Electrolysis of Aqueous Solutions (HL) (DP IB Chemistry): Revision Note

Electrolysis of aqueous solutions

• In molten binary compounds, the products of electrolysis are predictable using the ions present:

• At the cathode, positive metals ions (cations) are reduced to metal:

Pb²⁺ (l) + 2e^-  → Pb (l)

• At the anode, negative ions (anions) are oxidised to non-metals:

2Br^- (l) → Br₂ (g) + 2e^-

• In aqueous solutions, the situation is more complex because water introduces additional ions that can undergo oxidation or reduction

• 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

• The species discharged depends on:

  • The relative values of the standard electrode potentials, E^θ

  • The concentration of the ions

  • The identity of the electrode

Products of specified electrolytes

• The electrolysis of water, sodium chloride solution and copper sulfate solutions produces:

Table showing the electrolysis products of aqueous solutions

The influence of relative values of E^θ

• Electrolysis of pure water is slow due to low ion concentration

  • Adding acid or base increases ion availability and speeds up the reaction

• The overall products are the same in acidic and alkaline conditions

  • But, the half-equations differ depending on the conditions

In acidic conditions

• This could use dilute sulfuric acid as the electrolyte

• At the cathode:

2H₂O (l) + 2e^- →  H₂ (g) + 2OH^- (aq)              E^θ = -0.83V

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

• The E^θ value for H⁺ is more positive than for water, so it is preferentially reduced

  • This results in the discharge of hydrogen gas

• At the anode:

  • Although sulfate ions are present in the solution, they contain sulfur in its maximum oxidation state (+6) and cannot be further oxidised

  • So, only water can be oxidised

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

In alkaline conditions

• This could use dilute sodium hydroxide as the electrolyte

• At the cathode:

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

2H₂O (l) + 2e^- → H₂ (g) +  2OH^- (aq)         E^θ = -0.83 V

• Water has a more positive E^θ than sodium ions, so it is preferentially reduced

  • This results in the discharge of hydrogen gas

• 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

• In both acidic and alkaline conditions, the overall reaction is:

2H₂O (l)  →  2H₂ (g)   + O₂ (g)

The influence of concentration of the ions

• In the electrolysis of sodium chloride solution concentration affects the anode product

• At the cathode:

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

• At the anode:

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

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

• The E^θ values are similar

• So, either chlorine or oxygen can be discharged depending on conditions

  • At high chloride ion concentration (> 25%), chlorine is the main product

• The overall reaction 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 reactions are active electrodes

  • Inert electrodes, such as platinum and carbon, that do not take part in the redox reactions are passive electrodes

• In the electrolysis of copper sulfate solution, CuSO₄ (aq),  the electrode affects the products

Passive electrodes

• At the cathode:

      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:

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

• The overall 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:

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:

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

• This reaction is used to recycle / purify copper to a very high grade of copper for use in electrical wires

  • The anode is made of impure copper

    • The copper goes into solution as copper(II) ions

  • The cathode is made of pure copper

    • The copper(II) ions are deposited as copper on the cathode

  • Impurities from the anode fall to the bottom of the cell, as anode slime

    • They may contain precious metals

Diagram to show the purification of copper via electrolysis