• We have seen previously how simple binary compounds can be electrolysed when molten and the products of electrolysis can be predicted using our knowledge of the ions present
• At the?cathode, positive metals ions (cations) are discharged resulting in metals being deposited
• The cations are reduced by the electrons coming from the cathode:

Pb²⁺(l) + 2e^-? → Pb (l)

• Meanwhile, at the?anode, anions are discharged by oxidation:

2Br^-?(l) → Br₂?(g) + 2e^-

• However, when aqueous solutions of ionic compounds are electrolysed the products are a little more complicated to predict as there are additional ions present from the water
• Water can be oxidised to oxygen or reduced to hydrogen:
  • Oxidation reaction:

2H₂O (l)?^?→ 4H^+?(aq) + O₂?(g)?+ 4e^-?

• • Reduction reaction:

2H₂O (l)?+ 2e^-→ H_2?(g) + 2OH^-?(aq)

• At the cathode, either the metal ion M⁺?or water can be reduced
• At the anode, either the anion A^-?or water can be oxidized
• Which species is discharged depends on three things:
  • The?relative values of?E^θ
  • The?concentration?of the ions present
  • The?identity?of the electrode

Products of specified electrolytes

• The electrolysis of water, sodium chloride solution and copper sulfate solutions is as follows:

Table showing the electrolysis products of aqueous solutions

The influence of relative values of E^θ

• The electrolysis of water is very slow as there are few ions present, so a little acid or base can be added to increase the number of ions present and speed up the electrolysis
• Whether acid or base is added the products are the same, but the electrode reactions are slightly different
• Using dilute sulfuric acid as the electrolyte, the cathode reactions could be

2H₂O (l) + 2e^-?→??H_2?(g) + 2OH^-?(aq)?? ? ? ? ? ? ?E^θ?= -0.83V

2H^+?(aq) + 2e^-?→? H_2?(g)? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?E^θ?=? 0.00 V

• The?E^θ?is smaller for the hydrogen ion so it is preferentially reduced and H₂?(g) will be discharged
• At the anode, although sulfate ions are present in the solution, only water can be oxidised
• This is because the sulfate ion, SO₄^2-, contains sulfur in its maximum oxidation state (+6) so it cannot be further oxidised
• The oxidation of water produces oxygen gas:

2H₂O (l)? ?→? 4H^+?(aq) +? O₂?(g) +? 4e^-? ? ? ? ??E^θ?= -1.23 V

• If the water is made basic by the addition of dilute sodium hydroxide solution, the cathode reactions could be:

Na^+?(aq) + e^-?→ Na (s)? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??E^θ?=? -2.71 V

2H₂O (l) + 2e^-?→ H_2?(g) +? 2OH^-?(aq)? ? ? ? ?E^θ?= -0.83 V

• The?E^θ?is smaller for water than the sodium ion, so water is preferentially reduced and H₂?(g) will be discharged
• At the anode, either the hydroxide ion or water can be oxidised:

4OH^-?(aq)? →? 2H₂O (l) + O₂?(g)? +? 4e^-? ? ? ? ? ? ? ? ? ??E^θ?= -0.40 V

2H₂O (l)? ?→? 4H^+?(aq) + O₂?(g) +? 4e^-? ? ? ? ? ? ? ? ? ? ? ? ??E^θ?= -1.23 V

• Based on these values the hydroxide ion is preferentially oxidized and O₂?(g) will be discharged
• The overall reaction whether in acid or alkali conditions is:

2H₂O (l)? →? 2H_2?(g)? ?+ O₂ (g)

The influence of concentration of the ions

• The electrolysis of sodium chloride solution provides an illustration of the influence of concentration on the products discharged
• As before, we would expect hydrogen ion to be preferentially discharged at the cathode before the sodium ion:

2H^+?(aq) + 2e^-?→ H₂?(g)? ? ? ? ? ? ? ? ?E^θ?=? 0.00 V

• However at the anode, the relative proximity of the?E^θ?values allows the possibility of both reactions occurring:

2Cl^-??(aq) → Cl_2?(g)? + 2e^-? ? ? ? ? ? ??E^θ?= -1.36 V

2H₂O (l) → ?4H^+?(aq)? +? O_2?(g) +? 4e^-? ? ?E^θ?= -1.23 V

• In fact, when concentration of the sodium chloride increases to more than 25% the Cl^-?becomes preferentially discharged and chlorine gas is the main product of the reaction at the anode
• The overall reaction equation is:

2NaCl (aq)? + 2H₂O (l) →? ? 2NaOH (aq) +? H_2?(g)? +? Cl₂ (g)

Influence of the electrodes

• The products of electrolysis are influenced by the identity of the electrodes
• Electrodes that take part in the redox processes are know as?active electrodes?and inert electrodes such as platinum and carbon are called?passive electrodes
• The electrolysis of copper sulfate solution, CuSO_4?(aq),? is an example of where active and passive electrodes determine the products

Active electrodes

• At the cathode, the possible reactions that could take place are:

Cu^2+?(aq) + 2e^-?→ Cu (s)? ? ? ? ? ? ? ? ? ? ? ? ?E^θ?=? +0.34 V

2H₂O (l) + 2e^-?→ H_2?(g) +? 4OH^-?(aq)? ? ? ??E^θ?= -0.83 V

• Copper ions are preferentially reduced, so copper metal is deposited on the cathode
• At the anode, water is oxidised, so oxygen gas is produced (the sulfate ion cannot be oxidised):

2H₂O (l) → ?4H^+?(aq)? +? O_2?(g) +? 4e^-? ? ?E^θ?= -1.23 V

• The overall equation for the reaction is:

2CuSO_4?(aq) + 2H₂O (l) → 2Cu (s) + O_2?(g) + 2SO₄^2-?(aq)?+?4H^+?(aq)

OR

2CuSO_4?(aq) + 2H₂O (l) → 2Cu (s) + O_2?(g) + 2H₂SO_4?(aq)

Passive electrodes

• At the cathode, the reaction is the same as with inert electrodes:

Cu^2+?(aq) + 2e^-?→ Cu (s)? ? ? ? ? ? ? ? ? ? ? ? ?E^θ?=? +0.34 V

• However, at the anode the copper electrode is oxidised and dissolves to form copper ions

Cu (s)??→?Cu^2+?(aq) + 2e^-? ? ? ? ? ? ? ? ? ? ??E^θ?=? -0.34 V

• This reaction is used to purify copper, needed to produce a very high grade of copper for use in electrical wires
• The impure copper is made the anode, and the cathode is made of pure copper
• The impurities from the anode fall to the bottom of the cell

The purification of copper by electrolysis

• The anode slowly dissolves away and the cathode builds up pure copper
• The impurities form a slime under the anode which is actually quite valuable as it often contains significant quantities of precious metals like silver