According to the CSEC syllabus, you should be able to discuss the electrolysis of certain substances, especially in how electrolysis relates to the concentration of an electrolyte and how electrolysis is affected by the type of electrode, that is, active or inert. (Active means reactive and inert means unreactive). Active electrodes include electrodes made of copper. Inert electrodes include electrodes made of carbon
Electrolysis of Dilute Sulphuric Acid (Using Inert Electrodes)
Here, the electrolytic cell is filled with dilute sulphuric acid. There are gases liberated at each electrode, and the volume of gas at the cathode is twice the amount at the anode.
The gas at the cathode gives a pop with a lit splint (a piece of wood on fire) showing that it is hydrogen, and the gas at the anode relights a glowing splint, showing that it is oxygen.
The apparatus we will use to demonstrate this is known as a Hoffman Voltameter.
As this cell is run, the electrolyte becomes more concentrated over time. Why does this happen?
First, let's look at the ions in solution. First, there are the H⁺ and OH⁻ ions from the ionization of the water and the H⁺ and SO₄²⁻ ions from the ionization of the sulphuric acid.
The positive ions (cations), the H⁺ ions, move towards the cathode (negative electrode) and the negative ions (anions), the OH⁻ and SO₄²⁻ ions, move towards the anode (positive electrode).
So, the H⁺ ions are discharged at the cathode (by gaining electrons) like this:
H⁺(aq) + e⁻ → H⁰(g)
2H⁺(aq) + 2e⁻ → H₂(g)
There is only one type of cation at the cathode, so there is no preferential discharge.
However, there are two types of anions, the OH⁻ and SO₄²⁻ ions at the anode. So, there must be some form of preferential discharge. If you remember the electrochemical series:
Then you'll know that the hydroxide ion is easier to discharge than the sulphate ion, and is lower in the electrochemical series than the sulphate ion.
So, the hydroxide ion is discharged preferentially like this:
4OH⁻(aq) → 2H₂O(l) + O₂(g)
So, 1 mol of hydrogen molecules is produced for every 2 mol of electrons transferred. 1 mol of oxygen is produced for every 4 mol of electrons transferred. So, 2 volumes of hydrogen are produced for each volume of oxygen produced during electrolysis of dilute sulphuric acid. So, there is two times as much hydrogen produced as oxygen.
You've probably realized by now why exactly the electrolyte (dilute sulfuric acid) would become more concentrated over time. The acid is dilute because it has been mixed with water. The hydroxide ions from the water are discharged preferentially, along with hydrogen ions. And, as shown in this equation:
4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
Four hydroxide ions from four water molecules are converted to two water molecules, meaning that the amount of water in the solution lessens (halves) over time, making the acid more concentrated.
Electrolysis of Dilute or Concentrated Aqueous Sodium Chloride Using Inert Electrodes
When looking at problems involving electrolysis, we always start by looking at the ions in solution. Here:
H⁺(aq) and OH⁻(aq) ions are present from water and;
Na⁺(aq) and Cl⁻(aq) ions are present from sodium chloride.
H⁺(aq) and Na⁺(aq) ions move towards the cathode, and H⁺ is lower on the electrochemical series than Na⁺, so H⁺ is preferentially discharged like this:
2H⁺(aq) + 2e⁻ → H₂(g)
This is unaffected by the concentration of the solution.
Na⁺(aq) and Cl⁻(aq) ions move towards the anode. What happens here depends on whether the solution is dilute or concentrated. If the solution is dilute (dilute aqueous sodium chloride), then the hydroxide is preferentially discharged (as it is lower on the electrochemical series), resulting in the liberation of oxygen at the anode like this:
4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
Little to no chlorine is liberated here.
However, if the solution is concentrated, the chloride ions are preferentially discharged at the anode while very little oxygen is evolved, like this:
2Cl⁻(aq) → Cl₂(g) + 2e⁻
This can all be shown in this diagram:
Electrolysis of Aqueous Copper (II) Sulphate Using Inert Electrodes
Once again, we evaluate our ions in solution:
H⁺(aq) and OH⁻(aq) ions are present from water and;
Cu²⁺(aq) and SO₄²⁻(aq) ions are present from copper (II) sulphate.
(Remember that copper sulphate is blue, it will be useful in many questions later on)
H⁺(aq) and Cu²⁺(aq) ions go to the cathode.
OH⁻(aq) and SO₄²⁻ ions go to the anode.
Copper is lower than hydrogen on the electrochemical series, so it is preferentially discharged at the cathode.
Cu²⁺(aq) + 2e⁻ → Cu(s)
You see, because copper is a solid at room temperature, it isn't let off as a gas like the other instances we looked at. Instead, the copper is deposited at the cathode, and the characteristic blue colour of the electrolyte fades.
Hydroxide ions are lower than sulphate ions on the electrochemical series so they are preferentially discharged at the anode like this:
4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
The concentration of the electrolyte (in terms of copper) decreases and the electrolyte become more acidic, as there are more hydrogen ions not discharged (which, as you know, make a substance acidic) while the hydroxide ions are discharged (which, as you know, make a substance basic).
The reaction is the same if only a copper cathode is used.
Electrolysis of Aqueous Copper (II) Sulphate Using Active Copper Cathode and Anode
our ions in solution:
H⁺(aq) and OH⁻(aq) ions are present from water and;
Cu²⁺(aq) and SO₄²⁻(aq) ions are present from copper (II) sulphate.
The reaction at the cathode is this:
Cu²⁺(aq) + 2e⁻ → Cu(s)
(copper atoms are deposited at the cathode)
The reaction at the anode is this:
Cu(s) → Cu²⁺(aq) + 2e⁻
This is because the copper atoms in the anode leave the anode and enter the electrolyte as copper (II) ions, Cu²⁺(aq). As the mass of the anode decreases, the mass of the cathode increases by the same amount. The concentration of the electrolyte does not change, and the blue colour does not fade. This is what is used in electroplating and copper refining.
Electrolysis of Molten Lead (II) Bromide Using Inert Electrodes
The reaction at the cathode is as follows:
Pb²⁺(l) + 2e⁻ → Pb(s)
The reaction at the anode:
2Br⁻ (l) → Br₂ + 2e⁻
