ELECTROLYSIS | FACTORS THAT AFFECT THE IONS TO BE DISCHARGED AT ELECTRODES.

In electrolysis, there are cases where different ions of the same charge migrate to the same electrode, but it is only one type of ion gets to be discharged at the supposed electrode.

There are three factors that determine the type of ion to be discharged selectively, when different ions of the same charge migrate to the same electrode:

1. Position of the ion in the electrochemical series: Electrochemical series can be obtained when cations are arranged according to their ability to accept electrons easily (getting reduced), and also when anions are arranged according to their ability to lose electrons (getting oxidized).

In the diagram above, the cations (from top to bottom) were arranged in order of increasing reduction (ability to gain electrons), so, the ion lower in the series will be discharged at the cathode electrode.

Similarly, the anions in the diagram where arranged (from top to bottom) in order of increasing oxidation (ability to lose electrons), so the ion lower in the series will likely be discharged at the anode electrode.

For example, during the electrolysis of aqueous NaCl (NaCl dissolved in H2O), four different ions are formed in the electrolyte;

The salt ionizes; NaCl(aq) ➡️ Na+(aq) + Cl-(aq)

Water ionizes; H2O(l) ➡️ H+(aq) + OH-(aq)

Thus we have four different ions (Na+, Cl-, H+, and OH-) present in the electrolyte.

The two cations, Na+ and H+ moves to the cathode, but H+ is selectively discharged as hydrogen gas, H2, because it is lower than Na+ in the electrochemical series.

2H+ + 2e- ➡️ H2(g)

Na+ is left inside the electrolyte.

The two anions, Cl- and OH- move to the anode, but OH- is selectively discharged as oxygen gas, O2, because it is lower than Cl- in the electrochemical series.

4OH- ➡️ 2H2O + O2 + 4e-

Also, in the electrolysis of acidulated water (dilute H2SO4);

Acid ionizes; H2S04(aq) ➡️ 2H+(aq) + SO42-(aq)

Water ionizes; H2O(l) = H+(aq) + OH-(aq)

Thus we have three different ions (H+, SO42-, and OH-) present in the electrolyte. At the cathode, H+ is discharged because it is the only cation present in the electrolyte.

At the anode, SO42-, and OH- migrates and OH- is preferentially discharged being lower than SO42- in the electrochemical series.

2. Relative concentration of the ions:

Electrolysis of dilute and concentrated solutions of the same electrolyte would give different products respectively.

Usually, in the electrolysis of a dilute solution, the products/ions to be discharged depend on the position of the ions in the electrochemical series.

But conversely, in the electrolysis of a concentrated solution, the anion to be discharged at the anode could be altered if a halogen (Cl-, Fl-, Br- and I-) ion is present in the electrolyte. Concentration effect is limited to the anions of the halogens. Halogens ions tend to be preferentially discharged when in a concentrated solution because their ions are more concentrated than other anions in the electrolyte.

Consider the electrolysis of dilute and concentrated solutions of NaCl.

3. Nature of the electrodes: When inactive electrodes like graphite, an allotrope of carbon, and platinum, a metal, are used in electrolysis, they do not alter/affect the ions to be discharged at their respective electrodes; rather, the ions to be discharged could be affected by position of ions in the electrochemical series, or relative concentration of ions in the electrolyte.

When metals (except platinum) are used as the anode, and the electrolyte to be electrolyzed is a salt of the metal used as the anode, it does affect the ion to be discharged at the anode. Being that the metal anode is dipped into its metal salt (electrolyte), the metal anode loses electrons to the metal salt which leads to it dissolving into the electrolyte. With lesser energy, it is easier to remove electrons from the anode metal, than it is to remove electrons from the anions in the electrolyte which requires greater energy. Consider the electrolysis of aqueous CuSO4(aq) using copper electrode as the anode;

1. Salt ionizes; CuS04(aq) ➡️ Cu2+(aq) + SO42-(aq)

Water ionizes; H2O(l) ➡️ H+(aq) + OH-(aq)

At the anode, copper anode dissolves by electron loss;

Cu(s) ➡️ Cu2+(aq) + 2e-

At the cathode, Cu2+ is discharged, being lower than H+ in the electrochemical series;

Cu2+(aq) + 2e- ➡️ Cu(s)