Electrolysis of Aqueous Copper(II) Nitrate

Aqueous copper(II) nitrate contains Cu²⁺, NO₃⁻, and water. During electrolysis, the ions migrate toward oppositely charged electrodes, but migration alone does not determine which species reacts. The species preferentially discharged depends on thermodynamics and kinetics under the actual conditions.

The usual textbook interpretation assumes inert electrodes, such as platinum or graphite, and a sufficiently concentrated copper(II) nitrate solution. Under those conditions, copper is deposited at the cathode and oxygen gas is released at the anode.

The equations below use KaTeX’s mhchem extension for \ce{...} chemical notation.

What happens at the cathode?

The cathode is the negative electrode in an electrolytic cell. Reduction occurs there. Copper(II) ions accept electrons:

$$\ce{Cu^2+ + 2e^- -> Cu(s)}$$

Copper therefore forms as a solid coating on the cathode. Although water can also be reduced to hydrogen, copper deposition is favored while a useful concentration of Cu²⁺ is present:

$$\ce{2H2O + 2e^- -> H2(g) + 2OH^-}$$

In practice, the exact competition depends on electrode material, overpotential, current density, concentration, temperature, and cell design. For ordinary aqueous copper(II) nitrate electrolysis with inert electrodes, the expected cathode product is copper while appreciable Cu²⁺ remains available.

What happens at the anode?

The anode is the positive electrode in an electrolytic cell. Oxidation occurs there. Nitrate is usually not oxidized under these conditions; instead, water gives up electrons:

$$\ce{2H2O(l) -> O2(g) + 4H+ + 4e^-}$$

The anode consequently produces bubbles of oxygen gas. Nitrate ions remain dissolved as spectator ions. The positive ions in solution, especially Cu²⁺ and H⁺, are balanced by the nitrate ions, but nitrate itself is not consumed in the electrode reaction.

Overall equation with inert electrodes

Multiply the cathode half-reaction by two so that both half-reactions involve four electrons:

$$\ce{2Cu^2+ + 4e^- -> 2Cu}$$ $$\ce{2H2O -> O2 + 4H+ + 4e^-}$$

Adding the half-reactions gives:

$$\ce{2Cu^2+ + 2H2O -> 2Cu + O2 + 4H+}$$

Including the nitrate ions gives the molecular-form equation:

$$\ce{2Cu(NO3)2(aq) + 2H2O(l) -> 2Cu(s) + O2(g) + 4HNO3(aq)}$$

This equation is a convenient way to describe the changing composition of the solution. More precisely, nitrate remains a solvated ion and the newly formed H⁺ is solvated; writing \ce{HNO3} indicates the corresponding acid in solution.

Thus, as electrolysis proceeds, Cu²⁺ is removed from the solution, copper accumulates at the cathode, oxygen evolves at the anode, and the remaining solution becomes more acidic.

Worked quantitative example

Suppose a current of $1.00\ \mathrm{A}$ flows for $30.0$ minutes through the cell. The charge passed is

$$Q=It=(1.00\ \mathrm{C\,s^{-1}})(30.0\times60.0\ \mathrm{s})=1800\ \mathrm{C}$$

Using Faraday’s constant, $F=96485\ \mathrm{C\,mol^{-1}}$, the amount of electrons transferred is

$$n(e^-)=\frac{Q}{F}=\frac{1800}{96485}=0.0187\ \mathrm{mol}$$

Two electrons deposit one mole of copper, so

$$n(\ce{Cu})=\frac{0.0187}{2}=0.00933\ \mathrm{mol}$$

The deposited mass is therefore

$$m(\ce{Cu})=(0.00933\ \mathrm{mol})(63.546\ \mathrm{g\,mol^{-1}})=0.593\ \mathrm{g}$$

Four electrons produce one mole of oxygen, giving

$$n(\ce{O2})=\frac{0.0187}{4}=0.00466\ \mathrm{mol}$$

At approximately $25\ ^\circ\mathrm{C}$ and $1\ \mathrm{atm}$, this is about $0.114\ \mathrm{L}$, or $114\ \mathrm{mL}$, of oxygen. At $0\ ^\circ\mathrm{C}$ and $1\ \mathrm{atm}$, using $22.4\ \mathrm{L\,mol^{-1}}$, it would be about $104\ \mathrm{mL}$. The calculation assumes 100% current efficiency and no competing reactions.

The same electron amount produces $0.0187$ mol of H⁺ according to the anode half-reaction. If the solution volume is $1.00\ \mathrm{L}$ and its initial acidity is negligible, the idealized added hydrogen-ion concentration would be about $0.0187\ \mathrm{M}$, corresponding to a pH near $1.73$. Real pH values can differ because of initial acidity, activity effects, mixing, and side reactions.

What if the electrode is copper?

“Inert electrodes” is not a minor detail. If the anode is made of copper, the anode can oxidize instead of water:

$$\ce{Cu(s) -> Cu^2+ + 2e^-}$$

At the cathode:

$$\ce{Cu^2+ + 2e^- -> Cu(s)}$$

The net ionic reaction is then effectively zero: copper is transferred from the anode to the cathode. This is the principle of copper electrorefining and electroplating. Ideally, Cu²⁺ concentration stays nearly constant, and little or no oxygen is produced at the anode. Impurities in an industrial copper anode may fall as anode slime or enter the electrolyte depending on their chemistry.

Common mistakes

• Treating nitrate as the anode product. Nitrate is generally a spectator ion in this experiment; water is oxidized to oxygen.
• Producing hydrogen at the cathode automatically. Hydrogen is possible, but Cu²⁺ is preferentially reduced while it is available under typical conditions.
• Forgetting the electrode material. A copper anode dissolves, whereas graphite or platinum is treated as inert.
• Calling the cathode positive. In an electrolytic cell, the cathode is negative and the anode is positive. Reduction always occurs at the cathode and oxidation always occurs at the anode.
• Assuming the solution remains unchanged. With inert electrodes, Cu²⁺ is removed and H⁺ is generated, so the composition and acidity change.