Question

Write the balanced molecular and net ionic equations for the reaction between solid copper and nitric acid (6M) concentration. Identify the oxidation and reduction half-reactions. How would this change if the concetration was15.8M?

Balanced molecular equation:

Ionic Equation:

Net ionic equation:

Oxidation half-reaction:

Reduction half-reaction:

Answer 1

for 6M HNO3 , we have

Balanced molecular equation: 3Cu(s) + 8HNO₃(aq) ——> 3Cu(NO₃)₂(aq) + 2NO(g) + 4H₂O(l)

Ionic Equation: 3 Cu(s) + 8H+ (aq) + 8 NO₃^-(aq) ——> 3 Cu2+ (aq) + 6 NO₃^- (aq) + 2 NO(g) + 4H₂O(l)

Net ionic equation: 3 Cu(s) + 8H+ (aq) + 8 NO₃^-(aq) ——> 3 Cu2+ (aq) + 2 NO₃ ^- (aq)

Oxidation half-reaction: : Cu(s) ----------->   Cu2+ (aq) + 2e-

Reduction half-reaction:  8H+ (aq) + 2 NO₃^-(aq) + 6 e- ---------->    2 NO(g) + 4H₂O(l)

for 15.8 M HNO3

Balanced molecular equation: Cu(s) + 4HNO₃(aq) ——> Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

Ionic Equation: Cu(s) + 4 H+ (aq) + 4 NO₃^-(aq) ——> Cu2+ (aq) + 2 NO₃ ^- (aq) + 2 NO(g) + 2H₂O(l)

Net ionic equation: Cu(s) + 4H+ (aq) + 4 NO₃^-(aq) ——> Cu2+ (aq) + 2 NO₃ ^- (aq)

Oxidation half-reaction: : Cu(s) ----------->   Cu2+ (aq) + 2e-

Reduction half-reaction:  4H+ (aq) + 2 NO₃^-(aq) + 2e- ---------->    2 NO(g) + 2H₂O(l)