Introduction to electrolysis
Francine Taylor-Campbell, Contributor

YOU SHOULD BE ABLE TO:

• Define the terms electrolyte, electrode, strong and weak electrolytes and electrolysis
• Describe the factors which influence the discharge of the ions during electrolysis

POINTS TO NOTE

• Electrolysis is the decomposition of an electrolyte by an electric current.
  This is carried out in an electrolytic cell.
• An electrolyte is a molten compound or the solution of a compound which conducts an electric current and is decomposed by it.
• Electrolytes contain ions which are charged atoms or charged radicles (groups of atoms) which carry the electric current. Eg Na⁺, OH^-.

Strong electrolytes consist totally of ions, that is, they are fully ionized, for example all salts, strong acids and alkalis. These have high electrical conductivity. Weak electrolytes consist mainly of molecules with relatively few ions, for example weak acids and weak bases such as ethanoic acid and aqueous ammonia. In solution, a large proportion of the molecules remain undissociated.

NH₃ (aq) + H₂O (l) == NH₄⁺(aq) + OH^-(aq)

The electrodes in the electrolytic cell are the anode, which is the positive electrode and the cathode which is the negative electrode. During electrolysis, the anode gains electrons which flow from the anode to the cathode which loses electrons.

Anions are negative ions; they travel towards the anode during electrolysis, eg OH^-, Cl^- where they may give up electrons. (2Cl^-(aq) = Cl₂(g) + 2e)

Cations are positive ions; they travel towards the cathode during electrolysis, eg H⁺, Na⁺, Cu²⁺ where they may gain electrons. (Cu²⁺ + 2e = Cu)

To identify the ions present in electrolytes

(i) Write the formula of the compound eg Al₂(SO₄)₃

(ii) Write a balanced equation to separate it into atoms or radicles, eg

Al₂(SO₄)₃ = 2Al + 3SO₄

(iii) Put the appropriate charge on the atoms or radicles (numerically equal to the valency) ie Al₄(SO₄)₃ = 2Al³⁺ + 3SO₄^2-

(iv) Check to see that the total charge on the ions is zero (2x3+) + (3x2-) = 0

In the case of aqueous solutions, relatively very small numbers of H+ and OH- ions are present due to the extremely slight ionization of water

H₂O == OH^-(aq) + H⁺ (aq)

Q1 (a) Write equations for the ionization of (i) H₂SO₄ (ii) CuSO₄ (iii) FeCl₃ (iv) Al₂O₃ (v) AlPO₄ (vi) (NH₄)₃PO₄ (b) Give the formulae of the ions and or molecules present in (i) CuSO₄(aq) (ii) NaCl(aq) (iii) NH₃(aq) iv. CH₃CO₂H(aq).

Factors affecting the formation of the products during electrolysis

(i) The degree of electropositiveness and electronegativeness of the ions. At the cathode, if more than one type of positive ion arrives at the cathode, the one which gives up its charge most readily, that is, the least electropositive ion, is discharged, for example, H+ and Na+, H+ is preferentially discharged 2H+ + 2e = H2 Thus the ion derived from the element lower in the electrochemical series is discharged. At the anode, the least electronegative ion is discharged, that is, the ion which gives up its charge more readily. Thus, in the electrolysis of dilute aqueous sodium chloride OH- ions are discharged in preference to Cl- ions ie 4OH-(aq) = 2H2O(l) + O2(g) + 4e There are exceptions to this rule: (ii) 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) 2Cl- = Cl2 + 2e (iii) The type of electrode. In some cases, that is, 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.

Francine Taylor-Campbell teaches at Jamaica College. Send questions and comments to kerry-ann.hepburn@gleanerjm.com