1

Step 1: Identify the change in charge for atoms:

HNO₃ ⇒ NO; N goes from +5 to +2

Cu ⇒ Cu²⁺; Cu goes from 0 to +2

Half equations:

                                                                    N⁵⁺ + 3e^- → N²⁺

                                                                    Cu → Cu²⁺ + 2e^-

 

i)                                      Reduction: HNO₃ + 3e^- + 3H⁺ → 2H₂O + NO

                                         Oxidation: Cu → Cu²⁺ + 2e^-

iii)  Net Ionic Equation:

                                                            2N⁵⁺ + 3Cu → 2N²⁺ + 3Cu²⁺

 

      Net Equation (multiply the reduction by 2 and oxidation by 3)

                                               3Cu + 2HNO₃ + 6H⁺ → 2NO + 4H₂O + 3Cu²⁺

 

Take Aways:

1) Always identify which atom is transferring charge.

2) Make sure that you have balanced charges on both sides of oxidation and reduction half reactions. This typically can be balanced by protons, and thus lead to water in the product usually.