The Electrolysis Setup
Electrolysis is the decomposition of an electrolyte using an electric current. Unlike molten compounds, **aqueous solutions** contain water molecules ($\text{H}_2\text{O}$), which break down into hydrogen ions ($\text{H}^+$) and hydroxide ions ($\text{OH}^-$).
This means there are always four ions in solution competing to be discharged at the electrodes:
- Positive metal ions ($M^+$) and hydrogen ions ($\text{H}^+$) migrating to the negative **cathode**.
- Negative non-metal ions ($X^-$) and hydroxide ions ($\text{OH}^-$) migrating to the positive **anode**.
Discharge Rules at the Cathode (Negative Electrode)
At the cathode, reduction occurs (gaining electrons). The rule depends on the reactivity series:
If the metal is MORE reactive than hydrogen
Hydrogen gas ($\text{H}_2$) is produced at the cathode. The metal ions remain in solution.
Half-equation: $2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2$
If the metal is LESS reactive than hydrogen
The metal is deposited as a solid coating on the cathode. This applies only to **copper, silver, and gold** at GCSE.
Example (Copper): $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$
Discharge Rules at the Anode (Positive Electrode)
At the anode, oxidation occurs (losing electrons). The rule depends on the presence of halide ions ($\text{Cl}^-$, $\text{Br}^-$, $\text{I}^-$):
If halide ions are present
The respective halogen gas (chlorine, bromine, or iodine) is produced.
Half-equation (Chlorine): $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-$
If NO halide ions are present (e.g. sulfate $\text{SO}_4^{2-}$ or nitrate $\text{NO}_3^-$)
Oxygen gas ($\text{O}_2$) is produced from discharged hydroxide ions.
Half-equation: $4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-$
Worked Exam Example (Electrolysis of Brine)
Worked GCSE Example
Question: Predict the products at the cathode and anode during the electrolysis of concentrated sodium chloride solution ($\text{NaCl}$ in water). Write half-equations for each reaction.
Solution:
- Identify all ions present: $\text{Na}^+$ and $\text{Cl}^-$ from salt; $\text{H}^+$ and $\text{OH}^-$ from water.
- At the cathode, sodium ($\text{Na}$) is more reactive than hydrogen ($\text{H}$). Therefore, **hydrogen gas** is discharged.
Cathode Equation: $2\text{H}^+(aq) + 2\text{e}^- \rightarrow \text{H}_2(g)$ - At the anode, chloride ions ($\text{Cl}^-$) are halides and are present in high concentration. Therefore, **chlorine gas** is discharged.
Anode Equation: $2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2\text{e}^-$ - The ions left in the solution are $\text{Na}^+$ and $\text{OH}^-$, forming sodium hydroxide ($\text{NaOH}$), which makes the solution alkaline.
Products: Hydrogen gas at cathode, Chlorine gas at anode, Sodium Hydroxide in solution.
Frequently Asked Questions
Sodium is highly reactive and has a high tendency to remain as ions rather than returning to a neutral metallic form. Hydrogen is less reactive, meaning it is more easily reduced (gains electrons) than sodium under aqueous conditions.
For hydrogen, use a burning splint; it will make a 'squeaky pop' sound. For oxygen, use a glowing splint; it will relight. For chlorine, use damp blue litmus paper; it will turn red and then bleach white.
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