Problem: 2 NO3-(aq) + 8 H+(aq) + 3 Cu(s) → 2 NO(g) + 4 H2O(l) + 3 Cu2+(aq)1) Indicate the half-reaction occurring at Anode.Express your answer as a chemical equation. Identify all of the phases in your answer.2) Indicate the half-reaction occurring at Cathode.Express your answer as a chemical equation. Identify all of the phases in your answer.

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2 NO3-(aq) + 8 H+(aq) + 3 Cu(s) → 2 NO(g) + 4 H2O(l) + 3 Cu2+(aq)


Step 1. Separate the whole reaction into half-reactions

*ignore H2O(l) and H+

NO3-(aq) → NO(g)

Cu(s) → Cu2+(aq)


Step 2. Balance the non-hydrogen and non-oxygen elements first

NO3-(aq) → NO(g)

Cu(s) → Cu2+(aq)


Step 3. Balance oxygen by adding H2O to the side that needs oxygen (1 O: 1 H2O)

NO3-(aq) → NO(g) + 2 H2O(l)

Cu(s) → Cu2+(aq)


Step 4. Balance hydrogen by adding H+ to the side that needs hydrogen (1 H: 1 H+)

NO3-(aq) + 4 H+(aq) → NO(g) + 2 H2O(l)

Cu(s) → Cu2+(aq)


Step 5. Balance the charges: add electrons to the more positive side (or less negative side)

NO3-(aq) + 4 H+(aq) + 3 e → NO(g) + 2 H2O(l)

Cu(s) → Cu2+(aq) + 2 e


Step 6. Balance electrons on the two half-reactions

[ NO3-(aq) + 4 H+(aq) + 3 e → NO(g) + 2 H2O(l) ] x 2

Cu(s) → Cu2+(aq) + 2 e ] x 3


2 NO3-(aq) + 8 H+(aq) + 6 e → 2 NO(g) + 4 H2O(l)

                                  3 Cu(s) → 3 Cu2+(aq) + 6 e
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2 NO3-(aq) + 3 Cu(s) + 8 H+(aq) → 2 NO(g) + 3 Cu2+(aq) + 4 H2O(l)



Step 4. Determine the oxidation half-reaction and the reduction half-reaction. 

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Problem Details

2 NO3-(aq) + 8 H+(aq) + 3 Cu(s) → 2 NO(g) + 4 H2O(l) + 3 Cu2+(aq)

1) Indicate the half-reaction occurring at Anode.

Express your answer as a chemical equation. Identify all of the phases in your answer.

2) Indicate the half-reaction occurring at Cathode.

Express your answer as a chemical equation. Identify all of the phases in your answer.