Question

Redox Reactions: Less active metals like copper will reacti with concentrated nitric acid to produce nitric oxide gas and the Cu²⁺_(aq) ion at 25C.

A) Write balanced chemical equation for this process.

              6H⁺_(aq) + 2HNO_3(aq) + 3Cu_(s) --> 3Cu²⁺_(aq) + 2NO_(g) + 4H₂O _(aq)

B) Evaluate change in E for the reaction in (a).

Answer 1

Step 0: Write the “skeleton equation”

Cu(s) + H+(aq) + NO3 −(aq) −→ Cu2+(aq) + NO(g)

and determine the oxidation state (the oxidation state is per atom!)

Cu⁰ (s)=0

H+(aq)=+1

NO3 −(aq)=+5 −2

→ Cu2+(aq)=+2

NO(g)=   +2 −2

So we have

Cu : 0 → +2 ox

N : +5 → +2 red

Step 1: Write the “skeleton” half-reactions

ox: Cu → Cu²⁺

red: NO^{3 −} → NO

Step 2: Balance each half-reaction “atomically”

• all atoms other than H and O (You can use any of the species that appear in the skeleton equation—Step 0—for this purpose.)

• balance O atoms by adding H2O

• balance H atoms by adding H+

ox: Cu −→ Cu2+

red: NO3 − −→ NO + 2 H2O

red: NO3 − + 4 H⁺ −→ NO + 2 H2O

Step 3: Balance the electric charges by adding electrons

ox: Cu −→ Cu2+ + 2 e⁻

red: NO3 ⁻ + 4 H⁺ + 3 e⁻ −→ NO + 2 H2O

The electrons have to appear on the right hand side for the oxidation half-reaction, and on the left hand side for the reduction half-reaction.

Step 4: Prepare the two half-equations for summation by making the number of electrons the same in both, i.e, find the least common multiple

3 × ox: 3 Cu → 3 Cu2+ + 6 e⁻

2 × red: 2 NO3 ⁻ + 8 H+ + 6 e⁻ → 2 NO + 4 H2O

Step 5: Combine the two half-reactions

3 Cu + 2 NO3 ⁻ + 8 H⁺ + 6 e⁻ → 3 Cu²⁺ + 6 e− + 2 NO + 4 H2O

Step 6: Simplify

3 Cu + 2 NO3 − + 8 H+ −→ 3 Cu2+ + 2 NO + 4 H2O

Step 7: Indicate the state of each species

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

This is the fully balanced net ionic equation.