Electrolysis: Principles and Applications

TL;DR

Electrolysis uses electricity to split compounds into their elements. In molten compounds, positive ions go to the negative electrode and negative ions go to the positive electrode. For aqueous solutions, water's ions also compete, affecting the products formed.

1. The Mental Model

Think of electrolysis like an "electric fork" that pulls apart chemically bonded atoms. You're using electrical energy to force a non-spontaneous chemical reaction, like taking apart a LEGO model piece by piece.

2. The Core Material

Electrolysis is a process where electrical energy drives a non-spontaneous chemical reaction. You need an electrolyte (a molten ionic compound or an aqueous solution of an ionic compound) and two electrodes connected to a power supply.

How it Works:

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1. Electrolyte: Contains mobile ions that can conduct electricity.
2. Electrodes: Usually inert (like graphite or platinum) but can be reactive.
   • Anode: The positive electrode, where oxidation (loss of electrons) occurs. Negative ions are attracted here.
   • Cathode: The negative electrode, where reduction (gain of electrons) occurs. Positive ions are attracted here.
3. Power Supply: Provides the electrical energy to push electrons.

Electrolysis of Molten Compounds

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When an ionic compound is molten, its ions are free to move.
* Cations (positive ions) move to the cathode and gain electrons (reduction).
* Anions (negative ions) move to the anode and lose electrons (oxidation).

Example: Molten Lead(II) Bromide (PbBr₂)
* Ions present: Pb²⁺, Br⁻
* At the cathode (-): Pb²⁺ ions are attracted. Pb²⁺ + 2e⁻ → Pb (reduction, lead metal forms)
* At the anode (+): Br⁻ ions are attracted. 2Br⁻ → Br₂ + 2e⁻ (oxidation, bromine gas forms)

Electrolysis of Aqueous Solutions

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This is trickier because water (H₂O) is also present and can ionise to H⁺ and OH⁻ ions. You need to consider what gets discharged at each electrode.

At the Cathode (-):
* Competition: Positive metal ions (e.g., Cu²⁺, Na⁺) vs. H⁺ ions (from water).
* Rule: The less reactive ion will be discharged.
* If the metal is less reactive than hydrogen (e.g., Cu, Ag), the metal will be deposited.
* If the metal is more reactive than hydrogen (e.g., Na, K, Ca), hydrogen gas will be produced (2H⁺ + 2e⁻ → H₂ or 2H₂O + 2e⁻ → H₂ + 2OH⁻).

At the Anode (+):
* Competition: Negative non-metal ions (e.g., Cl⁻, SO₄²⁻, NO₃⁻) vs. OH⁻ ions (from water).
* Rule:
* If halide ions (Cl⁻, Br⁻, I⁻) are present in high concentration, the halogen gas will be produced (e.g., 2Cl⁻ → Cl₂ + 2e⁻).
* If other non-halide ions (e.g., SO₄²⁻, NO₃⁻) are present, or if halide concentration is low, then oxygen gas will be produced from the OH⁻ ions or water itself (4OH⁻ → O₂ + 2H₂O + 4e⁻ or 2H₂O → O₂ + 4H⁺ + 4e⁻). The non-halide ions remain in solution.

Here's a flowchart to help you decide the products:

graph TD
    A["Start Electrolysis"] --> B{Molten or Aqueous Solution?};
    B -- Molten --> C["Ions present (e.g., A+, B-)"];
    C --> D{"Cathode (-)"};
    D -- "A+ gains e-" --> E["A deposited/formed"];
    C --> F{"Anode (+)"};
    F -- "B- loses e-" --> G["B deposited/formed"];

    B -- Aqueous --> H{"Cathode (-)"};
    H --> I{"Metal less reactive than H?"};
    I -- Yes (e.g., Cu, Ag) --> J["Metal deposited"];
    I -- No (e.g., Na, K) --> K["Hydrogen gas (H₂) produced"];

    H --> L{"Anode (+)"};
    L --> M{"Halide ions (Cl-, Br-, I-) concentrated?"};
    M -- Yes --> N["Halogen gas (e.g., Cl₂) produced"];
    M -- No --> O["Oxygen gas (O₂) produced"];

Applications of Electrolysis:

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• Extraction of reactive metals: e.g., Aluminium from molten alumina.
• Electroplating: Coating a metal object with a thin layer of another metal (e.g., silver plating). The object to be plated is made the cathode.
• Purification of metals: Refining impure copper.
• Production of chemicals: e.g., Chlorine, sodium hydroxide, hydrogen.

3. Worked Example

Let's predict the products when aqueous copper(II) chloride (CuCl₂) is electrolysed using inert electrodes (like graphite).

1. Ions present in solution: Cu²⁺, Cl⁻ (from CuCl₂), H⁺, OH⁻ (from water).

2. At the Cathode (-):

   • Competing ions: Cu²⁺ and H⁺.
   • Reactivity: Copper is less reactive than hydrogen.
   • Result: Copper metal will be deposited.
   • Equation: Cu²⁺(aq) + 2e⁻ → Cu(s)

3. At the Anode (+):

   • Competing ions: Cl⁻ and OH⁻.
   • Concentration: We assume Cl⁻ ions are in high concentration here (since it's copper chloride).
   • Result: Chlorine gas will be produced.
   • Equation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻

Therefore, the products are copper metal at the cathode and chlorine gas at the anode.

4. Key Takeaways

• Electrolysis uses electrical energy to force non-spontaneous chemical reactions.
• The anode is positive (oxidation occurs), and the cathode is negative (reduction occurs).
• In molten compounds, the metal is reduced at the cathode, and the non-metal is oxidised at the anode.
• In aqueous solutions, water's ions (H⁺ and OH⁻) compete with the electrolyte's ions.
• At the cathode in aqueous solutions, the less reactive positive ion (metal or hydrogen) gets discharged.
• At the anode in aqueous solutions, halide ions are discharged over hydroxide ions (oxygen) if concentrated.

Common Mistakes to Avoid:
- Mixing up which electrode is positive/negative or where oxidation/reduction happens. Remember "AN OX" (anode, oxidation) and "RED CAT" (reduction, cathode).
- Forgetting about the H⁺ and OH⁻ ions from water in aqueous electrolysis.
- Incorrectly applying the reactivity series to determine cathode products in aqueous solutions.
- Assuming oxygen is always produced at the anode in aqueous solutions; concentrated halides take priority.

5. Now Try It

You're electrolysing aqueous sodium sulfate (Na₂SO₄) using inert electrodes. What are the products formed at the cathode and anode? Write down the half-equations for the reactions occurring at each electrode. What would you observe?