Electrolysis of aqueous solutions

YOU SHOULD BE ABLE TO

1. Identify ions present in electrolytes. 2. Predict the electrode to which an ion will drift. 3. Identify the products appearing during electrolysis and write relevant ionic equations.

4. Describe the electrolysis of certain substances: conc. hydrochloric and dil sulphuric acid; conc. aq. sodium chloride and dil. aq. sodium chloride; copper (II) sulphate using carbon or platinum and copper electrodes; and fused halide, e.g., lead bromide.

When electrolysing aqueous solutions, the ions of the substance being electrolysed and water are present. The discharge of ions at the electrodes will, therefore, be dependent on the position of the ion in the electrochemical series, the concentration of the solution and the type of electrode being used.

At the cathode, if more than one type of positive ion arrives at the cathode, the one which gives up its charge most readily – the least electropositive ion – is discharged, for example, H+ and Na+, H+ is preferentially discharged 2H+ (aq)+ 2e –– H2 (g). Thus, the ion derived from the element lower in the electrochemical series is discharged. At the anode, the least electronegative ion is discharged – the ion which gives up its charge more readily. Thus, in the electrolysis of dilute aqueous sodium chloride, OHions are discharged in preference to Cl- ions.

Due to high concentration of an electrolyte, its anion may be discharged in preference to a less electronegative ion. For example, with concentrated aqueous sodium chloride, Cl- ions are discharged in preference to OH- ions (from water).

ELECTROLYSIS OF DILUTE SULPHURIC ACID

Ions present are H+ and OH- from water, and H+ and SO42- from sulphuric acid.

At the cathode: H+ ions migrate towards the cathode and are discharged as hydrogen gas. 2H+(aq) + 2e –– H2 (g)

At the anode: OH- and SO42- migrate towards the anode, but the OH- ion is discharged in preference to the SO42- ion. 4OH- (aq) –– 2H2O (l) + O2(g) + 4e Comment: Since 4mol of electrons are produced at the anode, then these 4mol of electrons must be used up at the cathode. So in this reaction, for every 1mol of oxygen gas produced, 2mol of hydrogen gas are formed.

ELECTROLYSIS OF AQUEOUS SODIUM CHLORIDE USING PT/C ELECTRODE

Ions present: Na+, H+, OH- and ClAt the cathode: Na+ and H+ migrate towards the cathode, but H+ is discharged in preference to Na+ (lower in electrochemical series). 2H+(aq) + 2e –– H2(g) At the anode: OH- and Cl- ions migrate towards the anode. If the sodium chloride solution is dilute, then OH- ions are discharged in preference to Cl-. 4OH-(aq) –– 2H2O (l) + O2(g) + 4e If the solution is concentrated, then Cl- ions are discharged preferentially and in this case, a carbon electrode is used instead of platinum, as they are resistant to attack by chlorine. 2Cl- (aq) –– Cl2 (g) + 2e

In some cases, i.e. for active electrodes, the anode may dissolve to provide electrons, for example, a copper electrode in the electrolysis of aqueous copper sulphate. Cu(s) ? Cu2+(aq) + 2e (left at the anode) since this occurs more readily than the discharge of the anions. Platinum and carbon (graphite) are inert electrodes, they do not form ions during electrolysis.

ELECTROLYSIS OF AQUEOUS COPPER (II) SULPHATE SOLUTION

Ions present: H+, Cu2+, OH- and SO42At the cathode: Cu2+ and H+ ions migrate towards the cathode where the Cu2+ ions are discharged in preference to H+ ions, as it is lower in the series. Cu2+(aq) + 2e –– Cu (s). This copper is deposited on the cathode which gets thicker. The blue colour of the copper sulphate solution gradually fades.

At the anode: Two reactions can occur at the anode, depending on the type of electrode used. Using Pt/C electrodes, OH- ions are discharged in preference to SO42-. 4OH- (aq) –– 2H2O(l) + O2(g) + 4e. The solution becomes acidic as Cu2+ and OH- ions are removed, leaving H+ and SO42- ions.

Using copper electrode as the anode results in the copper from the electrode being converted to copper ions (a process which requires less energy) and the anode dissolves and gets thinner.

Cu(s) –– Cu2+ (aq) + 2e. The concentration of the solution is unchanged (the blue colour remains the same).