Electrolysis of Aqueous Solutions (HL) (DP IB Chemistry)

Electrolysis of Aqueous Solutions

• 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^-→ H₂ (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^- →  H₂ (g) + 2OH^- (aq)              E^θ = -0.83V

2H⁺ (aq) + 2e^- →  H₂ (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^- → H₂ (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)  →  2H₂ (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) → Cl₂ (g)  + 2e^-              E^θ = -1.36 V

    2H₂O (l) →  4H⁺ (aq)  +  O₂ (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) +  H₂ (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 known as active electrodes and inert electrodes such as platinum and carbon are called passive electrodes

• The electrolysis of copper sulfate solution, CuSO₄ (aq),  is an example of where active and passive electrodes determine the products

Passive electrodes

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

      Cu²⁺ (aq) + 2e^- → Cu (s)                         E^θ =  +0.34 V

  2H₂O (l) + 2e^- → H₂ (g) +  2OH^- (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)  +  O₂ (g) +  4e^-     E^θ = -1.23 V

• The overall equation for the reaction is:

2CuSO₄ (aq) + 2H₂O (l) → 2Cu (s) + O₂ (g) + 2SO₄^2- (aq) + 4H⁺ (aq) 

OR

2CuSO₄ (aq) + 2H₂O (l) → 2Cu (s) + O₂ (g) + 2H₂SO₄ (aq)

Active electrodes

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

Cu²⁺ (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)  → Cu²⁺ (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 anode is made of impure copper and the cathode is made of pure copper

• The impurities from the anode fall to the bottom of the cell

Diagram to show the purification of copper via electrolysis

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