Reactivity 3.2.15 — Electrolysis of Aqueous Solutions

Aqueous electrolysis involves both ionic solutes and water molecules.

• At the cathode (reduction), positively charged ions (cations) compete with water for reduction.
• At the anode (oxidation), negatively charged ions (anions) compete with water for oxidation.
• Which species is discharged depends on their standard electrode potentials (E^⦵).
• Water can act as both a reducing and oxidizing agent in electrolysis, producing hydrogen or oxygen gas.

Reactions of Water

• Reduction (at cathode): \( \text{H}_2\text{O} + e^- \rightarrow \tfrac{1}{2} \text{H}_2(g) + \text{OH}^- \quad E^\circ = -0.83\ \text{V} \)
• Oxidation (at anode): \( \text{H}_2\text{O} \rightarrow \tfrac{1}{2} \text{O}_2(g) + 2\text{H}^+ + 2e^- \quad E^\circ = -1.23\ \text{V} \)

Species with more positive (less negative) E^⦵ are preferentially discharged.

Electrolysis of Pure Water

• Pure water is a poor conductor → dilute sulfuric acid (or a small amount of salt) is often added to increase conductivity.
• Water itself is both oxidized and reduced.

Electrodes: Inert (e.g., platinum)

Half-equations:

At Cathode (Reduction):

$2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq)$
$E^\circ = -0.83\ \text{V}$

At Anode (Oxidation):

$2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-$
$E^\circ = +1.23\ \text{V}$

Overall reaction:

$2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g)$

pH effect: Solution becomes slightly basic due to formation of OH⁻ at the cathode.

Electrolysis of Aqueous Sodium Chloride (NaCl)

• Competing reactions between water and ions (Na⁺ and Cl⁻).
• Products depend on concentration of chloride.

Electrodes: Inert (e.g., carbon or platinum)

Half-equations for concentrated NaCl:

At Cathode (Reduction):

$2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq)$

At Anode (Oxidation):

$2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-$

Overall reaction (concentrated NaCl):

$2\text{Cl}^-(aq) + 2\text{H}_2\text{O}(l) \rightarrow \text{H}_2(g) + \text{Cl}_2(g) + 2\text{OH}^-(aq)$

Products: Hydrogen gas, chlorine gas, sodium hydroxide solution (NaOH)

Electrolysis of Aqueous Copper(II) Sulfate (CuSO₄)

• Cu²⁺ is more easily reduced than H₂O.
• H₂O is oxidized more easily than SO₄²⁻.

Electrodes:

Inert → Pt or graphite (used when Cu metal not involved)
Copper → allows Cu plating

Half-equations (inert electrodes):

At Cathode (Reduction):

$\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$

At Anode (Oxidation):

$2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-$

Overall reaction:

$2\text{Cu}^{2+}(aq) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{Cu}(s) + \text{O}_2(g) + 4\text{H}^+(aq)$

Products: Copper metal at cathode, oxygen gas at anode

If copper electrodes used:

Anode:

$\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-$

Overall reaction: No net change in Cu²⁺ — used for electroplating or purification

Effect of Concentration on Discharge

• In concentrated NaCl, Cl⁻ is discharged at the anode.
• In dilute NaCl, water may be oxidized instead (producing O₂).
• This happens because water has a higher E⦵ than Cl⁻, but Cl⁻ can still be discharged due to kinetic factors and overpotential differences.

Effect of Electrode Material

• Inert electrodes (e.g., graphite or platinum) do not participate in reactions.
• Active electrodes (e.g., copper electrodes in CuSO₄): copper anode dissolves into solution.
• This leads to a different reaction at the anode: \( \text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^- \)