Practice Problem 5

Use cell potential data to explain why copper metal does not dissolve in a typical strong acid, such as hydrochloric acid,

Cu(s) + 2 H+(aq)

but will dissolve in 1 M nitric acid.

3 Cu(s) + 2 HNO3(aq) + 6 H+(aq) 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l)

Answer

Copper does not dissolve in a typical strong acid because the overall cell potential for the oxidation of copper metal to Cu²⁺ ions coupled with the reduction of H⁺ ions to H₂ is negative.

Cu Cu2+ + 2 e-		Eoox = -(0.34 V)
+ 2 H+ + 2 e- H2		Eored = 0.000... V
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯		¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Cu + 2 H+ Cu2+ + H2		Eo = Eoox + Eored = -0.34 V

Copper dissolves in nitric acid because the reaction at the cathode now involves the reduction of nitric acid to NO gas, and the potential for this half-reaction is strong enough to overcome the half-cell potential for oxidation of copper metal to Cu²⁺ ions.

3 (Cu Cu2+ + 2 e-)		Eoox = -(0.34 V)
+ 2(HNO3 + 3 H+ + 3 e- NO + 2 H2O)		Eored = 0.96 V
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯		¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
3 Cu + 2 HNO3 + 6 H+ 3 Cu2+ + 2 NO + 4 H2O		Eo = Eoox + Eored = 0.62 V