Science - Combined (0653) | Electrochemistry

🔬 C4.1 Electrochemistry: The Power of Electricity in Chemistry

Hello future scientist! Welcome to Electrochemistry, a fascinating chapter where we learn how electricity can be used to drive chemical reactions that normally wouldn't happen. It’s like giving chemistry a jolt of energy!

Understanding electrolysis is crucial because it is used in real life for manufacturing essential elements like chlorine, and for plating expensive items like jewelry. Don't worry if this seems tricky at first; we will break down the process into simple steps.

1. Defining Electrolysis (The Core Concept)

What is Electrolysis?

Electrolysis is the decomposition (breaking down) of an ionic compound, either when molten (melted) or in aqueous solution (dissolved in water), by the passage of an electric current.

• The ionic compound must be able to move freely. This means it must be either molten or dissolved in water.
• If the ions cannot move (e.g., in a solid ionic crystal), the electricity cannot pass through the substance and electrolysis cannot occur.

The Electrolytic Cell Setup

For electrolysis to happen, you need a special setup called an electrolytic cell, which contains three key parts:

1. The Electrolyte: The molten or aqueous ionic compound that undergoes decomposition. It contains the moving ions.
2. The Electrodes: Two electrical conductors (usually rods of carbon/graphite or platinum, which are inert—meaning they don't react). These are connected to a power source (like a battery).
3. The Power Source: Provides the electric current.

Identifying the Electrodes: Anode and Cathode

In an electrolytic cell, the electrodes are defined by the charge of the power supply terminal they are connected to:

• Anode: The Positive (+) electrode.
• Cathode: The Negative (-) electrode.

The positive ions (Cations) are attracted to the negative electrode (Cathode).

The negative ions (Anions) are attracted to the positive electrode (Anode).

Key Takeaway: Electrolysis uses electricity to split ionic compounds into elements. It requires moving ions in a molten liquid or solution (the electrolyte) and two electrodes (anode and cathode).

2. Electrolysis of Molten Ionic Compounds

This is the simplest type of electrolysis because only two types of ions are present: the metal cation and the non-metal anion.

Example: Electrolysis of Molten Lead(II) Bromide (PbBr₂) (Core 3a & Supp 5)

Lead(II) bromide is an ionic compound. When molten (heated until it melts), the ions (Pb²⁺ and Br⁻) are free to move.

The products formed are always the pure elements that make up the compound.

Step-by-Step Process:

1. Ions present: Pb²⁺ (cations) and Br⁻ (anions).
2. At the Cathode (-) (Reduction occurs): The positive lead ions ($\text{Pb}^{2+}$) are attracted to the negative electrode. They gain electrons to form liquid lead metal.
   Product: Molten Lead (a metal).
   Observation: A silvery liquid (molten metal) forms at the bottom of the electrode (since lead is dense).
3. At the Anode (+) (Oxidation occurs): The negative bromide ions ($\text{Br}^{-}$) are attracted to the positive electrode. They lose electrons to form bromine gas ($\text{Br}_2$).
   Product: Bromine gas (a non-metal).
   Observation: Brown fumes (bromine gas) are seen bubbling up.

Prediction Rule for Molten Binary Compounds (Supplement 5):

• The metal is always formed at the Cathode (-).
• The non-metal is always formed at the Anode (+).

Key Takeaway: When a binary compound is molten, electrolysis simply separates the metal (at the cathode) and the non-metal (at the anode).

3. Electrolysis of Aqueous Solutions

When an ionic compound is dissolved in water ($\text{H}_2\text{O}$), the process becomes more complicated because water molecules can also be involved in the reactions.

In addition to the ions from the salt (the solute), you also have ions from the water: $\text{H}^+$ and $\text{OH}^-$ ions (produced from the slight ionisation of water).

At each electrode, there is a competition to discharge (react).

General Rules for Product Selection (Supplement 4)

At the Cathode (-) (where reduction occurs):

The products are always metals or hydrogen gas ($\text{H}_2$).

Rule: If the metal ion in the electrolyte is more reactive than hydrogen (e.g., Sodium, Potassium, Calcium), then Hydrogen gas ($\text{H}_2$) is produced. Otherwise, the metal is produced.

At the Anode (+) (where oxidation occurs):

The products are always non-metals (other than hydrogen) or Oxygen gas ($\text{O}_2$).

Rule: If the solution contains halide ions ($\text{Cl}^{-}$, $\text{Br}^{-}$, $\text{I}^{-}$), they are usually discharged. If no halide ions are present (or if the solution is very dilute), then Oxygen gas ($\text{O}_2$) is produced from the discharge of $\text{OH}^{-}$ ions.

Case Study 1: Concentrated Aqueous Sodium Chloride (Brine) (Core 3b)

Ions present: $\text{Na}^{+}$, $\text{Cl}^{-}$, $\text{H}^{+}$, $\text{OH}^{-}$

1. At the Cathode (-): $\text{Na}^{+}$ and $\text{H}^{+}$ compete. Since sodium ($\text{Na}$) is highly reactive (more reactive than hydrogen), hydrogen ions discharge.
   Product: Hydrogen gas ($\text{H}_2$).
   Observation: Colourless gas bubbles (Hydrogen) seen. The solution around the cathode becomes alkaline due to the remaining $\text{Na}^{+}$ and $\text{OH}^{-}$ ions.
2. At the Anode (+): $\text{Cl}^{-}$ and $\text{OH}^{-}$ compete. Because the solution is concentrated, the chloride ions discharge easily (even though $\text{OH}^{-}$ usually discharges to make $\text{O}_2$).
   Product: Chlorine gas ($\text{Cl}_2$).
   Observation: Pale yellow-green gas bubbles (Chlorine) seen.

Case Study 2: Dilute Sulfuric Acid ($\text{H}_2\text{SO}_4$) (Core 3c)

Sulfuric acid ionises heavily. Since it is dilute, water is the major component.

Ions present: $\text{H}^{+}$, $\text{SO}_4^{2-}$, $\text{OH}^{-}$ (from water)

1. At the Cathode (-): Only $\text{H}^{+}$ ions are present as cations.
   Product: Hydrogen gas ($\text{H}_2$).
   Observation: Colourless gas bubbles seen.
2. At the Anode (+): $\text{SO}_4^{2-}$ (sulfate) and $\text{OH}^{-}$ compete. Sulfate ions are difficult to discharge, so $\text{OH}^{-}$ ions (from water) discharge.
   Product: Oxygen gas ($\text{O}_2$).
   Observation: Colourless gas bubbles seen.

Note: The overall process in dilute sulfuric acid is essentially the electrolysis of water itself, yielding Hydrogen (2 volumes) and Oxygen (1 volume).

Key Takeaway: In aqueous electrolysis, water's ions ($\text{H}^{+}$ and $\text{OH}^{-}$) compete. Reactivity determines the cathode product (H₂ if metal is reactive), and concentration/halides determine the anode product ($\text{Cl}_2$ if concentrated, $\text{O}_2$ otherwise).